# Does Ohm's law apply in an RLC circuit?

Suppose I have the discharging free source RC circuit as shown below (I am referring to figure a, not figure b):

From this, I can easily calculate that the charge on the capacitor as a function of time is $$Q(t)=Q_0 e^{-\frac{t}{RC}}$$ where $$\Q_0\$$ is the initial charge on the capacitor.

I then note that $$\I=-\frac{dQ}{dt}\$$ and so taking the derivative of my calculated formula $$\Q(t)=Q_0 e^{-\frac{t}{RC}}\$$, I get that the current as a function of time is $$I(t)=\frac{Q_0}{RC}e^{-\frac{t}{RC}}$$

But now from Ohm's law ($$\V=I\cdot R\$$) I can get the voltage as a function of time simply by multiplying my function $$\I(t)=\frac{Q_0}{RC}e^{-\frac{t}{RC}}\$$ by $$\R\$$, so that I have $$V(t)=\frac{Q_0}{C}e^{-\frac{t}{RC}}$$ or since $$\V_0 =\frac{Q_0}{C}\$$ I get $$V(t)=V_0 e^{-\frac{t}{RC}}$$

I know that this is the correct answer since multiple sources corroborate it. But now suppose I have a source-free discharging RLC circuit like this one shown.

My textbook (Electricity and Magnetism, Purcell and Morin) calculates that the voltage with respect to time is of the form $$V(t)=e^{-\alpha t}(Acos(\omega t)+Bsin(\omega t))$$ and I can easily follow the steps as to why.

Then since for our circuit we have that $$\I=-C\frac{dV}{dt}\$$, we can calculate the current by deriving the voltage formula and we get that $$I(t)=-C\frac{dV}{dt}=AC\omega (\sin(\omega t)+\frac{\alpha}{\omega}\cos(\omega t))e^{-\alpha t}$$

But now if I simply applied Ohm's law by dividing the original equation for by voltage $$\V(t)=e^{-\alpha t}(Acos(\omega t)+Bsin(\omega t))\$$ by $$\R\$$ I get a totally different answer:

$$I(t)=\frac{1}{R}e^{-\alpha t}(Acos(\omega t)+Bsin(\omega t))$$

In this form of the answer, the phase difference between the current and voltage seems to have completely disappeared, so why does the straightforward application of Ohm's law not produce the correct result?

• In addition to the answers below, be aware that your differentiation is wrong: (i) you need to apply the product rule; and (ii) $\frac{d}{dt}\:cos\: \omega t = -\omega\: sin\:\omega t$
– Chu
Sep 1, 2020 at 16:33

The place where beginners usually screw up with Ohm's law is by not being clear about which voltage and/or which current they are referring to.

Ohm's law relates the voltage across the resistor to the current through the resistor.

"V" in your diagram is not the voltage across the resistor; it is the voltage across the capacitor.

• Okay this makes perfect sense. Just to be clear though: In my first diagram , the voltage across the resistor is always equal to the voltage across the capacitor so I can simply divide by the resistance of the resistor since from ohms law we have $V_{resistor}=IR$ but we know that $V_{resistor}=V_{capacitor}$ so we can just as well say that $V_{capacitor}=IR$. But in the case of my second diagram , the formula for the voltage is the voltage across the capacitor and is not always equal to the voltage across the resistor and so we cannot apply this same reasoning right? Sep 1, 2020 at 12:42
• Right (also be careful about signs, it's easy to screw them up) Sep 1, 2020 at 12:54

The voltage in your figure is not the one across the resistor, so it'll not work as intended and that yields incorrect results.

Remember as well that for capacitors and inductors, the use of phasors or the Laplace Transform may ease all your time domain calculations, a quick grasp of MATLAB functions can solve you lots of equations for bigger circuits, which would take lots of sines and cosines out, effort cutback and will leave only what is actually important in the end.

Ohm’s Law works best using Z(f) for each part with the signal spectrum rather than time domain.

ZL(f)=j 2pi * f L ,
Zc(f)=1/(j 2pi*f C)=-j/(2pi * f * C)

A step input is broad spectrum and so if you need time domain, use asymptotic Tau=RC as the 63% target voltage derived from (e-1)/e≈0.63. ￼

• What do you mean by "for each part with the signal spectrum"? Can you elaborate? E.g., what are parts? - real and imaginary (complex numbers)? Deriving the impedance from frequency measurements (amplitude and phase)? Sep 2, 2020 at 14:41
• commons.wikimedia.org/wiki/File:Time_constant_fr.svg Sep 2, 2020 at 14:58
The text passage discusses the application of Ohm's law in different types of circuits, specifically an RC circuit and an RLC circuit. It includes calculations and derivations of voltage and current as functions of time in these circuits and questions why a straightforward application of Ohm's law does not yield the correct result in the RLC circuit.
single_question_with_answer

the passage contains a single mathematical question about why the straightforward application of ohm's law does not produce the correct result in an rlc circuit. the answer is provided in the comments and the subsequent discussion, explaining that the voltage in the diagram is not the voltage across the resistor, and thus ohm's law cannot be directly applied in the same way as in the rc circuit.
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
# Does Ohm's law apply in an RLC circuit?

