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## Capacitors - Basic Electrical

Engineering Questions and
by Manish
4 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitors”.

## 1. What is the relation between current and voltage in a

capacitor?
a) I=1/C*integral(Vdt)
b) I=CdV/dt
c) I=1/CdV/dt
d) I=Ct

Explanation: Current=rate of change of charge=> I=dQ/dt.
Q=CV, hence I=CdQ/dt.

## 2. If 2V is supplied to a 3F capacitor, calculate the chance

stored in the capacitor.
a) 1.5C
b) 6C
c) 2C

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d) 3C

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV.
Q=3*2=6C.

## 3. Calculate the current in the capacitor having 2V supply

voltage and 3F capacitance in 2seconds.
a) 2A
b) 5A
c) 6A
d) 3A

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV.
Q=3*2=6C.
I=Q/t= 6/2=3A.

## 4. A 4microF capacitor is charged to 120V, the charge in

the capacitor would be?
a) 480C
b) 480microC
c) 30C
d) 30microC

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## Explanation: Q is directly proportional to V. The constant of

proportionality in this case is C, that is, the capacitance.
Hence Q=CV.
Q=4*120=480microC.

## 5. For high frequencies, capacitor acts as _________

a) Open circuit
b) Short circuit
c) Amplifier
d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very high frequencies, the
impedance is almost equal to zero, hence it acts as a short
circuit and there is no voltage across it.

## 6. For very low frequencies, capacitor acts as ________

a) Open circuit
b) Short circuit
c) Amplifier
d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very low frequencies the impedance
is almost infinity and hence acts as an open circuit and no
current flows through it.

## 7. A capacitor consists of_________

a) Two conductors

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b) Two semiconductors
c) Two dielectrics
d) Two insulators

Explanation: A capacitor consists of two conductors
connected in parallel to each other so that it can store
charge in between the plates.

## 8. Capacitor preferred when there is high frequency in the

circuits is __________
a) Electrolyte capacitor
b) Mica capacitor
c) Air capacitor
d) Glass capacitor

Explanation: Mica capacitors are preferred for high
frequency circuits because they have low ohmic losses and
less reactance.

## 9. Capacitance increases with ________

a) Increase in plate area
b) Decrease in plate area
c) Increase in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is directly proportional to plate
area. Hence as plate area increases, the capacitance also

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increases.

## 10. Capacitance decreases with __________

a) Increase in distance between the plates
b) Decrease in plate area
c) Decrease in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is inversely proportional to the
distance between the two parallel plates. Hence, as the
distance between the plate decreases, the capacitance
increases.

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Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:25 AM

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## Charge & Voltage - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Charge and
Voltage”.

## 1. Which among the following expressions relate charge,

voltage and capacitance of a capacitor?
a) Q=C/V
b) Q=V/C
c) Q=CV
d) C=Q2V

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV.

## 2. If a 2F capacitor has 1C charge, calculate the voltage

across its terminals.
a) 0.5V

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b) 2V
c) 1.5V
d) 1V

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence C/Q=V.
V=2/1=1V.

## 3. What is the voltage across a capacitor at the time of

switching, that is, when t=0?
a) Infinity
b) 0V
c) Cannot be determined
d) 1V

Explanation: At the time of switching, when t=0, the
capacitor acts as a short circuit. The voltage across a short
is always equal to zero hence the voltage across the
capacitor is equal to zero.

## 4. What is the voltage across the capacitor if the switch is

closed and steady state is reached?

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a) 8V
b) 0V
c) 10V
d) Infinity

Explanation: When steady state is reached, the capacitor
acts as a short circuit and the 10V is connected in parallel
to it. Hence Vc=10V.

## 5. If one plate of a parallel plate capacitor is charged to

positive charge the other plate is charged to?
a) Positive
b) Negative
c) Positive of negative
d) Not charged

Explanation: If one plate is charged to positive, the other
plate is automatically charges to negative so that it can
store electrical charge.

## 6. When voltage across a capacitor increases, what

happens to the charge stored in it?
a) Increases
b) Decreases
c) Becomes zero
d) Cannot be determined

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## Explanation: When voltage across a capacitor increases,

the charge stored in it also increases because charge is
directly proportional to voltage, capacitance being the
constant of proportionality.

## 7. When will capacitor fully charged?

a) When the voltage across its plates is half the voltage
from ground to one of its plates
b) When current through the capacitor is a 1/root2 time its
value
c) When the supply voltage is equal to the capacitor voltage
d) Never

Explanation: When the capacitor voltage is equal to the
supply voltage the current stops flowing through the circuit
and the charging phase is over.

## 8. What happens to the current flow in a fully charged

capacitor?
a) Current flow stops
b) Current flow doubles
c) Current flow becomes half its original value
d) Current flow becomes one-fourth its original value

Explanation: When a capacitor is fully charged, it does not
store any more charge. There is no change in charge with
time. Current is the rate of change of charge, hence it
becomes zero, or stops.

4 of 5 3/19/18, 8:25 AM

## 9. Calculate the capacitance of a capacitor that stores

40microC of charge and has a voltage of 2V.
a) 20F
b) 20microF
c) 10F
d) 10microF

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence C=Q/V.
C=40microC/2V=20microF.

## 10. What happens to the capacitance when the voltage

across the capacitor increases?
a) Decreases
b) Increases
c) Becomes 0
d) No affect

Explanation: The capacitance is inversely proportional to
the voltage across its terminals(C=Q/V). Hence as voltage
increases, capacitance decreases.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:25 AM

sanfoundry.com

## Capacitance - Basic Electrical

Engineering Questions and
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitance”.

## 1. Power factor of a circuit can be improved by placing

which, among the following, in a circuit?
a) Inductor
b) Capacitor
c) Resistor
d) Switch

Explanation: Raising the power factor on a kW load
decreases its kVA. Hence, by adding a capacitor in a circuit,
without altering the kVA. Hence, the power factor is
improved.

## 2. When the supply frequency increases, what happens to

the capacitive reactance in the circuit?

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a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The expression for capacitive reactance is:
Xc=1/(2*pi*f*C). This relation shows that frequency is
inversely related to capacitive reactance. Hence, as supply
frequency increases, the capacitive reactance increases.

## 3. Calculate the time constant of a series RC circuit

consisting of a 100microF capacitor in series with a 100ohm
resistor.
a) 0.1 sec
b) 0.1 msec
c) 0.01 sec
d) 0.01 msec

Explanation: The time constant of a RC circuit= R*C=
100*10-6*100=0.01 sec.

## 4. Capacitors charge and discharge in __________

manner.
a) Linear
b) Constant
c) Square
d) Exponential

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Explanation: Capacitors charge and discharge in an
exponential manner because of the relation:
C=1/(2*pi*f*Xc).

## 5. Air has a dielectric constant of ___________

a) Unity
b) Zero
c) Infinity

Explanation: Dielectris constant of air is the same as that of
vacuum which is equal to unity. Dielctric constant of air is
taken as the reference to measure the dielectric constant of
all other materials.

## 6. What is the value of capacitance of a capacitor which has

a voltage of 4V and ha 8C of charge.
a) 2F
b) 4F
c) 6F
d) 8F

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV. From the relation, C=Q/V= 8/4=2F.

## 7. Unit of capacitance is___________

a) Volts

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c) Henry
d) Newton

Explanation: Volts is the unit of voltage, Henry for
inductance and Newton for force. Hence the unit for

## 8. What will happen to the capacitor when the source is

removed?
a) It will not remain in its charged state
b) It will remain in its charged state
c) It will start discharging
d) It will become zero

Explanation: As soon as the source is removed, the
capacitor does not start discharging it remains in the same
charged state.

## 9. Which among the following equations is incorrect?

a) Q=CV
b) Q=C/V
c) V=Q/C
d) C=Q/V

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.

4 of 5 3/19/18, 8:25 AM

Hence Q=CV. From the given relation we can derive all the
equations except for Q=C/V.

## 10. Capacitance is directly proportional to__________

a) Area of cross section between the plates
b) Distance of separation between the plates
c) Both area and distance
d) Neither area nor distance

Explanation: The relation between capacitance, area and
distance between the plates is:
C=K*epsilon*A/D. According to this relation, capacitance is
proportional to area.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:25 AM

sanfoundry.com

## Capacitors in Parallel - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitors in
Parallel”.

## 1. What is the total capacitance when three capacitors, C1,

C2 and C3 are connected in parallel?
a) C1/(C2+C3)
b) C1+C2+C3
c) C2/(C1+C3)
d) 1/C1+1/C2+1/C3

Explanation: When capacitors are connected in parallel, the
total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence Ctotal=C1+C2+C3.

## 2. Calculate the total capacitance.

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a) 10F
b) 15F
c) 13F
d) 20F

Explanation: The equivalent capacitance when capacitors
are connected in parallel is the sum of all the capacitors=
1+2+10= 13F.

## 3. Calculate the voltage across AB if the total change stored

in the combination is 13C.

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a) 1V
b) 2V
c) 3V
d) 4V

Explanation: The equivalent capacitance when capacitors
are connected in parallel is the sum of all the capacitors=
1+2+10= 13F. V=Q/C= 13/13=1V.

## 4. Calculate the charge in the 2F capacitor.

a) 200C
b) 100C
c) 300C
d) 400C

3 of 7 3/19/18, 8:25 AM

Explanation: Since the capacitors are connected in parallel,
the voltage across each is the same, it does not get divided.
Q=CV= 2*100=200C.

## 5. Calculate the charge in the 1F capacitor.

a) 200C
b) 100C
c) 300C
d) 400C

Explanation: Since the capacitors are connected in parallel,
the voltage across each is the same, it does not get divided.
Q=CV= 1*100=100C.

## 6. Calculate the total charge of the system.

a) 200C
b) 100C

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c) 300C
d) 400C

Explanation: The equivalent capacitance when capacitors
are connected in parallel is the sum of all the
capacitors=1+2=3F. Q=CV= 3*100= 300V.

## 7. When capacitors are connected in parallel, the total

capacitance is always __________ the individual
capacitance values.
a) Greater than
b) Less than
c) Equal to
d) Cannot be determined

Explanation: When capacitors are connected in parallel, the
total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence Ctotal=C1+C2+C3. Since it is
the sum of all the capacitance values, total capacitance is
greater the the individual capacitance values.

## 8. When capacitors are connected in parallel, what happens

to the effective plate area?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

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Explanation: When capacitors are connected in parallel, the
top plates of each of the capacitors are connected together
while the bottom plates are connected to each other. This
effectively increases the top plate area and the bottom plate
area.

## 9. Three capacitors having capacitance equal to 2F, 4F and

6F are connected in parallel. Calculate the effective parallel.
a) 10F
b) 11F
c) 12F
d) 13F

Explanation: When capacitors are connected in parallel, the
total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence Ctotal=C1+C2+C3=
2+4+6=12F.

## 10. Two capacitors having capacitance value 4F, three

capacitors having capacitance value 2F and 5 capacitors
having capacitance value 1F are connected in parallel,
calculate the equivalent capacitance.
a) 20F
b) 19F
c) 18F
d) 17F

6 of 7 3/19/18, 8:25 AM

## Explanation: When capacitors are connected in parallel, the

total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence
Ctotal=4+4+2+2+2+1+1+1+1+1=19F.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

7 of 7 3/19/18, 8:25 AM

sanfoundry.com

## Capacitors in Series - Basic

Electrical Engineering Questions
by Manish
4 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitors in
Series”.

## 1. What is the total capacitance when two capacitors C1

and C2 are connected in series?
a) (C1+C2)/C1C2
b) 1/C1+1/C2
c) C1C2/(C1+C2)
d) C1+C2

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2, therefore Ctotal= C1C2/(C1+C2).

## 2. N capacitors having capacitance C are connected in

series, calculate the equivalent capacitance.
a) C/N

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b) C
c) CN
d) N/C

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal= 1/C+1/C+1/C+……..N times.
1/Ctotal=N/C.
Ctotal=C/N.

## 3. When capacitors are connected in series, the equivalent

capacitance is ___________ each individual capacitance.
a) Greater than
b) Less then
c) Equal to
d) Insufficient data provided

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2. Since we find the reciprocals of the
sum of the reciprocals, the equivalent capacitance is less
than the individual capacitance values.

## 4. What is the equivalent capacitance?

a) 1.5F

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b) 0.667F
c) 2.45F
d) 2.75F

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2= 1/2+1= 0.667F.

## 5. When capacitors are connected in series ___________

remains the same.
a) Voltage across each capacitor
b) Charge
c) Capacitance
d) Resistance

Explanation: When capacitors are connected in series, the
charge remains the same because the same amount of
current flow exists in each capacitor.

## 6. When capacitors are connected in series

_______________ Varies
a) Voltage across each capacitor
b) Charge
c) Capacitance
d) Resistance

Explanation: When capacitors are connected in series, the

3 of 6 3/19/18, 8:26 AM

## voltage varies because the voltage drop across each

capacitor is different.

## 7. Four 10F capacitors are connected in series, calculate

the equivalent capacitance.
a) 0.2F
b) 0.4F
c) 0.5F
d) 0.6F

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2+1/C3+1/C4=1/10+1/10+1
/10+1/10=0.4F.

## 8. Calculate the charge in the circuit.

a) 66.67C
b) 20.34C
c) 25.45C
d) 30/45C

Explanation: When capacitors are connected in series, the

4 of 6 3/19/18, 8:26 AM

## equivalent capacitance is:

1/Ctotal=1/C1+1/C2= 1/2+1=0.667F.
Q=CV= 1.5*100= 66.67C.

## 9. Calculate the voltage across the 1F capacitor.

a) 33.33V
b) 66.67V
c) 56.56V
d) 23.43V

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2= 1/2+1=0.667F.
Q=CV= 1.5*100= 66.67C.
V across the 1F capacitor= 66.67/1= 66.67V.

## 10. Calculate the voltage across the 2F capacitor.

5 of 6 3/19/18, 8:26 AM

a) 33.33V
b) 66.67V
c) 56.56V
d) 23.43V

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2= 1/2+1=0.667F.
Q=CV= 1.5*100= 66.67C.
V across the 1F capacitor= 66.67/2= 33.33V.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:26 AM

sanfoundry.com

Freshers
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Questions and

Answers for Freshers focuses on “Distribution of Voltage
Accross Capacitors in Series”.

## 1. The total voltage drop across a series of capacitors is

__________
a) The voltage drop across any one of the capacitors
b) The sum of the voltage drop across each of the
capacitors
c) The product of the voltage drop across each of the
capacitors
d) Zero

Explanation: The total voltage drop is equal to the sum of
the voltage drop across each off the capacitors because
when capacitors are connected in series, the voltage drops
across each capacitor.

1 of 6 3/19/18, 8:26 AM

## 2. Capacitors C1, C2 and C3 have voltage drops 2V, 3V

and 5V respectively. Calculate the total voltage in the
circuit.
a) 10V
b) 2V
c) 5V
d) 0V

Explanation: When capacitors are connected in series:
Vtotal=V12+V2+V3= 2+3+5=10V.

## 3. What is the voltage across the 2F capacitor?

a) 242V
b) 2V
c) 220V
d) 121V

Explanation: The equivalent capacitance is equal to:
1/C=1/2+1/4+1/6, therefore, C=1.1F.
Q=C*V= 220*1.1= 242C.

2 of 6 3/19/18, 8:26 AM

## 4. What is the voltage across the 4F capacitor?

a) 242V
b) 60.5V
c) 22.5V
d) 12.5V

Explanation: The equivalent capacitance is equal to:
1/C=1/2+1/4+1/6, therefore, C=1.1F.
Q=C*V= 220*1.1= 242C.
V across 4F capacitor = Q/C= 242/4= 60.5V.

## 5. Calculate the voltage across the 6F capacitor.

a) 242V

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b) 60.5V
c) 40.33V
d) 12.5V

Explanation: The equivalent capacitance is equal to:
1/C=1/2+1/4+1/6, therefore, C=1.1F.
Q=C*V= 220*1.1= 242C.
V across 6F capacitor = Q/C= 242/6= 40.33V.

## 6. When capacitors are connected in series, which of the

following rules are applied?
a) Voltage divider
b) Current divider
c) Both voltage divider and current divider
d) Neither voltage divider nor current divider

Explanation: Voltage divider is the rule applied when
capacitors are connected in series because when
capacitors are connected in series, the voltage is different
across each capacitor.

