# 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 '20 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? – SalahTheGoat Sep 1 '20 at 12:42
• Right (also be careful about signs, it's easy to screw them up) – Peter Green Sep 1 '20 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. ￼