Suppose I have the discharging free source RC circuit as shown below (I am referring to figure a, not figure b):

From this, I can easily calculate that the charge on the capacitor as a function of time is $$Q(t)=Q_0 e^{-\frac{t}{RC}}$$ where $$\Q_0\$$ is the initial charge on the capacitor.

I then note that $$\I=-\frac{dQ}{dt}\$$ and so taking the derivative of my calculated formula $$\Q(t)=Q_0 e^{-\frac{t}{RC}}\$$, I get that the current as a function of time is $$I(t)=\frac{Q_0}{RC}e^{-\frac{t}{RC}}$$

But now from Ohm's law ($$\V=I\cdot R\$$) I can get the voltage as a function of time simply by multiplying my function $$\I(t)=\frac{Q_0}{RC}e^{-\frac{t}{RC}}\$$ by $$\R\$$, so that I have $$V(t)=\frac{Q_0}{C}e^{-\frac{t}{RC}}$$ or since $$\V_0 =\frac{Q_0}{C}\$$ I get $$V(t)=V_0 e^{-\frac{t}{RC}}$$

I know that this is the correct answer since multiple sources corroborate it. But now suppose I have a source-free discharging RLC circuit like this one shown.

My textbook (Electricity and Magnetism, Purcell and Morin) calculates that the voltage with respect to time is of the form $$V(t)=e^{-\alpha t}(Acos(\omega t)+Bsin(\omega t))$$ and I can easily follow the steps as to why.

Then since for our circuit we have that $$\I=-C\frac{dV}{dt}\$$, we can calculate the current by deriving the voltage formula and we get that $$I(t)=-C\frac{dV}{dt}=AC\omega (\sin(\omega t)+\frac{\alpha}{\omega}\cos(\omega t))e^{-\alpha t}$$

But now if I simply applied Ohm's law by dividing the original equation for by voltage $$\V(t)=e^{-\alpha t}(Acos(\omega t)+Bsin(\omega t))\$$ by $$\R\$$ I get a totally different answer:

$$I(t)=\frac{1}{R}e^{-\alpha t}(Acos(\omega t)+Bsin(\omega t))$$

In this form of the answer, the phase difference between the current and voltage seems to have completely disappeared, so why does the straightforward application of Ohm's law not produce the correct result?

• In addition to the answers below, be aware that your differentiation is wrong: (i) you need to apply the product rule; and (ii) $\frac{d}{dt}\:cos\: \omega t = -\omega\: sin\:\omega t$
– Chu
Sep 1, 2020 at 16:33

The place where beginners usually screw up with Ohm's law is by not being clear about which voltage and/or which current they are referring to.

Ohm's law relates the voltage across the resistor to the current through the resistor.

"V" in your diagram is not the voltage across the resistor; it is the voltage across the capacitor.

• Okay this makes perfect sense. Just to be clear though: In my first diagram , the voltage across the resistor is always equal to the voltage across the capacitor so I can simply divide by the resistance of the resistor since from ohms law we have $V_{resistor}=IR$ but we know that $V_{resistor}=V_{capacitor}$ so we can just as well say that $V_{capacitor}=IR$. But in the case of my second diagram , the formula for the voltage is the voltage across the capacitor and is not always equal to the voltage across the resistor and so we cannot apply this same reasoning right? Sep 1, 2020 at 12:42
• Right (also be careful about signs, it's easy to screw them up) Sep 1, 2020 at 12:54

The voltage in your figure is not the one across the resistor, so it'll not work as intended and that yields incorrect results.

Remember as well that for capacitors and inductors, the use of phasors or the Laplace Transform may ease all your time domain calculations, a quick grasp of MATLAB functions can solve you lots of equations for bigger circuits, which would take lots of sines and cosines out, effort cutback and will leave only what is actually important in the end.

Ohm’s Law works best using Z(f) for each part with the signal spectrum rather than time domain.

ZL(f)=j 2pi * f L ,
Zc(f)=1/(j 2pi*f C)=-j/(2pi * f * C)

A step input is broad spectrum and so if you need time domain, use asymptotic Tau=RC as the 63% target voltage derived from (e-1)/e≈0.63. ￼

• What do you mean by "for each part with the signal spectrum"? Can you elaborate? E.g., what are parts? - real and imaginary (complex numbers)? Deriving the impedance from frequency measurements (amplitude and phase)? Sep 2, 2020 at 14:41
• commons.wikimedia.org/wiki/File:Time_constant_fr.svg Sep 2, 2020 at 14:58
user
The text passage discusses the application of Ohm's law in different types of circuits, specifically an RC circuit and an RLC circuit. It includes calculations and derivations of voltage and current as functions of time in these circuits and questions why a straightforward application of Ohm's law does not yield the correct result in the RLC circuit.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question about why the straightforward application of Ohm's law does not produce the correct result in an RLC circuit. The answer is provided in the comments and the subsequent discussion, explaining that the voltage in the diagram is not the voltage across the resistor, and thus Ohm's law cannot be directly applied in the same way as in the RC circuit.