## 7. A capacitor does not allow sudden changes in

_________
a) Current
b) Voltage
c) Resistance
d) Inductance

4 of 6 3/19/18, 8:26 AM

Explanation: Capacitor does not allow sudden changes in
voltage because these changes occur in zero time which
results in the current being infinity, which is not possible.

## 8. Which of the following expressions is correct with respect

to the voltage across capacitors in series?
a) V1/V2=C2/C1
b) V2/V1=C2/C1
c) V1*V2=C1*C2
d) V1/C1=V2/C2

Explanation: When capacitors are connected in series, the
charge across each capacitor remains the same whereas
the voltage across each varies. When two capacitors are
connected in series:
Q=V1C1; Q=V2C2. Thus: V1/V2=C2/C1.

## 9. Two 4F capacitors are connected in series, calculate the

voltage across each if the total voltage is 20V.
a) 10V
b) 5V
c) 20V
d) 0V

Explanation: The two capacitors have the same
capacitance, hence the voltage gets divides equally. V
across each=Total voltage/2= 20/2= 10V.

5 of 6 3/19/18, 8:26 AM

## 10. Two capacitors having voltage 2F and 4F are

connected in series. This combination is connected to a
100V supply, calculate the voltage across the 2F capacitor.
a) 66.67V b) 33.33V c) 100V d) 0V
Explanation: Using voltage divider rule:
V across 2F= 100*2/(2+4)= 33.33V.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering for

Freshers, here is complete set of 1000+ Multiple Choice

6 of 6 3/19/18, 8:26 AM

sanfoundry.com

by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitance and
the Capacitor”.

## 1. For high frequencies, capacitor acts as _______

a) Open circuit
b) Short circuit
c) Amplifier
d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very high frequencies, the
impedance is almost equal to zero, hence it acts as a short
circuit and there is no voltage across it.

## 2. For very low frequencies, capacitor acts as _______

a) Open circuit
b) Short circuit
c) Amplifier

1 of 5 3/19/18, 8:26 AM

d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very low frequencies the impedance
is almost infinity and hence acts as an open circuit and no
current flows through it.

## 3. A capacitor consists of___________

a) Two conductors
b) Two semiconductors
c) Two dielectrics
d) Two insulators

Explanation: A capacitor consists of two conductors
connected in parallel to each other so that it can store
charge in between the plates.

## 4. Capacitor preferred when there is high frequency in the

circuits is ______
a) Electrolyte capacitor
b) Mica capacitor
c) Air capacitor
d) Glass capacitor

Explanation: Mica capacitors are preferred for high
frequency circuits because they have low ohmic losses and
less reactance.

2 of 5 3/19/18, 8:26 AM

## 5. Capacitance increases with ______

a) Increase in plate area
b) Decrease in plate area
c) Increase in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is directly proportional to plate
area. Hence as plate area increases, the capacitance also
increases.

## 6. Capacitance decreases with _________

a) Increase in distance between the plates
b) Decrease in plate area
c) Decrease in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is inversely proportional to the
distance between the two parallel plates. Hence, as the
distance between the plates decreases, the capacitance
increases.

## 7. When the supply frequency increases, what happens to

the capacitive reactance in the circuit?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 5 3/19/18, 8:26 AM

Explanation: The expression for capacitive reactance is:
Xc=1/(2*pi*f*C). This relation shows that frequency is
inversely related to capacitive reactance. Hence, as supply
frequency increases, the capacitive reactance increases.

## 8. What is the value of capacitance of a capacitor which has

a voltage of 4V and ha 16C of charge.
a) 2F
b) 4F
c) 6F
d) 8F

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV. From the relation, C=Q/V= 16/4=4F.

## 9. Unit of capacitance is __________

a) Volts
c) Henry
d) Newton

Explanation: Volts is the unit of voltage, Henry for
inductance and Newton for force. Hence the unit for

## 10. What will happen to the capacitor when the source is

removed?

4 of 5 3/19/18, 8:26 AM

## a) It will not remain in its charged state

b) It will remain in its charged state
c) It will start discharging
d) It will become zero

Explanation: As soon as the source is removed, the
capacitor does not start discharging it remains in the same
charged state.

## Sanfoundry Global Education & Learning Series –

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## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:26 AM

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## Electric Fields - Basic Electrical

Engineering Questions and
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Electric Fields”.

## 1. The conventional direction of electric field is ________

a) Positive to negative
b) Negative to positive
c) No specific direction
d) Direction cannot be determined

Explanation: The conventional direction of field lines is from
positive to negative. The field lines originate at the positive
charge and terminate at the negative charge.

## 2. Electric field originates at __________

a) Positive charge
b) Negative charge
c) Neither positive nor negative
d) Both positive and negative

1 of 5 3/19/18, 8:26 AM

Explanation: Electric field originates at the positive charge
and terminates at the negative charge. The conventional
direction of field is from positive to negative.

## 3. Electric field terminates at ________

a) Positive charge
b) Negative charge
c) Neither positive nor negative
d) Both positive and negative

Explanation: Electric field originates at the positive charge
and terminates at the negative charge. The conventional
direction of field is from positive to negative.

## 4. Which among the following statements is true with regard

to electric field lines?
a) Electric field lines always intersect
b) Electric field lines may or may not intersect
c) Electric field lines can be seen
d) Electric field lines never intersect

Explanation: Electric field lines can never intersect because
the field lines represent the field strength. If the lines
intersect it means that at that point there are two different
values for electric field which is not possible.

2 of 5 3/19/18, 8:26 AM

## 5. Which, among the following, is the field where electric

charge experiences a force?
a) Electric field
b) Magnetic field
c) Gravitational field
d) Electric, magnetic or gravitational field

Explanation: Charges experience a force in an electric field
because charges come under the influence of a field which

## 6. A field that spreads outwards in all directions is

__________
a) Linear
c) Weak
d) Strong

directions. This field is known as the radial field because it

## 7. In uniform fields, all points have ________ field strength.

a) Zero
b) Same
c) Infinity
d) Different

3 of 5 3/19/18, 8:26 AM

Explanation: A uniform field is one-as the word suggests-in
which the field is spread over an area and at every point in
the field the, strength of the field is the same.

## 8. Which, among the following is the correct expression for

electric field?
a) E=F/C
b) E=F*C
c) E=F/Q
d) E=F*Q

Explanation: Electric field is the force per unit charge
hence, the correct expression among the following is:
E=F/Q.

## 9. What happens when one material is rubbed against

another?
a) The material becomes electrically neutral
b) The material becomes electrically charged
c) The material becomes negatively charged
d) The material becomes positively charged

Explanation: When one material is rubbed against another,
there is transfer of charges from one material to another
hence the material becomes electrically charged.

## 10. The insulant between the two plates of a capacitor is

called _______

4 of 5 3/19/18, 8:26 AM

a) Conductor
b) Semi-conductor
c) Dielectric
d) Superconductor

Explanation: The material between the two plates of a
capacitor is an insulator, more specifically known as a
dielectric.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:26 AM

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## Electric Field Strength & Electric

Flux Density Questions and
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Electric Field
Strength and Electric Flux Density”.

## 1. Gauss law is applicable for_________

a) Point charge
b) Sheet charge
c) Line charge
d) Point, sheet and line charge

Explanation: Gauss law states that the total electric flux
through any closed surface is equal to the charge enclosed
by that surface.Hence it is applicable for all point, surface
and volume.

## 2. “Total electric flux through any closed surface is equal to

the charge enclosed by that surface”. This is the statement
for?

1 of 5 3/19/18, 8:29 AM

a) Gauss law
b) Lenz law
c) Coloumb’s law

Explanation: Total electric flux through any closed surface is
equal to the charge enclosed by that surface is the
statement for Gauss law because among the four laws,
Gauss law deals with electric flux.

## 3. Electric flux density is a function of_______

a) Volume
b) Charge
c) Current
d) Voltage

Explanation: Electric flux density is the charge per unit area.
Hence it is a function of charge and not any of the other
values.

## 4. As charge increases, what happens to flux density?

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: Electric flux density is the charge per unit area.

2 of 5 3/19/18, 8:29 AM

## The expression for flux density is:

D=Q/A. Electric flux is directly proportional to charge, hence
as charge increases, electric flux also increases.

## 5. As area increases, what happens to electric flux density?

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: Electric flux density is the charge per unit area.
The expression for flux density is:
D=Q/A. Electric flux is inversely proportional to area, hence
as area increases, electric flux decreases.

## 6. Which, among the following, is the correct expression for

electric flux density?
a) D=epsilon*E
b) D=epsilon/E
c) D2=epsilon*E
d) D=epsilon*E2

Explanation: Electric flux density is directly proportional to
the electric field, epsilon(permittivity of free space) being the
constant of proportionality. Hence D=epsilon*E.

## 7. Strength of the electric field is ___________

a) Directly proportional to the force applied
b) Inversely proportional to the force applied

3 of 5 3/19/18, 8:29 AM

## c) Directly proportional to the square of the force applied

d) Inversely proportional to the square of the force applied

Explanation: Electric field intensity is the force per unit
charge, hence it id directly proportional to the force applied.

## 8. The force applied to a conductor is 10N, if the charge in

the conductor is 5C, what is the electric field intensity?
a) 10V/m
b) 2V/m
c) 3V/m
d) 15V/m

Explanation: Electric field intensity is the force per unit
charge. The formula is:
E=F/Q= 10/5= 2V/m.

## 9. What is the electric flux density in free space if the

electric field intensity is 1V/m?
a) 7.76*10-12C/m2
b) 8.85*10-12C /m2
c) 1.23*10-12C /m2
d) 3.43*10-12C /m2

Explanation: The formula for electric filed density is:
D=epsilon*E= 1*8.85*110-12= 8.85*10-12C /m2.

4 of 5 3/19/18, 8:29 AM

## 10. If the charge in a conductor is 16C and the area of

cross section is 4m2. Calculate the electric flux density.
a) 64C/m2
b) 16C/m2
c) 4C/m2
d) 2C/m2

Explanation: Flux density is the charge per unit area. The
formula is:
D=Q/A= 16/4= 4C/m2.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

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## Relative Permittivity - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Relative
Permittivity”.

## 1. In order to obtain a high value for capacitance the

permittivity of the dielectric medium should be?
a) Low
b) High
c) Zero
d) Unity

Explanation: Form the expression:
C=epsilon*A/d.
From this expression, it is seen that capacitance is directly
proportional to the permittivity, hence for capacitance value
to be high, permittivity value should be high.

## 2. Find the capacitance of a capacitor whose area of cross

1 of 5 3/19/18, 8:29 AM

## section of the plates is 4m2 and distance of separation

between the plates is 2m. The capacitor is placed in
vacuum.
a) 1.77*10-11F
b) 1.34*10-11F
c) 2.33*10-11F
d) 5.65*10-11F

Explanation: The expression for finding the value of
capacitance is:
C=epsilon*A/d.
The medium is free space hence, epsilon= 8.85*10-12.
Therefore, C=8.85*10-12*4/2= 1.77*10-12F.

## 3. What is relative permittivity?

a) Equal to absolute permittivity
b) Ratio of actual permittivity to absolute permittivity
c) Ratio of absolute permittivity to actual permittivity
d) Equal to actual permittivity

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium. As the
actual permittivity increases the relative permittivity also
increases.

## 4. What happens to relative permittivity when actual

permittivity decreases?
a) Increases

2 of 5 3/19/18, 8:29 AM

b) Decreases
c) Remains the same
d) Becomes zero

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium.
Relative permittivity is directly proportional to actual
permittivity. Hence, as actual permittivity increases, relative
permittivity also increases.

## 5. What is the relative permittivity when the actual

permittivity is 4F/m?
a) 4.57*10-11
b) 4.57*1012
c) 4.57*1011
d) 4.57*10-12

Explanation: Relative permittivity= Actual permittivity/
Absolute permittivity.
Relative permittivity= 4/(8.85*10-12)= 4.57*1011.

## 6. What happens to absolute permittivity when relative

permittivity increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 5 3/19/18, 8:29 AM

Explanation: Absolute permittivity does not depend on the
value of relative permittivity. Absolute permittivity is the
permittivity of free space and it is a constant value=
8.85*10-12F/m.

## 7. Calculate the actual permittivity of a medium whose

relative permittivity is 5.
a) 4.43*10-11F/m
b) 4.43*10-12F/m
c) 4.43*1011F/m
d) 4.43*1012F/m

Explanation: Actual permittivity= Relative
permittivity*absolute permittivity.
Actual permittivity= 5*8.85*10-12= 4.43*10-11F/m.

## 8. What is the unit for relative permittivity?

a) F/m
b) Fm
c) F/m2
d) No unit

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium. Since it
is a ratio, and we know that a ratio does not have any unit,
relative permittivity does not have any unit.

## 9. Which, among the following, will be unity in free space?

4 of 5 3/19/18, 8:29 AM

a) Absolute permittivity
b) Relative permittivity
c) Actual permittivity
d) Both absolute and relative permittivity

Explanation: Relative permittivity is constant for a particular
medium. For air or free space ,it is unity. Absolute
permittivity does not depend on the medium, its value is
always constant=8.85*10-12F/m. Actual permittivity is the
product of relative permittivity and absolute permittivity.

## 10. Which, among the following, do not have any unit?

a) Absolute permittivity
b) Relative permittivity
c) Actual permittivity
d) Both absolute and relative permittivity

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium. Since it
is a ratio, and we know that a ratio does not have any unit,
relative permittivity does not have any unit.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:29 AM

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by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitance of a
Multi Plate Capacitor”.

## 1. Which is the correct expression for capacitance of a multi

plate capacitor?
a) C=absolute permittivity*A/d
b) C=Actual permittivity*(n-1)*A/d
c) C=Actual permittivity*(n)*A/d
d) Actual permittivity*(n+1)*A/d

Explanation: The correct expression is: C=Actual
permittivity*(n-1)*A/d.
Where, n=number of plates, A=area of cross section of the
plates, d=distance of separation between the plates.

## 2. What happens to the capacitance of a multi plate

capacitor when the area of cross section of the plate
decreases?

1 of 6 3/19/18, 8:29 AM

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When the area of cross section decreases, the
capacitance also decreases since it is related by the
formula C=Actual permittivity*(n-1)*A/d. Here, we can see
that the capacitance is directly proportional to the area of
cross section.

## 3. What happens to the capacitance of a multi plate

capacitor when the distance of separation between the
plate increases?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When the distance of separation between the
plates decreases, the capacitance also decreases since it is
related by the formula C=Actual permittivity*(n-1)*A/d. Here,
we can see that the capacitance is inversely proportional to
the distance of separation.

## 4. What happens to the capacitance of a multi plate

capacitor when the number of plates increases?
a) Increases

2 of 6 3/19/18, 8:29 AM

b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When the number of capacitors increases, the
capacitance also increases since it is related by the formula
C=Actual permittivity*(n-1)*A/d. Here, we can see that the
capacitance is directly proportional to the number of
capacitors.

## 5. Find the capacitance of a multi plate capacitor whose

actual permittivity= 5F/m, n=3, A=4m2and d=2m.
a) 10F
b) 20F
c) 30F
d) 40F

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.
Thus, C=5*(3-1)*4/2= 20F.

## 6. Find the capacitance of a multi plate capacitor whose

relative permittivity=5, n=3, A=4m2 and d=2m.
a) 1.77*10-10 F
b) 1.77*1010 F
c) 1.77*10-11 F
d) 1.77*1011 F

3 of 6 3/19/18, 8:29 AM

Explanation: The formula for capacitance of a multi plate
capacitor: C=Relative permittivity*absolute permittivity*
(n-1)*A/d.
C= 5*8.85*10-12*(3-1)*4/2=1.77*10-10.

## 7. Find the number of plates in the multi plate capacitor

having C=20F absolute permittivity=5F/m, A=4m2 and
d=2m.
a) 1
b) 2
c) 3
d) 4

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.
Substituting the given values in the equation, we get n=3.

## 8. Calculate the distance between the plates of the

capacitor having C=20F, actual permittivity=F/m
n=3 and A=4m2.
a) 1m
b) 2m
c) 3m
d) 4m

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.

4 of 6 3/19/18, 8:29 AM

## 9. Calculate the area of cross section of the multi plate

capacitor having C=20F, actual permittivity=F/m n=3 and
d=2m.
a) 1m
b) 2m
c) 3m
d) 4m

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.
Substituting the given values in the equation, we get
A=4m2.

## 10. Calculate the number of plates in the multi plate

capacitor having C=1.77*10-10F relative permittivity=5,
A=4m2 and d=2m.
a) 1m
b) 2m
c) 3m
d) 4m

Explanation: The formula for capacitance of a multi plate
capacitor: C=Relative permittivity*absolute permittivity*
(n-1)*A/d.
Substituting the given values in the equation, we get n=3.

## Sanfoundry Global Education & Learning Series –

5 of 6 3/19/18, 8:29 AM

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:29 AM

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by Manish
5-6 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Composite
Dielectric Capacitor”.

## 1. Potential drop in a dielectric is equal to _______

a) Electric field strength*thickness
b) Electric field strength*area of cross section
c) Electric field strength
d) Zero

Explanation: When a dielectric is introduced between the
two plates of a parallel plate capacitor, the potential
difference decreases by the value of the product of electric
field strength*thickness which is the potential difference of
the dielectric.

## 2. The electric field strength is 10N/C and the thickness of

the dielectric is 3m. Calculate the potential drop in the
dielectric.

1 of 6 3/19/18, 8:30 AM

a) 10V
b) 20V
c) 30V
d) 40V

Explanation: The potential drop in a dielectric= electric field
strength*area of cross section= 10*3= 30V.

## 3. The electric fields of dielectrics having the same cross

sectional area in series is related to their relative
permittivities in which way?
a) Directly proportional
b) Inversely proportional
c) Equal
d) Not related

Explanation: Let us consider two plates having fields E1
and E2 and relative permittivities e1 and e2. Then,
E1=Q/(e0*e1*A) and E2=Q/(e0*e2*A), where e0=absolute
permittivity and A=area of cross section. From the given
expression, we see that E1/E2=e2/e1, hence the electric
field is inversely proportional to the relative permittivities.

## 4. What happens to the capacitance when a dielectric is

introduced between its plates?
a) Increases
b) Decreases
c) Remains the same

2 of 6 3/19/18, 8:30 AM

d) Becomes zero

Explanation: The capacitance of a capacitance increases
when a dielectric is introduced between its plates because
the capacitance is related to the dielectric constant k by the
equation:
C=ke0A/d.

## 5. Calculate the relative permittivity of the second dielectric

if the relative permittivity of the first is 4. The electric field
strength of the first dielectric is 8V/m and that of the second
is 2V/m.
a) 32
b) 4
c) 16
d) 8

Explanation: The relation between the two electric fields
and the relative permittivities is:
E1/E1=e2/e1. Substituting the given values, we get e2=16.

## 6. What happens to the potential drop between the two

plates of a capacitor when a dielectric is introduced
between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 6 3/19/18, 8:30 AM

Explanation: When a dielectric is introduced between the
two plates of a parallel plate capacitor, the potential
difference decreases because the potential difference of the
dielectric is subtracted from it.

## 7. If the potential difference across the plates of a capacitor

is 10V and a dielectric having thickness 2m is introduced
between the plates, calculate the potential difference after
introducing the dielectric. The electric field strength is 2V/m.
a) 4V
b) 6V
c) 8V
d) 10V

Explanation: When a dielectric is introduced between the
plates of a capacitor, its potential difference decreases.
New potential difference= potential difference without
dielectric-potential difference of dielectric= 10-2*2= 6V.

## 8. Calculate the capacitance if the dielectric constant=4,

area of cross section= 10m2 and the distance of separation
between the plates is 5m.
a) 7.08*10-11F
b) 7.08*1011F
c) 7.08*10-12F
d) 7.08*10-10F

4 of 6 3/19/18, 8:30 AM

Explanation: The expression to find capacitance when a
dielectric is introduced between the plates is:
C=ke0A/d. Substituting the given values in the equation, we
get C= 7.08*10-11F.

## 9. A dielectric is basically a ________

a) Capacitor
b) Conductor
c) Insulator
d) Semiconductor

Explanation: A dielectric is basically an insulator because it
has all the properties of an insulator.

## 10. What happens to the potential difference between the

plates of a capacitor as the thickness of the dielectric slab
increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a dielectric is introduced between the
plates of a capacitor, its potential difference decreases.
New potential difference= potential difference without
dielectric-potential difference of dielectric. Hence as the
thickness of the dielectric slab increases, a larger value is

5 of 6 3/19/18, 8:30 AM

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:30 AM

sanfoundry.com

by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Charging and
Discharging Currents”.

## 1. Which of the following charging and discharging of a

capacitor?
a) Time constant
b) Current
c) Power
d) Voltage

Explanation: The time constant in a circuit consisting of a
capacitor is the product of the resistance and the
capacitance. Smaller the time constant, faster is the
charging and discharging rate and vice versa.

## 2. What is the initial current while charging a capacitor?

a) High
b) Low

1 of 5 3/19/18, 8:30 AM

c) 0
d) Cannot be determined

Explanation: The initial current of a capacitor is very high
because the voltage source will transport charges from one
plate of the capacitor to the other plate.

## 3. What is the final current while charging a capacitor?

a) High
b) Zero
c) Infinity
d) Low

Explanation: The final current is almost equal to zero while
charging a capacitor because the capacitor is charged up to
the source voltage.

## 4. What happens to the current flow in a fully charged

capacitor?
a) Current flow stops
b) Current flow doubles
c) Current flow becomes half its original value
d) Current flow becomes one-fourth its original value

Explanation: When a capacitor is fully charged, it does not
store any more charge. There is no change in charge with
time. Current is the rate of change of charge, hence it

2 of 5 3/19/18, 8:30 AM

## 5. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the initial value of discharge
current.
a) 10A
b) 0A
c) Infinity
d) 20A

Explanation: When the capacitor is discharging the value of
initial current is a finite one. The finite initial current value is
found using ohm’s law: I=V/R= 400/20= 20A.

## 6. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the final value of discharge
current.
a) 10A
b) 0A
c) Infinity
d) 20A

Explanation: In a discharging circuit, the final voltage is
equal to zero as all the positive and negative charges have
combined. Since the voltage is equal to zero, the current is
also equal to zero as voltage is directly proportional to
current by ohm’s law.

## 7. When will be capacitors fully charged?

3 of 5 3/19/18, 8:30 AM

## a) When voltage is zero

b) When the supply voltage is equal to the capacitor voltage
c) When voltage is infinity
d) When capacitor voltage is equal to half the supply
voltage

Explanation: When the capacitor voltage is equal to the
source voltage, it means that all the charges have moved
from one plate of the capacitor to the other.

## 8. What happens to the capacitor when the capacitor

voltage is equal to the source voltage?
a) The charging phase of the capacitor is over
b) The discharging phase of the capacitor is over
c) The capacitor is switched off
d) The capacitor is switched on

Explanation: When the capacitor voltage is equal to the
source voltage, it means that all the charges have moved
from one plate of the capacitor to the other. Hence the
capacitor is charged and the charging phase is over.

## 9. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the final value of charging
current.
a) 10A
b) 0A
c) Infinity

4 of 5 3/19/18, 8:30 AM

d) 20A

Explanation: The final value of charging current in a
capacitor is equal to the initial value of discharging current
in it. Hence the final value of charging current is: Vc/R=20A.

## 10. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the initial value of charging
current.
a) 10A
b) 0A
c) Infinity
d) 20A

Explanation: Initially, there’s 0V voltage in a capacitor. As
the capacitor charges, the voltage increases. Since voltage
is proportional to current by ohm’s law, initial current is also
equal to zero.

## Sanfoundry Global Education & Learning Series –

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## Growth & Decay - Basic Electrical

Engineering Questions and
by Manish
5-6 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Growth and
Decay”.

## 1. The charging time constant of a circuit consisting of a

capacitor is the time taken for the charge in the capacitor to
become __________% of the initial charge.
a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(1-e-t/RC).
When RC=t, we have: Q=Q0(1-e-1)= 0.63*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.63 times its initial charge.

## 2. The discharging time constant of a circuit consisting of a

1 of 6 3/19/18, 8:30 AM

## capacitor is the time taken for the charge in the capacitor to

become __________% of the initial charge.
a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(e-t /RC).
When RC=t, we have: Q=Q0(e-1)= 0.37*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.37 times its initial charge.

## 3. A circuit has a resistance of 2 ohm connected in series

with a capacitance of 6F. Calculate the charging time
constant.
a) 3
b) 1
c) 12
d) 8

Explanation: The charging time constant in a circuit
consisting of a capacitor and resistor in series is the product
of the resistance and capacitance= 2*6=12.

## 4. A circuit has a resistance of 5 ohm connected in series

with a capacitance of 10F. Calculate the discharging time
constant.
a) 15

2 of 6 3/19/18, 8:30 AM

b) 50
c) 5
d) 10

Explanation: The discharging time constant in a circuit
consisting of a capacitor and resistor in se-ries is the
product of the resistance and capacitance= 5*10=50.

## 5. What is the value of current in a discharging capacitive

circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A

Explanation: At time t=RC, that is the time constant, we
know that the value of current at that time interval is equal
to 37% of the initial charge in the discharging circuit. Hence,
I=2*0.37= 0.74A.

## 6. What is the value of current in a charging capacitive

circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A

3 of 6 3/19/18, 8:30 AM

## Explanation: At time t=RC, that is the time constant, we

know that the value of current at that time interval is equal
to 63% of the initial charge in the charging circuit. Hence,
I=2*0.63= 1.26A.

## 7. While discharging, what happens to the current in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of current in a
discharging capacitive circuit is:
I=I0*e-t /RC. From this equation, we can see that the
current is exponentially decreasing since e is raised to a
negative power.

## 8. While discharging, what happens to the voltage in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of voltage in a
discharging capacitive circuit is:
V=V0*e-t /RC. From this equation, we can see that the

4 of 6 3/19/18, 8:30 AM

negative power.

## 9. While charging, what happens to the current in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of current in a
charging capacitive circuit is:
I=I0*(1-e-t /RC). From this equation, we can see that the
current is exponentially increasing since e is raised to a
negative power and we are subtracting it from 1. Hence as
the value of e-t /RC in-creases, the current increases
exponentially.

## 10. While charging, what happens to the voltage in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of voltage in a
charging capacitive circuit is:
V=V0*(1-e-t /RC). From this equation, we can see that the

5 of 6 3/19/18, 8:30 AM

## voltage is exponentially increasing since e is raised to a

negative power and we are subtracting it from 1. Hence as
the value of e-t /RC in-creases, the voltage increases
exponentially.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:30 AM

sanfoundry.com

for Freshers
by Manish
4-6 minutes

## This set of Basic Electrical Engineering Interview Questions

and Answers for freshers focuses on “Discharge of a
Capacitor Through a Resistor”.

## 1. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the time constant.
a) 1s
b) 2s
c) 3s
d) 4s

Explanation: The time constant is the product of the
resistance and capacitance in a series RC circuit.
Therefore, time constant= 8*10-6*4*106=4s.

## 2. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.

1 of 6 3/19/18, 8:30 AM

## Calculate the initial charging current.

a) 100 microA
b) 500 microA
c) 400 microA
d) 1000microA

Explanation: In a series RC circuit, the initial charging
current is:
I=V/R= 200/(0.5*106s)= 400*10-6A= 400 microA.

## 3. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the time taken for the potential difference across
the capacitor to gow to 160V.
a) 6.44s
b) 7.77s
c) 2.33s
d) 3.22s

Explanation: From the previous explanations, we know that
the initial current is 400mA and the time constant is 4s.
Substituting the values of capacitor voltage, initial voltage,
initial current and time constant in the equation: Vc=V(1-
e-t/RC), we get t=6.44s.

## 4. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the voltage in the capacitor 4s after the power is

2 of 6 3/19/18, 8:30 AM

supplied.
a) 123.4V
b) 126.4V
c) 124.5V
d) 132.5V

Explanation: We can get the value of the potential
difference across the capacitor in 4s, from the following
equation:
Vc=V(1-e-t /RC). Substituting the values in the given
equation, we get Vc= 126.4V.

## 5. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the current in the capacitor 4s after the power is
supplied.
a) 150 microA
b) 149 microA
c) 148 microA
d) 147 microA

Explanation: In the given question, the time constant is
equal to the time taken= 4s. Hence the value of current will
be 63% of its initial value= I=0.63*400= 147 microA.

## 6. The discharging time constant of a circuit consisting of a

capacitor is the time taken for the charge in the capacitor to
become __________% of the initial charge.

3 of 6 3/19/18, 8:30 AM

a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(e-t /RC).
When RC=t, we have: Q=Q0(e-1)= 0.37*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.37 times its initial charge.

## 7. The discharging time constant of a circuit consisting of a

capacitor is the time taken for the charge in the capacitor to
become __________% of the initial charge.
a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(e-t /RC).
When RC=t, we have: Q=Q0(e-1)= 0.37*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.37 times its initial charge.

## 8. A circuit has a resistance of 2 ohm connected in series

with a capacitance of 6F. Calculate the discharging time
constant.
a) 3
b) 1

4 of 6 3/19/18, 8:30 AM

c) 12
d) 8

Explanation: The discharging time constant in a circuit
consisting of a capacitor and resistor in series is the product
of the resistance and capacitance= 2*6=12.

## 9. What is the value of current in a discharging capacitive

circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A

Explanation: At time t=RC, that is the time constant, we
know that the value of current at that time interval is equal
to 37% of the initial charge in the discharging circuit. Hence,
I=2*0.37= 0.74A.

## 10. While discharging, what happens to the current in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of current in a

5 of 6 3/19/18, 8:30 AM

## discharging capacitive circuit is:

I=I0*e-t /RC. From this equation, we can see that the
current is exponentially decreasing since e is raised to a
negative power.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering for

Interviews, here is complete set of 1000+ Multiple Choice

6 of 6 3/19/18, 8:30 AM

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## CR Networks Transients - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Transients in CR
Networks”.

## 1. A CR network is one which consists of _________

a) A capacitor and resistor connected in parallel
b) A capacitor and resistor connected in series
c) A network consisting of a capacitor only
d) A network consisting of a resistor only

Explanation: A CR network is one which consists of a
capacitor connected in series with a resistor. The capacitor
discharges or charges through the resistor.

## 2. At DC, capacitor acts as _________

a) Open circuit
b) Short circuit
c) Resistor

1 of 7 3/19/18, 8:31 AM

d) Inductor

Explanation: At DC, the capacitor acts as open circuit
because the capacitive resistance is infinity. The frequency
of a DC circuit is 0. The capacitive resistance=1/(2*pi*f*C).
Therefore, if the frequency is 0, the capacitive resistance is
infinity and it acts as an open circuit.

## 3. In an RC series circuit, when the switch is closed and the

circuit is complete, what is the response?
a) Response does not vary with time
b) Decays with time
c) Increases with time
d) First increases, then decreases

Explanation: In an RC series circuit, the response decays
with time because according to the equation, there is an
exponential decrease in the response.

## 4. If the switch is clocked at t=0, what is the current in the

circuit?

a) 0A

2 of 7 3/19/18, 8:31 AM

b) 10A
c) 20A
d) Infinity

Explanation: As soon as the switch is closed at t=0, the
capacitor acts as a short circuit. The current in the circuit is:
I=V/R= 100/10= 10A.

## 5. Calculate the voltage across the capacitor at t=0.

a) 0A
b) 10A
c) 20A
d) Infinity

Explanation: When the switch is closed at t=0, the capacitor
has no voltage across it since it has not been charged. The
capacitor acts as a short circuit and the voltage across it is
zero.

## 6. Calculate di(0)/dt if the switch is closed at t=0.

3 of 7 3/19/18, 8:31 AM

a) -9.9A/s
b) -10A/s
c) 0A/s
d) -0.1A/s

Explanation: Applying KVL to the given circuit, we get:
100+10i(0)+1/10*integral(i(0)dt)=0
Differentiating once, we get:
10di(0)/dt+1/10*i.
From the previous questions, we know that i(0)=10A.
Substituting the values in the equation, we get di(0)/dt=-
0.1A/s.

## 7. Calculate d2i(0)/dt2 from the given circuit.

a) 10-6A/s2
b) 10-3A/s2
c) 106A/s2

4 of 7 3/19/18, 8:31 AM

d) 103A/s2

Explanation: Applying KVL to the given circuit, we get:
100+10i(0)+1/10*integral(i(0)dt)=0
Differentiating once, we get:
10di(0)/dt+1/10*i.
Differentiating once again, we get:
10d2i(0)/dt2+10di(0)/dt=0.
Substituting the values of di/dt from the previous
explanation, we get d2i(0)/dt2=10-3A/s2.

## 8. The current equation for the given circuit is?

a) i=10e(-0.01)t A
b) i=10e(0.01)t A
c) i=10e(-0.001)t A
d) i=100e(-0.01)t A

Explanation: The KVL equation is:
100+10i(0)+1/10*integral(i(0)dt)=0
On applying Laplace transform to this equation, we get:
100/s=I(s)/10s+10I(s)

5 of 7 3/19/18, 8:31 AM

i=10e(-0.01)t A.

## 9. The expression for current in an RC circuit is?

a) i=(V/R)e(t/RC )
b) i=(V/R)e(-t/RC )
c) i=(V/R)/e(t/RC )
d) i=(V/R)/e(-t/RC )

Explanation: The particular solution of the current equation
is zero. Hence, the expression for the current in an RC
circuit is:
i=(V/R)e(-t/RC ).

## 10. What is the voltage in the resistor as soon as the switch

is closed at t=0.

a) 0V
b) Infinity
c) 220V
d) Insufficient information provided

Explanation: As soon as the switch is closes at t=0, there is

6 of 7 3/19/18, 8:31 AM

## no charge in the capacitor, hence the voltage across the

capacitor os zero and all the 220V voltage is the voltage
across the resistor.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

7 of 7 3/19/18, 8:31 AM

sanfoundry.com

by Manish
3-4 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Energy Stored in
a Charged Capacitor”.

## 1. Work done in charging a capacitor is ____________

a) QV
b) 1/2QV
c) 2QV
d) QV2

Explanation: We know that work done= Q2/2C.
Substituting C as Q/V, we get work done= Q/2V.

## 2. Energy stored in 2000mF capacitor charged to a

potential difference of 10V is?
a) 0.1J
b) 0.2J
c) 0.3J
d) 0.4J

1 of 6 3/19/18, 8:31 AM

Explanation: From the expression:
WD= CV2/2= 0.1J.

## 3. When do we get maximum energy from a set of

capacitors?
a) When they are connected in parallel
b) When they are connected in series
d) Both in series and parallel
d) Insufficient information provided

Explanation: We get maximum energy when capacitors are
connected in parallel because the equivalent capacitance is
larger than the largest individual capacitance when
connected in parallel. The relation between capacitance
and energy is:
Energy=CV2/2, hence as the capacitance increases, the
energy stored in it also increases.

## 4. If the charge stored in a capacitor is 4C and the value of

capacitance is 2F, calculate the energy stored in it.
a) 2J
b) 4J
c) 8J
d) 16J

Explanation: The expression for finding the value of energy

2 of 6 3/19/18, 8:31 AM

is:
U=Q2/2C= 4*4/(2*2)= 4J.

## 5. If the charge in a capacitor is 4C and the energy stored in

it is 4J, find the value of capacitance.
a) 2F
b) 4F
c) 8F
d) 16F

Explanation: The expression for finding the value of energy
is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.

## 6. If the charge in a capacitor is 4C and the energy stored in

it is 4J, calculate the voltage across its plates.
a) 2V
b) 4V
c) 8V
d) 16V

Explanation: The expression for finding the value of energy
is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.
V=Q/C, hence V=4/2=2V.

## 7. Calculate the energy in the 2F capacitor.

3 of 6 3/19/18, 8:31 AM

a) 8.6kJ
b) 64kJ
c) 64J
d) 6.4kJ

Explanation: From the expression:
WD= CV2/2= 2*802/2=6400J=6.4kJ.

## 8. Calculate the energy in the 4F capacitor.

a) 128kJ
b) 1.28kJ

4 of 6 3/19/18, 8:31 AM

c) 12.8kJ
d) 1280J

Explanation: From the expression:
WD= CV2/2= 4*802/2=12800J=12.8kJ.

## 9. Calculate the energy stored in the combination of the

capacitors.

a) 192kJ
b) 1.92kJ
c) 19.2kJ
d) 1920J

Explanation: The equivalent capacitance is: Ceq=4+2=6F.
From the expression:
WD= CV2/2= 6*802/2=19200J=19.2kJ.

## 10. What is the energy in a capacitor if the voltage is 5V

and the charge is10C?

5 of 6 3/19/18, 8:31 AM

a) 6.25J
b) 2.35J
c) 6.54J
d) 4.55J

Explanation: We know that Q/V=C. Hence the value of
capacitance is 2F.
From the expression:
U=Q2/2C, we get U= 6.25J.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:31 AM

sanfoundry.com

Experienced
by Manish
5-6 minutes

## This set of Basic Electrical Engineering Questions and

Answers for Experienced people focuses on “Force of
Attraction Between Oppositely Charged Plates”.

## 1. Which among the following is the correct expression for

force between the plates of a parallel plate capacitor?
a) F=epsilon*(V/x)2/2
b) F=epsilon*(V*x)2/2
c) F=epsilon/(V/x)2/2
d) F=epsilon*(V/x)2/3

Explanation: The force is proportional to the square of the
potential gradient per metre and the area. Hence the force
F=epsilon*(V/x)2/2.

## 2. When the area of cross section of the plate increases,

what happens to the force between the plates?
a) Increases

1 of 6 3/19/18, 8:31 AM

b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is directly proportional to the area of cross
section of the plates, hence as area of cross section
increases, the force of attraction also increases.

## 3. When the potential gradient increases, what happens to

the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is directly proportional to the potential
also increases.

## 4. In which of the following mediums, will the force of

attraction between the plates of a capacitor be greater?
a) Air
b) Water
c) Does not depend on medium
d) Cannot be determined

2 of 6 3/19/18, 8:31 AM

Explanation: The absolute permittivity(epsilon) of water is
greater than that of air. The expression relating F and
epsilon is: F=epsilon*(V/x)2/2. From this expression we can
see that as epsilon increases, the force of attraction also
increases.

## 5. A metal parallel plate capacitor has 100mm diameter and

the distance between the plates is 1mm. The capacitor is
placed in air. Calculate the force on each plate if the
potential difference between the plates is 1kV.
a) 350N
b) 0.035kN
c) 0.035N
d) 3.35kN

Explanation: From the given data:
A=pi*d2/4=0.007854m2
F=epsilon*(V/x)2/2
Therefore, F=0.035N.

## 6. A metal parallel plate capacitor has 100mm diameter and

the distance between the plates is ‘a’ mm. The capacitor is
placed in air. Force on each plate is 0.035N and the
potential difference between the plates is 1kV. Find ‘a’.
a) 1m
b) 1cm
c) 10cm

3 of 6 3/19/18, 8:31 AM

d) 1mm

Explanation: From the given data:
A=pi*d2/4=0.007854m2
F=epsilon*(V/x)2/2
Substituting the given values, we find a=1mm.

## 7. A metal parallel plate capacitor has ‘a’mm diameter and

the distance between the plates is 1mm. The capacitor is
placed in air. Force on each plate is 0.035N and the
potential difference between the plates is 1kV. Find ‘a’.
a) 10mm
b) 100mm
c) 1000m
d) 1000cm

Explanation: From the given data:
A=pi*d2/4=pi*a2/4
F=epsilon*(V/x)2/2
Substituting the given values, we get d=100mm.

## 8. A metal parallel plate capacitor has 100mm diameter and

the distance between the plates is 1mm. The capacitor is
placed in air. Calculate the potential difference between the
plates if the force on each plate is 0.035N.
a) 1kV

4 of 6 3/19/18, 8:31 AM

b) 1V
c) 2kV
d) 2V

Explanation: From the given data:
A=pi*d2/4=0.007854m2
F=epsilon*(V/x)2/2
Substituting the given values in the above expression, we
get V=1kV.

## 9. What happens to the force of attraction between the

capacitors when the potential difference between the plates
decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is directly proportional to the potential
difference between the plates, hence as potential difference
decreases, the force of attraction also decreases.

## 10. What happens to the force of attraction between the

capacitors when the distance of separation between the
plates increases?
a) Increases

5 of 6 3/19/18, 8:31 AM

b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is inversely proportional to the distance
between the plates, hence as distance increases, the force
of attraction decreases.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering for

Experienced people, here is complete set of 1000+ Multiple

6 of 6 3/19/18, 8:31 AM

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## Dielectric Strength - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Dielectric
Strength”.

## 1. The unit for dielectric strength is ____________

a) V/m2
b) MV/m2
c) MV/m
d) Vm

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and length, its unit is MV/m.

## 2. If the Voltage increases, what happens to dielectric

strength?
a) Increases
b) Decreases

1 of 6 3/19/18, 8:31 AM

## c) Remains the same

d) Becomes zero

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and length. Hence as
potential increases, dielectric strength also increases.

## 3. The electric fields of dielectrics having the same cross

sectional area in series is related to their relative
permittivities in which way?
a) Directly proportional
b) Inversely proportional
c) Equal
d) Not related

Explanation: Let us consider two plates having fields E1
and E2 and relative permittivities e1 and e2. Then,
E1=Q/(e0*e1*A) and E2=Q/(e0*e2*A), where e0=absolute
permittivity and A=area of cross section. From the given
expression, we see that E1/E2=e2/e1, hence the electric
field is inversely proportional to the relative permittivities.

## 4. If the potential difference in a material is 4MV and the

thickness of the material is 2m, calculate the dielectric
strength.
a) 2MV/m
b) 4MV/m

2 of 6 3/19/18, 8:31 AM

c) 6MV/m
d) 8MV/m

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness.
Dielectric strength= V/t= 4/2= 2MV/m.

## 5. If the dielectric strength of a material is 4MV/m and its

potential difference is 28MV, calculate the thickness of the
material.
a) 4m
b) 7m
c) 5m
d) 11m

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness.
V/dielectric strength= t= 28/4=7m.

## 6. If the thickness of the material increases, what happens

to the dielectric strength?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 6 3/19/18, 8:31 AM

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness. Hence as
thickness increases, dielectric strength decreases.

## 7. The thickness of a material having dielectric strength

10MV/m is 5m, calculate the potential difference.
a) 2MV
b) 10MV
c) 50MV
d) 100MV

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness.
V=t*dielectric strength= 5*10=50MV.

## 8. What happens to the potential drop between the two

plates of a capacitor when a dielectric is introduced
between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a dielectric is introduced between the
two plates of a parallel plate capacitor, the potential

4 of 6 3/19/18, 8:31 AM

## difference decreases because the potential difference of the

dielectric is subtracted from it.

## 9. A dielectric is basically a ____________

a) Capacitor
b) Conductor
c) Insulator
d) Semiconductor

Explanation: A dielectric is basically an insulator because it
has all the properties of an insulator.

## 10. What happens to the potential difference between the

plates of a capacitor as the thickness of the dielectric slab
increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a dielectric is introduced between the
plates of a capacitor, its potential difference decreases.
New potential difference= potential difference without
dielectric-potential difference of dielectric. Hence as the
thickness of the dielectric slab increases, a larger value is
subtracted fro the original potential difference.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

5 of 6 3/19/18, 8:31 AM

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:31 AM

sanfoundry.com

by Manish
4-6 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Leakage and
Conduction Currents in Capacitors”.

## 1. Leakage in capacitors is primarily caused by _________

a) Transistors
b) Resistors
c) Inductors
d) DC motors

Explanation: Leakage is primarily caused due to electronic
devices, such as transistors, connected to the capacitors.
Transistors conduct a small amount of current even when
they are turned off, hence they are responsible for leakage
current.

## 2. What is the conduction current when a capacitor is fully

charged?
a) Infinity

1 of 5 3/19/18, 8:31 AM

b) Zero
c) 100A
d) 1000A

Explanation: When a capacitor is fully charged, there is no
conduction of electrons from one plate of the capacitor to
another, hence there is no conduction current and
conduction current is equal to zero.

## 3. The flow of electrons in dielectric is due to _________

a) Conduction
b) Potential difference
c) Breakdown
d) Resistance

Explanation: There is, under normal circumstance, no flow
of electrons in a dielectric since a dielectric is basically an
insulator. Hence, there is flow of electrons in a dielectric
only at breakdown voltage.

## 4. The flow of electrons which does not pass through the

battery is known as ________
a) Conduction current
b) Leakage current
c) Either conduction or leakage current
d) Neither conduction nor leakage current

2 of 5 3/19/18, 8:31 AM

## Explanation: Conduction current is the flow of electrons

from the positive plate of the capacitor to the negative plate
of the capacitor, not through the battery. Hence the type of
current which flows without passing through the battery is
conduction current.

## 5. The free electrons in practical dielectrics is due to

_________
a) There are no free electrons
b) Conductors
c) Impurities
d) Both conductors and impurities

Explanation: Ideally, dielectrics are insulators and do not
contain any free electrons. But no dielectric is a perfect
dielectric, hence the free electrons are due to impurities
present in each dielectric.

## 6. The current in conductors connecting the voltage source

to the plates of a capacitor is ______
a) Conduction current
b) Leakage current
c) Charging current
d) Zero

Explanation: The current in conductors connecting the
voltage source to the plates of a capacitor is the charging
current and not the conduction or leakage current.

3 of 5 3/19/18, 8:31 AM

## 7. What is the type of current where the electrons actually

move?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

Explanation: Conduction current is the current caused by
the actual flow of electrons and displacement current is the
current where no charge carriers are involved.

## 8. What is the type of current caused due to variations in

the field?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

Explanation: Displacement current is the current where no
charge carriers are involved. It is caused due to variations
in the electric field.

## 9. Under normal conditions capacitors have _________

a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

4 of 5 3/19/18, 8:31 AM

Explanation: Under normal conditions capacitors contain an
insulating material called dielectric sandwiched between the
plates of the capacitor. Since insulators can carry only an
electric field but not moving carriers, therefore normally a
capacitor has displacement current and not conduction
current.

## 10. If a large amount of voltage is applied to a capacitor,

what is the current that flows through it?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

Explanation: When a large amount of voltage is applied
between the plates of a capacitor, the dielectric between
the plates does not behave as an insulator anymore and
starts conducting and conduction currents flow through it.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and
### Content Structure
The provided text is a series of multiple-choice questions (MCQs) and answers related to basic electrical engineering, specifically focusing on capacitors, electric fields, and related concepts. Each question is followed by multiple options, and the correct answer is provided along with an explanation.

### Analysis
multiple_questions_with_answers

### explanation
1. **multiple questions**: the text contains multiple questions, each with multiple-choice options.
2. **answers provided**: for each question, the correct answer is provided along with an explanation.

### example questions and answers
1. **question**: what is the relation between current and voltage in a capacitor?
   - **options**:
     - a) i=1/c*integral(vdt)
     - b) i=cdv/dt
     - c) i=1/cdv/dt
     - d) i=ct
   - **answer**: b) i=cdv/dt
   - **explanation**: current=rate of change of charge=> i=dq/dt. q=cv, hence i=cdq/dt.

2. **question**: if 2v is supplied to a 3f capacitor, calculate the charge stored
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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## Capacitors - Basic Electrical

Engineering Questions and
by Manish
4 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitors”.

## 1. What is the relation between current and voltage in a

capacitor?
a) I=1/C*integral(Vdt)
b) I=CdV/dt
c) I=1/CdV/dt
d) I=Ct

Explanation: Current=rate of change of charge=> I=dQ/dt.
Q=CV, hence I=CdQ/dt.

## 2. If 2V is supplied to a 3F capacitor, calculate the chance

stored in the capacitor.
a) 1.5C
b) 6C
c) 2C

1 of 5 3/19/18, 8:25 AM

d) 3C

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV.
Q=3*2=6C.

## 3. Calculate the current in the capacitor having 2V supply

voltage and 3F capacitance in 2seconds.
a) 2A
b) 5A
c) 6A
d) 3A

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV.
Q=3*2=6C.
I=Q/t= 6/2=3A.

## 4. A 4microF capacitor is charged to 120V, the charge in

the capacitor would be?
a) 480C
b) 480microC
c) 30C
d) 30microC

2 of 5 3/19/18, 8:25 AM

## Explanation: Q is directly proportional to V. The constant of

proportionality in this case is C, that is, the capacitance.
Hence Q=CV.
Q=4*120=480microC.

