# 19.4 Sources of electromotive force

## Presentation on theme: "19.4 Sources of electromotive force"— Presentation transcript:

19.4 Sources of electromotive force
(m) define e.m.f. in terms of the energy transferred by a source in driving unit charge round a complete circuit (n) distinguish between e.m.f. and p.d. in terms of energy considerations (o) show an understanding of the effects of the internal resistance of a source of e.m.f. on the terminal potential difference and output power.

19.4 Sources of electromotive force
(m) define e.m.f. in terms of the energy transferred by a source in driving unit charge round a complete circuit.

19.4 Sources of electromotive force
The voltage produced by the cell is called the electromotive force or e.m.f. for short. The e.m.f. is defined as the energy transferred by a source in driving unit charge round a complete circuit. Explain what is meant that the e.m.f of a cell is 1.5 volt.

19.4 Sources of electromotive force
distinguish between e.m.f. and p.d. in terms of energy considerations.

19.4 Sources of electromotive force
Potential difference and voltage are the same thing. P.d is the technical term Voltage is the common use, since potential difference is measured in Volts It would be the same as if one referred to distance as "meterage" or speed as the "kilometers per hourage”. Both are the amount of energy per charge that will be dissipated (or gained) between two points. e.g. 1 Volt means that 1 Coulomb of charge will lose or gain 1 Joule of energy.

19.4 Sources of electromotive force
EMF is a little different, since it refers to the actual source of potential difference. Often in terms of a battery, but also other sources like a change in magnetic field So the EMF refers to the specific mechanism where that charge gains its energy.

19.4 Sources of electromotive force
The best way to think of them is: Emf - is the amount of energy of any form that is changed into electrical energy per coulomb of charge. pd - is the amount of electrical energy that is changed into other forms of energy per coulomb of charge.

19.4 Sources of electromotive force
Sources of emf: Cell, battery (a combination of cells), solar cell, generator, dynamo, thermocouple.

19.4 Sources of electromotive force
show an understanding of the effects of the internal resistance of a source of e.m.f. on the terminal potential difference and output power.

19.4 Sources of electromotive force
Internal Resistance Cells and batteries are not perfect. Use them for a while and you will notice they get hot. Where is the heat energy coming from? It's from the current moving through the inside of the cell. The resistance inside the cell turns some of the electrical energy it produced to heat energy as the electrons move through it.

19.4 Sources of electromotive force
Imagine that each cell is perfect except that for some bizarre reason the manufacturers put a resistor in series with the cell inside the casing. Therefore, inside the cell, energy is put into the circuit by the cell (the emf). But some of this energy is taken out of the circuit by the internal resistor (a pd). So the pd available to the rest of the circuit (the external circuit) is the emf minus the pd lost inside the cell.

19.4 Sources of electromotive force
emf and internal resistance The voltage produced by the cell is called the electromotive force or e.m.f for short and this produces a p.d across the cell and across the external resistor (R). E = IR + Ir = V + Ir

19.4 Sources of electromotive force
The e.m.f (E) of the cell can be described as the maximum p.d that the cell can produce across its terminals, or the open circuit p.d since when no current flows from the cell no electrical energy can be lost within it.

19.4 Sources of electromotive force
The quantity of useful electrical energy per unit charge available outside the cell is IR and Ir is the energy per unit charge transformed to other forms within the cell itself.

19.4 Sources of electromotive force
Emf and internal resistance experiment

19.4 Sources of electromotive force
A 9.0 V battery has an internal resistance of 12.0 .   (a) What is the potential difference across its terminals when it is supplying a current of 50.0 mA? (b) What is the maximum current this battery could supply? 2. A cell in a deaf aid supplies a current of 25.0 mA through a resistance of 400 . When the wearer turns up the volume, the resistance is changed to 100  and the current rises to 60 mA. What is the emf and internal resistance of the cell? A battery is connected in series with a variable resistor and an ammeter. When the resistance of the resistor is 10  the current is 2.0 A. When the resistance is 5  the current is 3.8 A. Find the emf and the internal resistance of the battery. When a cell is connected directly across a high resistance voltmeter the reading is V. When the cell is shorted through a low resistance ammeter the current is 2.5 A. What is the emf and internal resistance of the cell?