## 5. For high frequencies, capacitor acts as _________

a) Open circuit
b) Short circuit
c) Amplifier
d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very high frequencies, the
impedance is almost equal to zero, hence it acts as a short
circuit and there is no voltage across it.

## 6. For very low frequencies, capacitor acts as ________

a) Open circuit
b) Short circuit
c) Amplifier
d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very low frequencies the impedance
is almost infinity and hence acts as an open circuit and no
current flows through it.

## 7. A capacitor consists of_________

a) Two conductors

3 of 5 3/19/18, 8:25 AM

b) Two semiconductors
c) Two dielectrics
d) Two insulators

Explanation: A capacitor consists of two conductors
connected in parallel to each other so that it can store
charge in between the plates.

## 8. Capacitor preferred when there is high frequency in the

circuits is __________
a) Electrolyte capacitor
b) Mica capacitor
c) Air capacitor
d) Glass capacitor

Explanation: Mica capacitors are preferred for high
frequency circuits because they have low ohmic losses and
less reactance.

## 9. Capacitance increases with ________

a) Increase in plate area
b) Decrease in plate area
c) Increase in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is directly proportional to plate
area. Hence as plate area increases, the capacitance also

4 of 5 3/19/18, 8:25 AM

increases.

## 10. Capacitance decreases with __________

a) Increase in distance between the plates
b) Decrease in plate area
c) Decrease in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is inversely proportional to the
distance between the two parallel plates. Hence, as the
distance between the plate decreases, the capacitance
increases.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:25 AM

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## Charge & Voltage - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Charge and
Voltage”.

## 1. Which among the following expressions relate charge,

voltage and capacitance of a capacitor?
a) Q=C/V
b) Q=V/C
c) Q=CV
d) C=Q2V

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV.

## 2. If a 2F capacitor has 1C charge, calculate the voltage

across its terminals.
a) 0.5V

1 of 5 3/19/18, 8:25 AM

b) 2V
c) 1.5V
d) 1V

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence C/Q=V.
V=2/1=1V.

## 3. What is the voltage across a capacitor at the time of

switching, that is, when t=0?
a) Infinity
b) 0V
c) Cannot be determined
d) 1V

Explanation: At the time of switching, when t=0, the
capacitor acts as a short circuit. The voltage across a short
is always equal to zero hence the voltage across the
capacitor is equal to zero.

## 4. What is the voltage across the capacitor if the switch is

closed and steady state is reached?

2 of 5 3/19/18, 8:25 AM

a) 8V
b) 0V
c) 10V
d) Infinity

Explanation: When steady state is reached, the capacitor
acts as a short circuit and the 10V is connected in parallel
to it. Hence Vc=10V.

## 5. If one plate of a parallel plate capacitor is charged to

positive charge the other plate is charged to?
a) Positive
b) Negative
c) Positive of negative
d) Not charged

Explanation: If one plate is charged to positive, the other
plate is automatically charges to negative so that it can
store electrical charge.

## 6. When voltage across a capacitor increases, what

happens to the charge stored in it?
a) Increases
b) Decreases
c) Becomes zero
d) Cannot be determined

3 of 5 3/19/18, 8:25 AM

## Explanation: When voltage across a capacitor increases,

the charge stored in it also increases because charge is
directly proportional to voltage, capacitance being the
constant of proportionality.

## 7. When will capacitor fully charged?

a) When the voltage across its plates is half the voltage
from ground to one of its plates
b) When current through the capacitor is a 1/root2 time its
value
c) When the supply voltage is equal to the capacitor voltage
d) Never

Explanation: When the capacitor voltage is equal to the
supply voltage the current stops flowing through the circuit
and the charging phase is over.

## 8. What happens to the current flow in a fully charged

capacitor?
a) Current flow stops
b) Current flow doubles
c) Current flow becomes half its original value
d) Current flow becomes one-fourth its original value

Explanation: When a capacitor is fully charged, it does not
store any more charge. There is no change in charge with
time. Current is the rate of change of charge, hence it
becomes zero, or stops.

4 of 5 3/19/18, 8:25 AM

## 9. Calculate the capacitance of a capacitor that stores

40microC of charge and has a voltage of 2V.
a) 20F
b) 20microF
c) 10F
d) 10microF

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence C=Q/V.
C=40microC/2V=20microF.

## 10. What happens to the capacitance when the voltage

across the capacitor increases?
a) Decreases
b) Increases
c) Becomes 0
d) No affect

Explanation: The capacitance is inversely proportional to
the voltage across its terminals(C=Q/V). Hence as voltage
increases, capacitance decreases.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:25 AM

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## Capacitance - Basic Electrical

Engineering Questions and
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitance”.

## 1. Power factor of a circuit can be improved by placing

which, among the following, in a circuit?
a) Inductor
b) Capacitor
c) Resistor
d) Switch

Explanation: Raising the power factor on a kW load
decreases its kVA. Hence, by adding a capacitor in a circuit,
without altering the kVA. Hence, the power factor is
improved.

## 2. When the supply frequency increases, what happens to

the capacitive reactance in the circuit?

1 of 5 3/19/18, 8:25 AM

a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The expression for capacitive reactance is:
Xc=1/(2*pi*f*C). This relation shows that frequency is
inversely related to capacitive reactance. Hence, as supply
frequency increases, the capacitive reactance increases.

## 3. Calculate the time constant of a series RC circuit

consisting of a 100microF capacitor in series with a 100ohm
resistor.
a) 0.1 sec
b) 0.1 msec
c) 0.01 sec
d) 0.01 msec

Explanation: The time constant of a RC circuit= R*C=
100*10-6*100=0.01 sec.

## 4. Capacitors charge and discharge in __________

manner.
a) Linear
b) Constant
c) Square
d) Exponential

2 of 5 3/19/18, 8:25 AM

Explanation: Capacitors charge and discharge in an
exponential manner because of the relation:
C=1/(2*pi*f*Xc).

## 5. Air has a dielectric constant of ___________

a) Unity
b) Zero
c) Infinity

Explanation: Dielectris constant of air is the same as that of
vacuum which is equal to unity. Dielctric constant of air is
taken as the reference to measure the dielectric constant of
all other materials.

## 6. What is the value of capacitance of a capacitor which has

a voltage of 4V and ha 8C of charge.
a) 2F
b) 4F
c) 6F
d) 8F

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV. From the relation, C=Q/V= 8/4=2F.

## 7. Unit of capacitance is___________

a) Volts

3 of 5 3/19/18, 8:25 AM

c) Henry
d) Newton

Explanation: Volts is the unit of voltage, Henry for
inductance and Newton for force. Hence the unit for

## 8. What will happen to the capacitor when the source is

removed?
a) It will not remain in its charged state
b) It will remain in its charged state
c) It will start discharging
d) It will become zero

Explanation: As soon as the source is removed, the
capacitor does not start discharging it remains in the same
charged state.

## 9. Which among the following equations is incorrect?

a) Q=CV
b) Q=C/V
c) V=Q/C
d) C=Q/V

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.

4 of 5 3/19/18, 8:25 AM

Hence Q=CV. From the given relation we can derive all the
equations except for Q=C/V.

## 10. Capacitance is directly proportional to__________

a) Area of cross section between the plates
b) Distance of separation between the plates
c) Both area and distance
d) Neither area nor distance

Explanation: The relation between capacitance, area and
distance between the plates is:
C=K*epsilon*A/D. According to this relation, capacitance is
proportional to area.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:25 AM

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## Capacitors in Parallel - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitors in
Parallel”.

## 1. What is the total capacitance when three capacitors, C1,

C2 and C3 are connected in parallel?
a) C1/(C2+C3)
b) C1+C2+C3
c) C2/(C1+C3)
d) 1/C1+1/C2+1/C3

Explanation: When capacitors are connected in parallel, the
total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence Ctotal=C1+C2+C3.

## 2. Calculate the total capacitance.

1 of 7 3/19/18, 8:25 AM

a) 10F
b) 15F
c) 13F
d) 20F

Explanation: The equivalent capacitance when capacitors
are connected in parallel is the sum of all the capacitors=
1+2+10= 13F.

## 3. Calculate the voltage across AB if the total change stored

in the combination is 13C.

2 of 7 3/19/18, 8:25 AM

a) 1V
b) 2V
c) 3V
d) 4V

Explanation: The equivalent capacitance when capacitors
are connected in parallel is the sum of all the capacitors=
1+2+10= 13F. V=Q/C= 13/13=1V.

## 4. Calculate the charge in the 2F capacitor.

a) 200C
b) 100C
c) 300C
d) 400C

3 of 7 3/19/18, 8:25 AM

Explanation: Since the capacitors are connected in parallel,
the voltage across each is the same, it does not get divided.
Q=CV= 2*100=200C.

## 5. Calculate the charge in the 1F capacitor.

a) 200C
b) 100C
c) 300C
d) 400C

Explanation: Since the capacitors are connected in parallel,
the voltage across each is the same, it does not get divided.
Q=CV= 1*100=100C.

## 6. Calculate the total charge of the system.

a) 200C
b) 100C

4 of 7 3/19/18, 8:25 AM

c) 300C
d) 400C

Explanation: The equivalent capacitance when capacitors
are connected in parallel is the sum of all the
capacitors=1+2=3F. Q=CV= 3*100= 300V.

## 7. When capacitors are connected in parallel, the total

capacitance is always __________ the individual
capacitance values.
a) Greater than
b) Less than
c) Equal to
d) Cannot be determined

Explanation: When capacitors are connected in parallel, the
total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence Ctotal=C1+C2+C3. Since it is
the sum of all the capacitance values, total capacitance is
greater the the individual capacitance values.

## 8. When capacitors are connected in parallel, what happens

to the effective plate area?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

5 of 7 3/19/18, 8:25 AM

Explanation: When capacitors are connected in parallel, the
top plates of each of the capacitors are connected together
while the bottom plates are connected to each other. This
effectively increases the top plate area and the bottom plate
area.

## 9. Three capacitors having capacitance equal to 2F, 4F and

6F are connected in parallel. Calculate the effective parallel.
a) 10F
b) 11F
c) 12F
d) 13F

Explanation: When capacitors are connected in parallel, the
total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence Ctotal=C1+C2+C3=
2+4+6=12F.

## 10. Two capacitors having capacitance value 4F, three

capacitors having capacitance value 2F and 5 capacitors
having capacitance value 1F are connected in parallel,
calculate the equivalent capacitance.
a) 20F
b) 19F
c) 18F
d) 17F

6 of 7 3/19/18, 8:25 AM

## Explanation: When capacitors are connected in parallel, the

total capacitance is equal to the sum of the capacitance of
each of the capacitors. Hence
Ctotal=4+4+2+2+2+1+1+1+1+1=19F.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

7 of 7 3/19/18, 8:25 AM

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## Capacitors in Series - Basic

Electrical Engineering Questions
by Manish
4 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitors in
Series”.

## 1. What is the total capacitance when two capacitors C1

and C2 are connected in series?
a) (C1+C2)/C1C2
b) 1/C1+1/C2
c) C1C2/(C1+C2)
d) C1+C2

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2, therefore Ctotal= C1C2/(C1+C2).

## 2. N capacitors having capacitance C are connected in

series, calculate the equivalent capacitance.
a) C/N

1 of 6 3/19/18, 8:26 AM

b) C
c) CN
d) N/C

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal= 1/C+1/C+1/C+……..N times.
1/Ctotal=N/C.
Ctotal=C/N.

## 3. When capacitors are connected in series, the equivalent

capacitance is ___________ each individual capacitance.
a) Greater than
b) Less then
c) Equal to
d) Insufficient data provided

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2. Since we find the reciprocals of the
sum of the reciprocals, the equivalent capacitance is less
than the individual capacitance values.

## 4. What is the equivalent capacitance?

a) 1.5F

2 of 6 3/19/18, 8:26 AM

b) 0.667F
c) 2.45F
d) 2.75F

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2= 1/2+1= 0.667F.

## 5. When capacitors are connected in series ___________

remains the same.
a) Voltage across each capacitor
b) Charge
c) Capacitance
d) Resistance

Explanation: When capacitors are connected in series, the
charge remains the same because the same amount of
current flow exists in each capacitor.

## 6. When capacitors are connected in series

_______________ Varies
a) Voltage across each capacitor
b) Charge
c) Capacitance
d) Resistance

Explanation: When capacitors are connected in series, the

3 of 6 3/19/18, 8:26 AM

## voltage varies because the voltage drop across each

capacitor is different.

## 7. Four 10F capacitors are connected in series, calculate

the equivalent capacitance.
a) 0.2F
b) 0.4F
c) 0.5F
d) 0.6F

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2+1/C3+1/C4=1/10+1/10+1
/10+1/10=0.4F.

## 8. Calculate the charge in the circuit.

a) 66.67C
b) 20.34C
c) 25.45C
d) 30/45C

Explanation: When capacitors are connected in series, the

4 of 6 3/19/18, 8:26 AM

## equivalent capacitance is:

1/Ctotal=1/C1+1/C2= 1/2+1=0.667F.
Q=CV= 1.5*100= 66.67C.

## 9. Calculate the voltage across the 1F capacitor.

a) 33.33V
b) 66.67V
c) 56.56V
d) 23.43V

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2= 1/2+1=0.667F.
Q=CV= 1.5*100= 66.67C.
V across the 1F capacitor= 66.67/1= 66.67V.

## 10. Calculate the voltage across the 2F capacitor.

5 of 6 3/19/18, 8:26 AM

a) 33.33V
b) 66.67V
c) 56.56V
d) 23.43V

Explanation: When capacitors are connected in series, the
equivalent capacitance is:
1/Ctotal=1/C1+1/C2= 1/2+1=0.667F.
Q=CV= 1.5*100= 66.67C.
V across the 1F capacitor= 66.67/2= 33.33V.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:26 AM

sanfoundry.com

Freshers
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Questions and

Answers for Freshers focuses on “Distribution of Voltage
Accross Capacitors in Series”.

## 1. The total voltage drop across a series of capacitors is

__________
a) The voltage drop across any one of the capacitors
b) The sum of the voltage drop across each of the
capacitors
c) The product of the voltage drop across each of the
capacitors
d) Zero

Explanation: The total voltage drop is equal to the sum of
the voltage drop across each off the capacitors because
when capacitors are connected in series, the voltage drops
across each capacitor.

1 of 6 3/19/18, 8:26 AM

## 2. Capacitors C1, C2 and C3 have voltage drops 2V, 3V

and 5V respectively. Calculate the total voltage in the
circuit.
a) 10V
b) 2V
c) 5V
d) 0V

Explanation: When capacitors are connected in series:
Vtotal=V12+V2+V3= 2+3+5=10V.

## 3. What is the voltage across the 2F capacitor?

a) 242V
b) 2V
c) 220V
d) 121V

Explanation: The equivalent capacitance is equal to:
1/C=1/2+1/4+1/6, therefore, C=1.1F.
Q=C*V= 220*1.1= 242C.

2 of 6 3/19/18, 8:26 AM

## 4. What is the voltage across the 4F capacitor?

a) 242V
b) 60.5V
c) 22.5V
d) 12.5V

Explanation: The equivalent capacitance is equal to:
1/C=1/2+1/4+1/6, therefore, C=1.1F.
Q=C*V= 220*1.1= 242C.
V across 4F capacitor = Q/C= 242/4= 60.5V.

## 5. Calculate the voltage across the 6F capacitor.

a) 242V

3 of 6 3/19/18, 8:26 AM

b) 60.5V
c) 40.33V
d) 12.5V

Explanation: The equivalent capacitance is equal to:
1/C=1/2+1/4+1/6, therefore, C=1.1F.
Q=C*V= 220*1.1= 242C.
V across 6F capacitor = Q/C= 242/6= 40.33V.

## 6. When capacitors are connected in series, which of the

following rules are applied?
a) Voltage divider
b) Current divider
c) Both voltage divider and current divider
d) Neither voltage divider nor current divider

Explanation: Voltage divider is the rule applied when
capacitors are connected in series because when
capacitors are connected in series, the voltage is different
across each capacitor.

## 7. A capacitor does not allow sudden changes in

_________
a) Current
b) Voltage
c) Resistance
d) Inductance

4 of 6 3/19/18, 8:26 AM

Explanation: Capacitor does not allow sudden changes in
voltage because these changes occur in zero time which
results in the current being infinity, which is not possible.

## 8. Which of the following expressions is correct with respect

to the voltage across capacitors in series?
a) V1/V2=C2/C1
b) V2/V1=C2/C1
c) V1*V2=C1*C2
d) V1/C1=V2/C2

Explanation: When capacitors are connected in series, the
charge across each capacitor remains the same whereas
the voltage across each varies. When two capacitors are
connected in series:
Q=V1C1; Q=V2C2. Thus: V1/V2=C2/C1.

## 9. Two 4F capacitors are connected in series, calculate the

voltage across each if the total voltage is 20V.
a) 10V
b) 5V
c) 20V
d) 0V

Explanation: The two capacitors have the same
capacitance, hence the voltage gets divides equally. V
across each=Total voltage/2= 20/2= 10V.

5 of 6 3/19/18, 8:26 AM

## 10. Two capacitors having voltage 2F and 4F are

connected in series. This combination is connected to a
100V supply, calculate the voltage across the 2F capacitor.
a) 66.67V b) 33.33V c) 100V d) 0V
Explanation: Using voltage divider rule:
V across 2F= 100*2/(2+4)= 33.33V.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering for

Freshers, here is complete set of 1000+ Multiple Choice

6 of 6 3/19/18, 8:26 AM

sanfoundry.com

by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitance and
the Capacitor”.

## 1. For high frequencies, capacitor acts as _______

a) Open circuit
b) Short circuit
c) Amplifier
d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very high frequencies, the
impedance is almost equal to zero, hence it acts as a short
circuit and there is no voltage across it.

## 2. For very low frequencies, capacitor acts as _______

a) Open circuit
b) Short circuit
c) Amplifier

1 of 5 3/19/18, 8:26 AM

d) Rectifier

Explanation: Capacitive impedance is inversely proportional
to frequency. Hence at very low frequencies the impedance
is almost infinity and hence acts as an open circuit and no
current flows through it.

## 3. A capacitor consists of___________

a) Two conductors
b) Two semiconductors
c) Two dielectrics
d) Two insulators

Explanation: A capacitor consists of two conductors
connected in parallel to each other so that it can store
charge in between the plates.

## 4. Capacitor preferred when there is high frequency in the

circuits is ______
a) Electrolyte capacitor
b) Mica capacitor
c) Air capacitor
d) Glass capacitor

Explanation: Mica capacitors are preferred for high
frequency circuits because they have low ohmic losses and
less reactance.

2 of 5 3/19/18, 8:26 AM

## 5. Capacitance increases with ______

a) Increase in plate area
b) Decrease in plate area
c) Increase in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is directly proportional to plate
area. Hence as plate area increases, the capacitance also
increases.

## 6. Capacitance decreases with _________

a) Increase in distance between the plates
b) Decrease in plate area
c) Decrease in distance between the plates
d) Increase in density of the material

Explanation: Capacitance is inversely proportional to the
distance between the two parallel plates. Hence, as the
distance between the plates decreases, the capacitance
increases.

## 7. When the supply frequency increases, what happens to

the capacitive reactance in the circuit?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 5 3/19/18, 8:26 AM

Explanation: The expression for capacitive reactance is:
Xc=1/(2*pi*f*C). This relation shows that frequency is
inversely related to capacitive reactance. Hence, as supply
frequency increases, the capacitive reactance increases.

## 8. What is the value of capacitance of a capacitor which has

a voltage of 4V and ha 16C of charge.
a) 2F
b) 4F
c) 6F
d) 8F

Explanation: Q is directly proportional to V. The constant of
proportionality in this case is C, that is, the capacitance.
Hence Q=CV. From the relation, C=Q/V= 16/4=4F.

## 9. Unit of capacitance is __________

a) Volts
c) Henry
d) Newton

Explanation: Volts is the unit of voltage, Henry for
inductance and Newton for force. Hence the unit for

## 10. What will happen to the capacitor when the source is

removed?

4 of 5 3/19/18, 8:26 AM

## a) It will not remain in its charged state

b) It will remain in its charged state
c) It will start discharging
d) It will become zero

Explanation: As soon as the source is removed, the
capacitor does not start discharging it remains in the same
charged state.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:26 AM

sanfoundry.com

## Electric Fields - Basic Electrical

Engineering Questions and
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Electric Fields”.

## 1. The conventional direction of electric field is ________

a) Positive to negative
b) Negative to positive
c) No specific direction
d) Direction cannot be determined

Explanation: The conventional direction of field lines is from
positive to negative. The field lines originate at the positive
charge and terminate at the negative charge.

## 2. Electric field originates at __________

a) Positive charge
b) Negative charge
c) Neither positive nor negative
d) Both positive and negative

1 of 5 3/19/18, 8:26 AM

Explanation: Electric field originates at the positive charge
and terminates at the negative charge. The conventional
direction of field is from positive to negative.

## 3. Electric field terminates at ________

a) Positive charge
b) Negative charge
c) Neither positive nor negative
d) Both positive and negative

Explanation: Electric field originates at the positive charge
and terminates at the negative charge. The conventional
direction of field is from positive to negative.

## 4. Which among the following statements is true with regard

to electric field lines?
a) Electric field lines always intersect
b) Electric field lines may or may not intersect
c) Electric field lines can be seen
d) Electric field lines never intersect

Explanation: Electric field lines can never intersect because
the field lines represent the field strength. If the lines
intersect it means that at that point there are two different
values for electric field which is not possible.

2 of 5 3/19/18, 8:26 AM

## 5. Which, among the following, is the field where electric

charge experiences a force?
a) Electric field
b) Magnetic field
c) Gravitational field
d) Electric, magnetic or gravitational field

Explanation: Charges experience a force in an electric field
because charges come under the influence of a field which

## 6. A field that spreads outwards in all directions is

__________
a) Linear
c) Weak
d) Strong

directions. This field is known as the radial field because it

## 7. In uniform fields, all points have ________ field strength.

a) Zero
b) Same
c) Infinity
d) Different

3 of 5 3/19/18, 8:26 AM

Explanation: A uniform field is one-as the word suggests-in
which the field is spread over an area and at every point in
the field the, strength of the field is the same.

## 8. Which, among the following is the correct expression for

electric field?
a) E=F/C
b) E=F*C
c) E=F/Q
d) E=F*Q

Explanation: Electric field is the force per unit charge
hence, the correct expression among the following is:
E=F/Q.

## 9. What happens when one material is rubbed against

another?
a) The material becomes electrically neutral
b) The material becomes electrically charged
c) The material becomes negatively charged
d) The material becomes positively charged

Explanation: When one material is rubbed against another,
there is transfer of charges from one material to another
hence the material becomes electrically charged.

## 10. The insulant between the two plates of a capacitor is

called _______

4 of 5 3/19/18, 8:26 AM

a) Conductor
b) Semi-conductor
c) Dielectric
d) Superconductor

Explanation: The material between the two plates of a
capacitor is an insulator, more specifically known as a
dielectric.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:26 AM

sanfoundry.com

## Electric Field Strength & Electric

Flux Density Questions and
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Electric Field
Strength and Electric Flux Density”.

## 1. Gauss law is applicable for_________

a) Point charge
b) Sheet charge
c) Line charge
d) Point, sheet and line charge

Explanation: Gauss law states that the total electric flux
through any closed surface is equal to the charge enclosed
by that surface.Hence it is applicable for all point, surface
and volume.

## 2. “Total electric flux through any closed surface is equal to

the charge enclosed by that surface”. This is the statement
for?

1 of 5 3/19/18, 8:29 AM

a) Gauss law
b) Lenz law
c) Coloumb’s law

Explanation: Total electric flux through any closed surface is
equal to the charge enclosed by that surface is the
statement for Gauss law because among the four laws,
Gauss law deals with electric flux.

## 3. Electric flux density is a function of_______

a) Volume
b) Charge
c) Current
d) Voltage

Explanation: Electric flux density is the charge per unit area.
Hence it is a function of charge and not any of the other
values.

## 4. As charge increases, what happens to flux density?

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: Electric flux density is the charge per unit area.

2 of 5 3/19/18, 8:29 AM

## The expression for flux density is:

D=Q/A. Electric flux is directly proportional to charge, hence
as charge increases, electric flux also increases.

## 5. As area increases, what happens to electric flux density?

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: Electric flux density is the charge per unit area.
The expression for flux density is:
D=Q/A. Electric flux is inversely proportional to area, hence
as area increases, electric flux decreases.

## 6. Which, among the following, is the correct expression for

electric flux density?
a) D=epsilon*E
b) D=epsilon/E
c) D2=epsilon*E
d) D=epsilon*E2

Explanation: Electric flux density is directly proportional to
the electric field, epsilon(permittivity of free space) being the
constant of proportionality. Hence D=epsilon*E.

## 7. Strength of the electric field is ___________

a) Directly proportional to the force applied
b) Inversely proportional to the force applied

3 of 5 3/19/18, 8:29 AM

## c) Directly proportional to the square of the force applied

d) Inversely proportional to the square of the force applied

Explanation: Electric field intensity is the force per unit
charge, hence it id directly proportional to the force applied.

## 8. The force applied to a conductor is 10N, if the charge in

the conductor is 5C, what is the electric field intensity?
a) 10V/m
b) 2V/m
c) 3V/m
d) 15V/m

Explanation: Electric field intensity is the force per unit
charge. The formula is:
E=F/Q= 10/5= 2V/m.

## 9. What is the electric flux density in free space if the

electric field intensity is 1V/m?
a) 7.76*10-12C/m2
b) 8.85*10-12C /m2
c) 1.23*10-12C /m2
d) 3.43*10-12C /m2

Explanation: The formula for electric filed density is:
D=epsilon*E= 1*8.85*110-12= 8.85*10-12C /m2.

4 of 5 3/19/18, 8:29 AM

## 10. If the charge in a conductor is 16C and the area of

cross section is 4m2. Calculate the electric flux density.
a) 64C/m2
b) 16C/m2
c) 4C/m2
d) 2C/m2

Explanation: Flux density is the charge per unit area. The
formula is:
D=Q/A= 16/4= 4C/m2.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:29 AM

sanfoundry.com

## Relative Permittivity - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Relative
Permittivity”.

## 1. In order to obtain a high value for capacitance the

permittivity of the dielectric medium should be?
a) Low
b) High
c) Zero
d) Unity

Explanation: Form the expression:
C=epsilon*A/d.
From this expression, it is seen that capacitance is directly
proportional to the permittivity, hence for capacitance value
to be high, permittivity value should be high.

## 2. Find the capacitance of a capacitor whose area of cross

1 of 5 3/19/18, 8:29 AM

## section of the plates is 4m2 and distance of separation

between the plates is 2m. The capacitor is placed in
vacuum.
a) 1.77*10-11F
b) 1.34*10-11F
c) 2.33*10-11F
d) 5.65*10-11F

Explanation: The expression for finding the value of
capacitance is:
C=epsilon*A/d.
The medium is free space hence, epsilon= 8.85*10-12.
Therefore, C=8.85*10-12*4/2= 1.77*10-12F.

## 3. What is relative permittivity?

a) Equal to absolute permittivity
b) Ratio of actual permittivity to absolute permittivity
c) Ratio of absolute permittivity to actual permittivity
d) Equal to actual permittivity

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium. As the
actual permittivity increases the relative permittivity also
increases.

## 4. What happens to relative permittivity when actual

permittivity decreases?
a) Increases

2 of 5 3/19/18, 8:29 AM

b) Decreases
c) Remains the same
d) Becomes zero

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium.
Relative permittivity is directly proportional to actual
permittivity. Hence, as actual permittivity increases, relative
permittivity also increases.

## 5. What is the relative permittivity when the actual

permittivity is 4F/m?
a) 4.57*10-11
b) 4.57*1012
c) 4.57*1011
d) 4.57*10-12

Explanation: Relative permittivity= Actual permittivity/
Absolute permittivity.
Relative permittivity= 4/(8.85*10-12)= 4.57*1011.

## 6. What happens to absolute permittivity when relative

permittivity increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 5 3/19/18, 8:29 AM

Explanation: Absolute permittivity does not depend on the
value of relative permittivity. Absolute permittivity is the
permittivity of free space and it is a constant value=
8.85*10-12F/m.

## 7. Calculate the actual permittivity of a medium whose

relative permittivity is 5.
a) 4.43*10-11F/m
b) 4.43*10-12F/m
c) 4.43*1011F/m
d) 4.43*1012F/m

Explanation: Actual permittivity= Relative
permittivity*absolute permittivity.
Actual permittivity= 5*8.85*10-12= 4.43*10-11F/m.

## 8. What is the unit for relative permittivity?

a) F/m
b) Fm
c) F/m2
d) No unit

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium. Since it
is a ratio, and we know that a ratio does not have any unit,
relative permittivity does not have any unit.

## 9. Which, among the following, will be unity in free space?

4 of 5 3/19/18, 8:29 AM

a) Absolute permittivity
b) Relative permittivity
c) Actual permittivity
d) Both absolute and relative permittivity

Explanation: Relative permittivity is constant for a particular
medium. For air or free space ,it is unity. Absolute
permittivity does not depend on the medium, its value is
always constant=8.85*10-12F/m. Actual permittivity is the
product of relative permittivity and absolute permittivity.

## 10. Which, among the following, do not have any unit?

a) Absolute permittivity
b) Relative permittivity
c) Actual permittivity
d) Both absolute and relative permittivity

Explanation: Relative permittivity is the ratio of actual
permittivity to the relative permittivity of the medium. Since it
is a ratio, and we know that a ratio does not have any unit,
relative permittivity does not have any unit.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

5 of 5 3/19/18, 8:29 AM

sanfoundry.com

by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Capacitance of a
Multi Plate Capacitor”.

## 1. Which is the correct expression for capacitance of a multi

plate capacitor?
a) C=absolute permittivity*A/d
b) C=Actual permittivity*(n-1)*A/d
c) C=Actual permittivity*(n)*A/d
d) Actual permittivity*(n+1)*A/d

Explanation: The correct expression is: C=Actual
permittivity*(n-1)*A/d.
Where, n=number of plates, A=area of cross section of the
plates, d=distance of separation between the plates.

## 2. What happens to the capacitance of a multi plate

capacitor when the area of cross section of the plate
decreases?

1 of 6 3/19/18, 8:29 AM

a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When the area of cross section decreases, the
capacitance also decreases since it is related by the
formula C=Actual permittivity*(n-1)*A/d. Here, we can see
that the capacitance is directly proportional to the area of
cross section.

## 3. What happens to the capacitance of a multi plate

capacitor when the distance of separation between the
plate increases?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When the distance of separation between the
plates decreases, the capacitance also decreases since it is
related by the formula C=Actual permittivity*(n-1)*A/d. Here,
we can see that the capacitance is inversely proportional to
the distance of separation.

## 4. What happens to the capacitance of a multi plate

capacitor when the number of plates increases?
a) Increases

2 of 6 3/19/18, 8:29 AM

b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When the number of capacitors increases, the
capacitance also increases since it is related by the formula
C=Actual permittivity*(n-1)*A/d. Here, we can see that the
capacitance is directly proportional to the number of
capacitors.

## 5. Find the capacitance of a multi plate capacitor whose

actual permittivity= 5F/m, n=3, A=4m2and d=2m.
a) 10F
b) 20F
c) 30F
d) 40F

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.
Thus, C=5*(3-1)*4/2= 20F.

## 6. Find the capacitance of a multi plate capacitor whose

relative permittivity=5, n=3, A=4m2 and d=2m.
a) 1.77*10-10 F
b) 1.77*1010 F
c) 1.77*10-11 F
d) 1.77*1011 F

3 of 6 3/19/18, 8:29 AM

Explanation: The formula for capacitance of a multi plate
capacitor: C=Relative permittivity*absolute permittivity*
(n-1)*A/d.
C= 5*8.85*10-12*(3-1)*4/2=1.77*10-10.

## 7. Find the number of plates in the multi plate capacitor

having C=20F absolute permittivity=5F/m, A=4m2 and
d=2m.
a) 1
b) 2
c) 3
d) 4

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.
Substituting the given values in the equation, we get n=3.