19.4 Sources of electromotive force
1. (a) pd = E – I r = 9 – (50 x 10-3 x 12) = 8.4 V (b) Max current = E/r = 9 / 12 = 0.75 A 2. E = I(R +r) E = 25 x 10-3 (400 + r) and E = 60 x 10-3 (100 + r) So 25 x 10-3 (400 + r) = 60 x 10-3 (100 + r) so r =  E = 10 + (25 x 10-3 x 114.3) = V 3. E = I(R +r) E = 2 (10 + r) and E = 3.8 (5 + r) so r= 0.56  E = 20 + (2 x 0.56) = 21.1 V 4. E = 1.5 V E = I r so 1.5 V = 2.5 A r and r = 0.6 

19.4 Sources of electromotive force
The headlamps of a car are connected in parallel across a twelve-volt battery. The starter motor is also in parallel controlled by the ignition switch. Since the starter motor has a low resistance it demands a very high current (say 60 A). The battery itself has a low internal resistance (say 0.01 Ω). The headlamps themselves draw a much lower current. What happens when the engine is started (switch to starter motor closed for a short time).

19.4 Sources of electromotive force
sudden demand for more current large lost volts (around 0.01 Ω  60 A = 6 V) terminal voltage drops to 12 V – 6 V = 6 V headlamps dim When the engine fires, the starter motor switch is opened and the current drops. The terminal voltage rises and the headlamps return to normal. It’s better to turn the headlamps off when starting the car.

19.4 Sources of electromotive force
Further Questions P217 q11-15 P275 q7 Further reading P212-3 & 215 Web links
The content structure consists of a series of slides or sections from a presentation on electromotive force (emf), including definitions, distinctions between emf and potential difference (p.d.), and examples of sources of emf. The latter part of the text includes several mathematical questions and their solutions related to emf and internal resistance.
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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>
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- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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# 19.4 Sources of electromotive force

## Presentation on theme: "19.4 Sources of electromotive force"— Presentation transcript:

19.4 Sources of electromotive force
(m) define e.m.f. in terms of the energy transferred by a source in driving unit charge round a complete circuit (n) distinguish between e.m.f. and p.d. in terms of energy considerations (o) show an understanding of the effects of the internal resistance of a source of e.m.f. on the terminal potential difference and output power.

19.4 Sources of electromotive force
(m) define e.m.f. in terms of the energy transferred by a source in driving unit charge round a complete circuit.

19.4 Sources of electromotive force
The voltage produced by the cell is called the electromotive force or e.m.f. for short. The e.m.f. is defined as the energy transferred by a source in driving unit charge round a complete circuit. Explain what is meant that the e.m.f of a cell is 1.5 volt.

19.4 Sources of electromotive force
distinguish between e.m.f. and p.d. in terms of energy considerations.

19.4 Sources of electromotive force
Potential difference and voltage are the same thing. P.d is the technical term Voltage is the common use, since potential difference is measured in Volts It would be the same as if one referred to distance as "meterage" or speed as the "kilometers per hourage”. Both are the amount of energy per charge that will be dissipated (or gained) between two points. e.g. 1 Volt means that 1 Coulomb of charge will lose or gain 1 Joule of energy.

19.4 Sources of electromotive force
EMF is a little different, since it refers to the actual source of potential difference. Often in terms of a battery, but also other sources like a change in magnetic field So the EMF refers to the specific mechanism where that charge gains its energy.

19.4 Sources of electromotive force
The best way to think of them is: Emf - is the amount of energy of any form that is changed into electrical energy per coulomb of charge. pd - is the amount of electrical energy that is changed into other forms of energy per coulomb of charge.

19.4 Sources of electromotive force
Sources of emf: Cell, battery (a combination of cells), solar cell, generator, dynamo, thermocouple.

19.4 Sources of electromotive force
show an understanding of the effects of the internal resistance of a source of e.m.f. on the terminal potential difference and output power.