## 8. Calculate the distance between the plates of the

capacitor having C=20F, actual permittivity=F/m
n=3 and A=4m2.
a) 1m
b) 2m
c) 3m
d) 4m

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.

4 of 6 3/19/18, 8:29 AM

## 9. Calculate the area of cross section of the multi plate

capacitor having C=20F, actual permittivity=F/m n=3 and
d=2m.
a) 1m
b) 2m
c) 3m
d) 4m

Explanation: The formula for capacitance of a multi plate
capacitor: C=Actual permittivity*(n-1)*A/d.
Substituting the given values in the equation, we get
A=4m2.

## 10. Calculate the number of plates in the multi plate

capacitor having C=1.77*10-10F relative permittivity=5,
A=4m2 and d=2m.
a) 1m
b) 2m
c) 3m
d) 4m

Explanation: The formula for capacitance of a multi plate
capacitor: C=Relative permittivity*absolute permittivity*
(n-1)*A/d.
Substituting the given values in the equation, we get n=3.

## Sanfoundry Global Education & Learning Series –

5 of 6 3/19/18, 8:29 AM

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:29 AM

sanfoundry.com

by Manish
5-6 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Composite
Dielectric Capacitor”.

## 1. Potential drop in a dielectric is equal to _______

a) Electric field strength*thickness
b) Electric field strength*area of cross section
c) Electric field strength
d) Zero

Explanation: When a dielectric is introduced between the
two plates of a parallel plate capacitor, the potential
difference decreases by the value of the product of electric
field strength*thickness which is the potential difference of
the dielectric.

## 2. The electric field strength is 10N/C and the thickness of

the dielectric is 3m. Calculate the potential drop in the
dielectric.

1 of 6 3/19/18, 8:30 AM

a) 10V
b) 20V
c) 30V
d) 40V

Explanation: The potential drop in a dielectric= electric field
strength*area of cross section= 10*3= 30V.

## 3. The electric fields of dielectrics having the same cross

sectional area in series is related to their relative
permittivities in which way?
a) Directly proportional
b) Inversely proportional
c) Equal
d) Not related

Explanation: Let us consider two plates having fields E1
and E2 and relative permittivities e1 and e2. Then,
E1=Q/(e0*e1*A) and E2=Q/(e0*e2*A), where e0=absolute
permittivity and A=area of cross section. From the given
expression, we see that E1/E2=e2/e1, hence the electric
field is inversely proportional to the relative permittivities.

## 4. What happens to the capacitance when a dielectric is

introduced between its plates?
a) Increases
b) Decreases
c) Remains the same

2 of 6 3/19/18, 8:30 AM

d) Becomes zero

Explanation: The capacitance of a capacitance increases
when a dielectric is introduced between its plates because
the capacitance is related to the dielectric constant k by the
equation:
C=ke0A/d.

## 5. Calculate the relative permittivity of the second dielectric

if the relative permittivity of the first is 4. The electric field
strength of the first dielectric is 8V/m and that of the second
is 2V/m.
a) 32
b) 4
c) 16
d) 8

Explanation: The relation between the two electric fields
and the relative permittivities is:
E1/E1=e2/e1. Substituting the given values, we get e2=16.

## 6. What happens to the potential drop between the two

plates of a capacitor when a dielectric is introduced
between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

3 of 6 3/19/18, 8:30 AM

Explanation: When a dielectric is introduced between the
two plates of a parallel plate capacitor, the potential
difference decreases because the potential difference of the
dielectric is subtracted from it.

## 7. If the potential difference across the plates of a capacitor

is 10V and a dielectric having thickness 2m is introduced
between the plates, calculate the potential difference after
introducing the dielectric. The electric field strength is 2V/m.
a) 4V
b) 6V
c) 8V
d) 10V

Explanation: When a dielectric is introduced between the
plates of a capacitor, its potential difference decreases.
New potential difference= potential difference without
dielectric-potential difference of dielectric= 10-2*2= 6V.

## 8. Calculate the capacitance if the dielectric constant=4,

area of cross section= 10m2 and the distance of separation
between the plates is 5m.
a) 7.08*10-11F
b) 7.08*1011F
c) 7.08*10-12F
d) 7.08*10-10F

4 of 6 3/19/18, 8:30 AM

Explanation: The expression to find capacitance when a
dielectric is introduced between the plates is:
C=ke0A/d. Substituting the given values in the equation, we
get C= 7.08*10-11F.

## 9. A dielectric is basically a ________

a) Capacitor
b) Conductor
c) Insulator
d) Semiconductor

Explanation: A dielectric is basically an insulator because it
has all the properties of an insulator.

## 10. What happens to the potential difference between the

plates of a capacitor as the thickness of the dielectric slab
increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a dielectric is introduced between the
plates of a capacitor, its potential difference decreases.
New potential difference= potential difference without
dielectric-potential difference of dielectric. Hence as the
thickness of the dielectric slab increases, a larger value is

5 of 6 3/19/18, 8:30 AM

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:30 AM

sanfoundry.com

by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Charging and
Discharging Currents”.

## 1. Which of the following charging and discharging of a

capacitor?
a) Time constant
b) Current
c) Power
d) Voltage

Explanation: The time constant in a circuit consisting of a
capacitor is the product of the resistance and the
capacitance. Smaller the time constant, faster is the
charging and discharging rate and vice versa.

## 2. What is the initial current while charging a capacitor?

a) High
b) Low

1 of 5 3/19/18, 8:30 AM

c) 0
d) Cannot be determined

Explanation: The initial current of a capacitor is very high
because the voltage source will transport charges from one
plate of the capacitor to the other plate.

## 3. What is the final current while charging a capacitor?

a) High
b) Zero
c) Infinity
d) Low

Explanation: The final current is almost equal to zero while
charging a capacitor because the capacitor is charged up to
the source voltage.

## 4. What happens to the current flow in a fully charged

capacitor?
a) Current flow stops
b) Current flow doubles
c) Current flow becomes half its original value
d) Current flow becomes one-fourth its original value

Explanation: When a capacitor is fully charged, it does not
store any more charge. There is no change in charge with
time. Current is the rate of change of charge, hence it

2 of 5 3/19/18, 8:30 AM

## 5. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the initial value of discharge
current.
a) 10A
b) 0A
c) Infinity
d) 20A

Explanation: When the capacitor is discharging the value of
initial current is a finite one. The finite initial current value is
found using ohm’s law: I=V/R= 400/20= 20A.

## 6. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the final value of discharge
current.
a) 10A
b) 0A
c) Infinity
d) 20A

Explanation: In a discharging circuit, the final voltage is
equal to zero as all the positive and negative charges have
combined. Since the voltage is equal to zero, the current is
also equal to zero as voltage is directly proportional to
current by ohm’s law.

## 7. When will be capacitors fully charged?

3 of 5 3/19/18, 8:30 AM

## a) When voltage is zero

b) When the supply voltage is equal to the capacitor voltage
c) When voltage is infinity
d) When capacitor voltage is equal to half the supply
voltage

Explanation: When the capacitor voltage is equal to the
source voltage, it means that all the charges have moved
from one plate of the capacitor to the other.

## 8. What happens to the capacitor when the capacitor

voltage is equal to the source voltage?
a) The charging phase of the capacitor is over
b) The discharging phase of the capacitor is over
c) The capacitor is switched off
d) The capacitor is switched on

Explanation: When the capacitor voltage is equal to the
source voltage, it means that all the charges have moved
from one plate of the capacitor to the other. Hence the
capacitor is charged and the charging phase is over.

## 9. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the final value of charging
current.
a) 10A
b) 0A
c) Infinity

4 of 5 3/19/18, 8:30 AM

d) 20A

Explanation: The final value of charging current in a
capacitor is equal to the initial value of discharging current
in it. Hence the final value of charging current is: Vc/R=20A.

## 10. A capacitor is charged to a voltage of 400V and has a

resistance of 20ohm. Calculate the initial value of charging
current.
a) 10A
b) 0A
c) Infinity
d) 20A

Explanation: Initially, there’s 0V voltage in a capacitor. As
the capacitor charges, the voltage increases. Since voltage
is proportional to current by ohm’s law, initial current is also
equal to zero.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

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5 of 5 3/19/18, 8:30 AM

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## Growth & Decay - Basic Electrical

Engineering Questions and
by Manish
5-6 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Growth and
Decay”.

## 1. The charging time constant of a circuit consisting of a

capacitor is the time taken for the charge in the capacitor to
become __________% of the initial charge.
a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(1-e-t/RC).
When RC=t, we have: Q=Q0(1-e-1)= 0.63*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.63 times its initial charge.

## 2. The discharging time constant of a circuit consisting of a

1 of 6 3/19/18, 8:30 AM

## capacitor is the time taken for the charge in the capacitor to

become __________% of the initial charge.
a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(e-t /RC).
When RC=t, we have: Q=Q0(e-1)= 0.37*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.37 times its initial charge.

## 3. A circuit has a resistance of 2 ohm connected in series

with a capacitance of 6F. Calculate the charging time
constant.
a) 3
b) 1
c) 12
d) 8

Explanation: The charging time constant in a circuit
consisting of a capacitor and resistor in series is the product
of the resistance and capacitance= 2*6=12.

## 4. A circuit has a resistance of 5 ohm connected in series

with a capacitance of 10F. Calculate the discharging time
constant.
a) 15

2 of 6 3/19/18, 8:30 AM

b) 50
c) 5
d) 10

Explanation: The discharging time constant in a circuit
consisting of a capacitor and resistor in se-ries is the
product of the resistance and capacitance= 5*10=50.

## 5. What is the value of current in a discharging capacitive

circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A

Explanation: At time t=RC, that is the time constant, we
know that the value of current at that time interval is equal
to 37% of the initial charge in the discharging circuit. Hence,
I=2*0.37= 0.74A.

## 6. What is the value of current in a charging capacitive

circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A

3 of 6 3/19/18, 8:30 AM

## Explanation: At time t=RC, that is the time constant, we

know that the value of current at that time interval is equal
to 63% of the initial charge in the charging circuit. Hence,
I=2*0.63= 1.26A.

## 7. While discharging, what happens to the current in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of current in a
discharging capacitive circuit is:
I=I0*e-t /RC. From this equation, we can see that the
current is exponentially decreasing since e is raised to a
negative power.

## 8. While discharging, what happens to the voltage in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of voltage in a
discharging capacitive circuit is:
V=V0*e-t /RC. From this equation, we can see that the

4 of 6 3/19/18, 8:30 AM

negative power.

## 9. While charging, what happens to the current in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of current in a
charging capacitive circuit is:
I=I0*(1-e-t /RC). From this equation, we can see that the
current is exponentially increasing since e is raised to a
negative power and we are subtracting it from 1. Hence as
the value of e-t /RC in-creases, the current increases
exponentially.

## 10. While charging, what happens to the voltage in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of voltage in a
charging capacitive circuit is:
V=V0*(1-e-t /RC). From this equation, we can see that the

5 of 6 3/19/18, 8:30 AM

## voltage is exponentially increasing since e is raised to a

negative power and we are subtracting it from 1. Hence as
the value of e-t /RC in-creases, the voltage increases
exponentially.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

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for Freshers
by Manish
4-6 minutes

## This set of Basic Electrical Engineering Interview Questions

and Answers for freshers focuses on “Discharge of a
Capacitor Through a Resistor”.

## 1. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the time constant.
a) 1s
b) 2s
c) 3s
d) 4s

Explanation: The time constant is the product of the
resistance and capacitance in a series RC circuit.
Therefore, time constant= 8*10-6*4*106=4s.

## 2. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.

1 of 6 3/19/18, 8:30 AM

## Calculate the initial charging current.

a) 100 microA
b) 500 microA
c) 400 microA
d) 1000microA

Explanation: In a series RC circuit, the initial charging
current is:
I=V/R= 200/(0.5*106s)= 400*10-6A= 400 microA.

## 3. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the time taken for the potential difference across
the capacitor to gow to 160V.
a) 6.44s
b) 7.77s
c) 2.33s
d) 3.22s

Explanation: From the previous explanations, we know that
the initial current is 400mA and the time constant is 4s.
Substituting the values of capacitor voltage, initial voltage,
initial current and time constant in the equation: Vc=V(1-
e-t/RC), we get t=6.44s.

## 4. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the voltage in the capacitor 4s after the power is

2 of 6 3/19/18, 8:30 AM

supplied.
a) 123.4V
b) 126.4V
c) 124.5V
d) 132.5V

Explanation: We can get the value of the potential
difference across the capacitor in 4s, from the following
equation:
Vc=V(1-e-t /RC). Substituting the values in the given
equation, we get Vc= 126.4V.

## 5. An 8microF capacitor is connected in series with a 0.5

megaohm resistor. The DC voltage supply is 200V.
Calculate the current in the capacitor 4s after the power is
supplied.
a) 150 microA
b) 149 microA
c) 148 microA
d) 147 microA

Explanation: In the given question, the time constant is
equal to the time taken= 4s. Hence the value of current will
be 63% of its initial value= I=0.63*400= 147 microA.

## 6. The discharging time constant of a circuit consisting of a

capacitor is the time taken for the charge in the capacitor to
become __________% of the initial charge.

3 of 6 3/19/18, 8:30 AM

a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(e-t /RC).
When RC=t, we have: Q=Q0(e-1)= 0.37*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.37 times its initial charge.

## 7. The discharging time constant of a circuit consisting of a

capacitor is the time taken for the charge in the capacitor to
become __________% of the initial charge.
a) 33
b) 63
c) 37
d) 36

Explanation: We know that: Q=Q0(e-t /RC).
When RC=t, we have: Q=Q0(e-1)= 0.37*Q0.
Hence the time constant is the time taken for the charge in
a capacitive circuit to become 0.37 times its initial charge.

## 8. A circuit has a resistance of 2 ohm connected in series

with a capacitance of 6F. Calculate the discharging time
constant.
a) 3
b) 1

4 of 6 3/19/18, 8:30 AM

c) 12
d) 8

Explanation: The discharging time constant in a circuit
consisting of a capacitor and resistor in series is the product
of the resistance and capacitance= 2*6=12.

## 9. What is the value of current in a discharging capacitive

circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A

Explanation: At time t=RC, that is the time constant, we
know that the value of current at that time interval is equal
to 37% of the initial charge in the discharging circuit. Hence,
I=2*0.37= 0.74A.

## 10. While discharging, what happens to the current in the

capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially

Explanation: The equation for the value of current in a

5 of 6 3/19/18, 8:30 AM

## discharging capacitive circuit is:

I=I0*e-t /RC. From this equation, we can see that the
current is exponentially decreasing since e is raised to a
negative power.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering for

Interviews, here is complete set of 1000+ Multiple Choice

6 of 6 3/19/18, 8:30 AM

sanfoundry.com

## CR Networks Transients - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Transients in CR
Networks”.

## 1. A CR network is one which consists of _________

a) A capacitor and resistor connected in parallel
b) A capacitor and resistor connected in series
c) A network consisting of a capacitor only
d) A network consisting of a resistor only

Explanation: A CR network is one which consists of a
capacitor connected in series with a resistor. The capacitor
discharges or charges through the resistor.

## 2. At DC, capacitor acts as _________

a) Open circuit
b) Short circuit
c) Resistor

1 of 7 3/19/18, 8:31 AM

d) Inductor

Explanation: At DC, the capacitor acts as open circuit
because the capacitive resistance is infinity. The frequency
of a DC circuit is 0. The capacitive resistance=1/(2*pi*f*C).
Therefore, if the frequency is 0, the capacitive resistance is
infinity and it acts as an open circuit.

## 3. In an RC series circuit, when the switch is closed and the

circuit is complete, what is the response?
a) Response does not vary with time
b) Decays with time
c) Increases with time
d) First increases, then decreases

Explanation: In an RC series circuit, the response decays
with time because according to the equation, there is an
exponential decrease in the response.

## 4. If the switch is clocked at t=0, what is the current in the

circuit?

a) 0A

2 of 7 3/19/18, 8:31 AM

b) 10A
c) 20A
d) Infinity

Explanation: As soon as the switch is closed at t=0, the
capacitor acts as a short circuit. The current in the circuit is:
I=V/R= 100/10= 10A.

## 5. Calculate the voltage across the capacitor at t=0.

a) 0A
b) 10A
c) 20A
d) Infinity

Explanation: When the switch is closed at t=0, the capacitor
has no voltage across it since it has not been charged. The
capacitor acts as a short circuit and the voltage across it is
zero.

## 6. Calculate di(0)/dt if the switch is closed at t=0.

3 of 7 3/19/18, 8:31 AM

a) -9.9A/s
b) -10A/s
c) 0A/s
d) -0.1A/s

Explanation: Applying KVL to the given circuit, we get:
100+10i(0)+1/10*integral(i(0)dt)=0
Differentiating once, we get:
10di(0)/dt+1/10*i.
From the previous questions, we know that i(0)=10A.
Substituting the values in the equation, we get di(0)/dt=-
0.1A/s.

## 7. Calculate d2i(0)/dt2 from the given circuit.

a) 10-6A/s2
b) 10-3A/s2
c) 106A/s2

4 of 7 3/19/18, 8:31 AM

d) 103A/s2

Explanation: Applying KVL to the given circuit, we get:
100+10i(0)+1/10*integral(i(0)dt)=0
Differentiating once, we get:
10di(0)/dt+1/10*i.
Differentiating once again, we get:
10d2i(0)/dt2+10di(0)/dt=0.
Substituting the values of di/dt from the previous
explanation, we get d2i(0)/dt2=10-3A/s2.

## 8. The current equation for the given circuit is?

a) i=10e(-0.01)t A
b) i=10e(0.01)t A
c) i=10e(-0.001)t A
d) i=100e(-0.01)t A

Explanation: The KVL equation is:
100+10i(0)+1/10*integral(i(0)dt)=0
On applying Laplace transform to this equation, we get:
100/s=I(s)/10s+10I(s)

5 of 7 3/19/18, 8:31 AM

i=10e(-0.01)t A.

## 9. The expression for current in an RC circuit is?

a) i=(V/R)e(t/RC )
b) i=(V/R)e(-t/RC )
c) i=(V/R)/e(t/RC )
d) i=(V/R)/e(-t/RC )

Explanation: The particular solution of the current equation
is zero. Hence, the expression for the current in an RC
circuit is:
i=(V/R)e(-t/RC ).

## 10. What is the voltage in the resistor as soon as the switch

is closed at t=0.

a) 0V
b) Infinity
c) 220V
d) Insufficient information provided

Explanation: As soon as the switch is closes at t=0, there is

6 of 7 3/19/18, 8:31 AM

## no charge in the capacitor, hence the voltage across the

capacitor os zero and all the 220V voltage is the voltage
across the resistor.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

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7 of 7 3/19/18, 8:31 AM

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by Manish
3-4 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Energy Stored in
a Charged Capacitor”.

## 1. Work done in charging a capacitor is ____________

a) QV
b) 1/2QV
c) 2QV
d) QV2

Explanation: We know that work done= Q2/2C.
Substituting C as Q/V, we get work done= Q/2V.

## 2. Energy stored in 2000mF capacitor charged to a

potential difference of 10V is?
a) 0.1J
b) 0.2J
c) 0.3J
d) 0.4J

1 of 6 3/19/18, 8:31 AM

Explanation: From the expression:
WD= CV2/2= 0.1J.

## 3. When do we get maximum energy from a set of

capacitors?
a) When they are connected in parallel
b) When they are connected in series
d) Both in series and parallel
d) Insufficient information provided

Explanation: We get maximum energy when capacitors are
connected in parallel because the equivalent capacitance is
larger than the largest individual capacitance when
connected in parallel. The relation between capacitance
and energy is:
Energy=CV2/2, hence as the capacitance increases, the
energy stored in it also increases.

## 4. If the charge stored in a capacitor is 4C and the value of

capacitance is 2F, calculate the energy stored in it.
a) 2J
b) 4J
c) 8J
d) 16J

Explanation: The expression for finding the value of energy

2 of 6 3/19/18, 8:31 AM

is:
U=Q2/2C= 4*4/(2*2)= 4J.

## 5. If the charge in a capacitor is 4C and the energy stored in

it is 4J, find the value of capacitance.
a) 2F
b) 4F
c) 8F
d) 16F

Explanation: The expression for finding the value of energy
is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.

## 6. If the charge in a capacitor is 4C and the energy stored in

it is 4J, calculate the voltage across its plates.
a) 2V
b) 4V
c) 8V
d) 16V

Explanation: The expression for finding the value of energy
is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.
V=Q/C, hence V=4/2=2V.

## 7. Calculate the energy in the 2F capacitor.

3 of 6 3/19/18, 8:31 AM

a) 8.6kJ
b) 64kJ
c) 64J
d) 6.4kJ

Explanation: From the expression:
WD= CV2/2= 2*802/2=6400J=6.4kJ.

## 8. Calculate the energy in the 4F capacitor.

a) 128kJ
b) 1.28kJ

4 of 6 3/19/18, 8:31 AM

c) 12.8kJ
d) 1280J

Explanation: From the expression:
WD= CV2/2= 4*802/2=12800J=12.8kJ.

## 9. Calculate the energy stored in the combination of the

capacitors.

a) 192kJ
b) 1.92kJ
c) 19.2kJ
d) 1920J

Explanation: The equivalent capacitance is: Ceq=4+2=6F.
From the expression:
WD= CV2/2= 6*802/2=19200J=19.2kJ.

## 10. What is the energy in a capacitor if the voltage is 5V

and the charge is10C?

5 of 6 3/19/18, 8:31 AM

a) 6.25J
b) 2.35J
c) 6.54J
d) 4.55J

Explanation: We know that Q/V=C. Hence the value of
capacitance is 2F.
From the expression:
U=Q2/2C, we get U= 6.25J.

## Sanfoundry Global Education & Learning Series –

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Experienced
by Manish
5-6 minutes

## This set of Basic Electrical Engineering Questions and

Answers for Experienced people focuses on “Force of
Attraction Between Oppositely Charged Plates”.

## 1. Which among the following is the correct expression for

force between the plates of a parallel plate capacitor?
a) F=epsilon*(V/x)2/2
b) F=epsilon*(V*x)2/2
c) F=epsilon/(V/x)2/2
d) F=epsilon*(V/x)2/3

Explanation: The force is proportional to the square of the
potential gradient per metre and the area. Hence the force
F=epsilon*(V/x)2/2.

## 2. When the area of cross section of the plate increases,

what happens to the force between the plates?
a) Increases

1 of 6 3/19/18, 8:31 AM

b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is directly proportional to the area of cross
section of the plates, hence as area of cross section
increases, the force of attraction also increases.

## 3. When the potential gradient increases, what happens to

the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is directly proportional to the potential
also increases.

## 4. In which of the following mediums, will the force of

attraction between the plates of a capacitor be greater?
a) Air
b) Water
c) Does not depend on medium
d) Cannot be determined

2 of 6 3/19/18, 8:31 AM

Explanation: The absolute permittivity(epsilon) of water is
greater than that of air. The expression relating F and
epsilon is: F=epsilon*(V/x)2/2. From this expression we can
see that as epsilon increases, the force of attraction also
increases.

## 5. A metal parallel plate capacitor has 100mm diameter and

the distance between the plates is 1mm. The capacitor is
placed in air. Calculate the force on each plate if the
potential difference between the plates is 1kV.
a) 350N
b) 0.035kN
c) 0.035N
d) 3.35kN

Explanation: From the given data:
A=pi*d2/4=0.007854m2
F=epsilon*(V/x)2/2
Therefore, F=0.035N.

## 6. A metal parallel plate capacitor has 100mm diameter and

the distance between the plates is ‘a’ mm. The capacitor is
placed in air. Force on each plate is 0.035N and the
potential difference between the plates is 1kV. Find ‘a’.
a) 1m
b) 1cm
c) 10cm

3 of 6 3/19/18, 8:31 AM

d) 1mm

Explanation: From the given data:
A=pi*d2/4=0.007854m2
F=epsilon*(V/x)2/2
Substituting the given values, we find a=1mm.

## 7. A metal parallel plate capacitor has ‘a’mm diameter and

the distance between the plates is 1mm. The capacitor is
placed in air. Force on each plate is 0.035N and the
potential difference between the plates is 1kV. Find ‘a’.
a) 10mm
b) 100mm
c) 1000m
d) 1000cm

Explanation: From the given data:
A=pi*d2/4=pi*a2/4
F=epsilon*(V/x)2/2
Substituting the given values, we get d=100mm.

## 8. A metal parallel plate capacitor has 100mm diameter and

the distance between the plates is 1mm. The capacitor is
placed in air. Calculate the potential difference between the
plates if the force on each plate is 0.035N.
a) 1kV

4 of 6 3/19/18, 8:31 AM

b) 1V
c) 2kV
d) 2V

Explanation: From the given data:
A=pi*d2/4=0.007854m2
F=epsilon*(V/x)2/2
Substituting the given values in the above expression, we
get V=1kV.

## 9. What happens to the force of attraction between the

capacitors when the potential difference between the plates
decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is directly proportional to the potential
difference between the plates, hence as potential difference
decreases, the force of attraction also decreases.

## 10. What happens to the force of attraction between the

capacitors when the distance of separation between the
plates increases?
a) Increases

5 of 6 3/19/18, 8:31 AM

b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The force of attraction between the two plates
of the capacitor is inversely proportional to the distance
between the plates, hence as distance increases, the force
of attraction decreases.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering for

Experienced people, here is complete set of 1000+ Multiple

6 of 6 3/19/18, 8:31 AM

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## Dielectric Strength - Basic

Electrical Engineering Questions
by Manish
4-5 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Dielectric
Strength”.

## 1. The unit for dielectric strength is ____________

a) V/m2
b) MV/m2
c) MV/m
d) Vm

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and length, its unit is MV/m.

## 2. If the Voltage increases, what happens to dielectric

strength?
a) Increases
b) Decreases

1 of 6 3/19/18, 8:31 AM

## c) Remains the same

d) Becomes zero

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and length. Hence as
potential increases, dielectric strength also increases.

## 3. The electric fields of dielectrics having the same cross

sectional area in series is related to their relative
permittivities in which way?
a) Directly proportional
b) Inversely proportional
c) Equal
d) Not related

Explanation: Let us consider two plates having fields E1
and E2 and relative permittivities e1 and e2. Then,
E1=Q/(e0*e1*A) and E2=Q/(e0*e2*A), where e0=absolute
permittivity and A=area of cross section. From the given
expression, we see that E1/E2=e2/e1, hence the electric
field is inversely proportional to the relative permittivities.

## 4. If the potential difference in a material is 4MV and the

thickness of the material is 2m, calculate the dielectric
strength.
a) 2MV/m
b) 4MV/m

2 of 6 3/19/18, 8:31 AM

c) 6MV/m
d) 8MV/m

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness.
Dielectric strength= V/t= 4/2= 2MV/m.

## 5. If the dielectric strength of a material is 4MV/m and its

potential difference is 28MV, calculate the thickness of the
material.
a) 4m
b) 7m
c) 5m
d) 11m

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness.
V/dielectric strength= t= 28/4=7m.

## 6. If the thickness of the material increases, what happens

to the dielectric strength?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

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Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness. Hence as
thickness increases, dielectric strength decreases.

## 7. The thickness of a material having dielectric strength

10MV/m is 5m, calculate the potential difference.
a) 2MV
b) 10MV
c) 50MV
d) 100MV

Explanation: Dielectric strength is the potential gradient
required to cause a breakdown in the material. Potential
gradient is the ratio of voltage and thickness.
V=t*dielectric strength= 5*10=50MV.

## 8. What happens to the potential drop between the two

plates of a capacitor when a dielectric is introduced
between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a dielectric is introduced between the
two plates of a parallel plate capacitor, the potential

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## difference decreases because the potential difference of the

dielectric is subtracted from it.

## 9. A dielectric is basically a ____________

a) Capacitor
b) Conductor
c) Insulator
d) Semiconductor

Explanation: A dielectric is basically an insulator because it
has all the properties of an insulator.

## 10. What happens to the potential difference between the

plates of a capacitor as the thickness of the dielectric slab
increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a dielectric is introduced between the
plates of a capacitor, its potential difference decreases.
New potential difference= potential difference without
dielectric-potential difference of dielectric. Hence as the
thickness of the dielectric slab increases, a larger value is
subtracted fro the original potential difference.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

5 of 6 3/19/18, 8:31 AM

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and

6 of 6 3/19/18, 8:31 AM

sanfoundry.com

by Manish
4-6 minutes

## This set of Basic Electrical Engineering Multiple Choice

Questions & Answers (MCQs) focuses on “Leakage and
Conduction Currents in Capacitors”.

## 1. Leakage in capacitors is primarily caused by _________

a) Transistors
b) Resistors
c) Inductors
d) DC motors

Explanation: Leakage is primarily caused due to electronic
devices, such as transistors, connected to the capacitors.
Transistors conduct a small amount of current even when
they are turned off, hence they are responsible for leakage
current.

## 2. What is the conduction current when a capacitor is fully

charged?
a) Infinity

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b) Zero
c) 100A
d) 1000A

Explanation: When a capacitor is fully charged, there is no
conduction of electrons from one plate of the capacitor to
another, hence there is no conduction current and
conduction current is equal to zero.

## 3. The flow of electrons in dielectric is due to _________

a) Conduction
b) Potential difference
c) Breakdown
d) Resistance

Explanation: There is, under normal circumstance, no flow
of electrons in a dielectric since a dielectric is basically an
insulator. Hence, there is flow of electrons in a dielectric
only at breakdown voltage.

## 4. The flow of electrons which does not pass through the

battery is known as ________
a) Conduction current
b) Leakage current
c) Either conduction or leakage current
d) Neither conduction nor leakage current

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## Explanation: Conduction current is the flow of electrons

from the positive plate of the capacitor to the negative plate
of the capacitor, not through the battery. Hence the type of
current which flows without passing through the battery is
conduction current.

## 5. The free electrons in practical dielectrics is due to

_________
a) There are no free electrons
b) Conductors
c) Impurities
d) Both conductors and impurities

Explanation: Ideally, dielectrics are insulators and do not
contain any free electrons. But no dielectric is a perfect
dielectric, hence the free electrons are due to impurities
present in each dielectric.

## 6. The current in conductors connecting the voltage source

to the plates of a capacitor is ______
a) Conduction current
b) Leakage current
c) Charging current
d) Zero

Explanation: The current in conductors connecting the
voltage source to the plates of a capacitor is the charging
current and not the conduction or leakage current.

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## 7. What is the type of current where the electrons actually

move?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

Explanation: Conduction current is the current caused by
the actual flow of electrons and displacement current is the
current where no charge carriers are involved.

## 8. What is the type of current caused due to variations in

the field?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

Explanation: Displacement current is the current where no
charge carriers are involved. It is caused due to variations
in the electric field.

## 9. Under normal conditions capacitors have _________

a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

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Explanation: Under normal conditions capacitors contain an
insulating material called dielectric sandwiched between the
plates of the capacitor. Since insulators can carry only an
electric field but not moving carriers, therefore normally a
capacitor has displacement current and not conduction
current.

## 10. If a large amount of voltage is applied to a capacitor,

what is the current that flows through it?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current

Explanation: When a large amount of voltage is applied
between the plates of a capacitor, the dielectric between
the plates does not behave as an insulator anymore and
starts conducting and conduction currents flow through it.

## Sanfoundry Global Education & Learning Series –

Basic Electrical Engineering.

## To practice all areas of Basic Electrical Engineering, here is

complete set of 1000+ Multiple Choice Questions and
user
### Content Structure
The provided text is a series of multiple-choice questions (MCQs) and answers related to basic electrical engineering, specifically focusing on capacitors, electric fields, and related concepts. Each question is followed by multiple options, and the correct answer is provided along with an explanation.

### Analysis
TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

### Explanation
1. **Multiple Questions**: The text contains multiple questions, each with multiple-choice options.
2. **Answers Provided**: For each question, the correct answer is provided along with an explanation.

### Example Questions and Answers
1. **Question**: What is the relation between current and voltage in a capacitor?
   - **Options**:
     - a) I=1/C*integral(Vdt)
     - b) I=CdV/dt
     - c) I=1/CdV/dt
     - d) I=Ct
   - **Answer**: b) I=CdV/dt
   - **Explanation**: Current=rate of change of charge=> I=dQ/dt. Q=CV, hence I=CdQ/dt.

2. **Question**: If 2V is supplied to a 3F capacitor, calculate the charge stored