19.4 Sources of electromotive force
Internal Resistance Cells and batteries are not perfect. Use them for a while and you will notice they get hot. Where is the heat energy coming from? It's from the current moving through the inside of the cell. The resistance inside the cell turns some of the electrical energy it produced to heat energy as the electrons move through it.

19.4 Sources of electromotive force
Imagine that each cell is perfect except that for some bizarre reason the manufacturers put a resistor in series with the cell inside the casing. Therefore, inside the cell, energy is put into the circuit by the cell (the emf). But some of this energy is taken out of the circuit by the internal resistor (a pd). So the pd available to the rest of the circuit (the external circuit) is the emf minus the pd lost inside the cell.

19.4 Sources of electromotive force
emf and internal resistance The voltage produced by the cell is called the electromotive force or e.m.f for short and this produces a p.d across the cell and across the external resistor (R). E = IR + Ir = V + Ir

19.4 Sources of electromotive force
The e.m.f (E) of the cell can be described as the maximum p.d that the cell can produce across its terminals, or the open circuit p.d since when no current flows from the cell no electrical energy can be lost within it.

19.4 Sources of electromotive force
The quantity of useful electrical energy per unit charge available outside the cell is IR and Ir is the energy per unit charge transformed to other forms within the cell itself.

19.4 Sources of electromotive force
Emf and internal resistance experiment

19.4 Sources of electromotive force
A 9.0 V battery has an internal resistance of 12.0 .   (a) What is the potential difference across its terminals when it is supplying a current of 50.0 mA? (b) What is the maximum current this battery could supply? 2. A cell in a deaf aid supplies a current of 25.0 mA through a resistance of 400 . When the wearer turns up the volume, the resistance is changed to 100  and the current rises to 60 mA. What is the emf and internal resistance of the cell? A battery is connected in series with a variable resistor and an ammeter. When the resistance of the resistor is 10  the current is 2.0 A. When the resistance is 5  the current is 3.8 A. Find the emf and the internal resistance of the battery. When a cell is connected directly across a high resistance voltmeter the reading is V. When the cell is shorted through a low resistance ammeter the current is 2.5 A. What is the emf and internal resistance of the cell?

19.4 Sources of electromotive force
1. (a) pd = E – I r = 9 – (50 x 10-3 x 12) = 8.4 V (b) Max current = E/r = 9 / 12 = 0.75 A 2. E = I(R +r) E = 25 x 10-3 (400 + r) and E = 60 x 10-3 (100 + r) So 25 x 10-3 (400 + r) = 60 x 10-3 (100 + r) so r =  E = 10 + (25 x 10-3 x 114.3) = V 3. E = I(R +r) E = 2 (10 + r) and E = 3.8 (5 + r) so r= 0.56  E = 20 + (2 x 0.56) = 21.1 V 4. E = 1.5 V E = I r so 1.5 V = 2.5 A r and r = 0.6 

19.4 Sources of electromotive force
The headlamps of a car are connected in parallel across a twelve-volt battery. The starter motor is also in parallel controlled by the ignition switch. Since the starter motor has a low resistance it demands a very high current (say 60 A). The battery itself has a low internal resistance (say 0.01 Ω). The headlamps themselves draw a much lower current. What happens when the engine is started (switch to starter motor closed for a short time).

19.4 Sources of electromotive force
sudden demand for more current large lost volts (around 0.01 Ω  60 A = 6 V) terminal voltage drops to 12 V – 6 V = 6 V headlamps dim When the engine fires, the starter motor switch is opened and the current drops. The terminal voltage rises and the headlamps return to normal. It’s better to turn the headlamps off when starting the car.

19.4 Sources of electromotive force
Further Questions P217 q11-15 P275 q7 Further reading P212-3 & 215 Web links
user
The content structure consists of a series of slides or sections from a presentation on electromotive force (emf), including definitions, distinctions between emf and potential difference (p.d.), and examples of sources of emf. The latter part of the text includes several mathematical questions and their solutions related to emf and internal resistance.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS