# How to calculate the values of $R$ and $C$ in a series $RC$ circuit?

The exercise I'm trying to solve reads as follows:

A series $$RC$$ circuit is driven by $$v_{s}(t)= 25u(t)\: V$$. Given that $$i(0)=2 \:mA$$, $$v_c(0)=4\:V$$, and $$v_c=10\: V$$ at $$t=80\:ms$$, determine the values of $$R$$ and $$C$$

I'm trying to guess that there should be another voltage source, because from the equation that the textbook provides $$v_c (t)= V_B(1-e^{-t/RC})u(t)$$, where $$v_{s}(t)= 25u(t)\: V$$ is our $$V_Bu(t)$$ time dependent voltage source, I'd assume that $$v_c (0)=0 V$$ at $$t=0$$, but apparently the inicial voltage $$v_c(0)=4 V$$.

So my first question is: does it mean that there exists another voltage source that isn't time dependent? If not, I'd appreciate if you could explain what it means, since the only scenario where the book shows that $$v_c (0)\neq0 V$$ has another source that doesn't depend on time.

Also, I'm guessing that the only way to obtain the values of R and C is related to the $$e^{-t/RC}$$ section in the $$v_c (t)= V_B(1-e^{-t/RC})u(t)$$ and $$i(t)=(\frac{V_B}{R})e^{\frac{-t}{RC}}u(t)$$ equations, but if I'm missing another equation, I'd appreciate if you could help me as well.

• What is the meaning of u(t) here?
– nasu
Nov 12 '20 at 23:59
• Unit step function. Nov 13 '20 at 0:14

So, this is a fairly simple RC circuit where the capacitor has an initial charge. No, there is not an additional independent voltage source, just a voltage on the pre-charged capacitor. You are on the right path - you just need to develop an equation for the current as below, $$i(t) = \frac{V_S-V_c(0)}{R}e^{\frac{-t}{RC}}$$

From that, you can ferret out the requisite answers i think.

UPDATE: Showing Laplace solution approach to this circuit.

You can solve this circuit fairly easily with common differential equation methods. I prefer Laplace Transform methods and will show you how to solve this circuit below with that approach. You can use any of the normal circuit analysis approaches to solve the transformed circuit (superposition, source transformation etc. etc.).

The switch is closed at time t = zero. Prior to the switch closing, the capacitor holds an initial charge (Q=CV) that results in a voltage of $$V_C(0)=4V$$ across the capacitor as shown.

Below is the Laplace Transformed circuit (s-domain or frequency domain). Notice that the voltage across the capacitor, $$V_C(s)$$, is a combination of the voltage drop across the $$\frac{1}{sC}$$ impedance plus the voltage source representing the initial cap voltage.

Solving for $$I(s)$$,

$$I(s)=\frac{\frac{V_S}{s}-\frac{V_C(0)}{s}}{R+\frac{1}{sC}}=\frac{V_S-V_C(0)}{R(s+\frac{1}{RC})}$$

Now, we just need to inverse Laplace Transform to go back from the s-domain to the time domain. Here is a good table of transform pairs to save you time.

$$\mathscr{L}^{-1}[I(s)]=i(t)$$

$$i(t) = \mathscr{L}^{-1}\left(\frac{V_S-V_C(0)}{R(s+\frac{1}{RC})}\right)=\frac{V_S-V_C(0)}{R}e^{\frac{-t}{RC}}$$

Now, we can find the voltage across the capacitor just as easily using KVL,

$$V_C(s)=\frac{V_C(0)}{s}+I(s)\frac{1}{sC}$$

$$V_C(s)=\frac{V_C(0)}{s}+\frac{V_S-V_C(0)}{sCR(s+\frac{1}{RC})}$$

Now, we can look at the table of transform pairs and find that the simple unit step pair fits our first term, and the asymptotic exponential pair fits our second term.

$$\mathscr{L}^{-1}[V_C(s)]=v_c(t)=V_C(0)+\frac{V_S-V_C(0)}{RC}\frac{1}{\frac{1}{RC}}(1-e^{\frac{-t}{RC}})$$

Which simplifies nicely to, $$v_c(t)=V_C(0)+(V_S-V_C(0))(1-e^{\frac{-t}{RC}})$$

A little practice and you can handle any simple problem like this. For much bigger problems Laplace can get a little unwieldly and you would tend toward using a simulation tool like ATPDraw/ATP or such (e.g. LTSpice etc.).

Also, here is a simple table showing how initial conditions are introduced into the circuit for capacitor (initial voltage) and inductor (initial current).

• If it has an initial charge, does that modifies the equation $v_c (t)= V_B(1-e^{-t/RC})u(t)$ as well? Nov 12 '20 at 23:47
• Hi @racc_p, yes it would. The equation you show above is only valid when capacitor initially uncharged. If you let VB=Vs-Vc(0) and add a + Vc(0) you will have it for case with initial charge. What class is this? I can show you how to solve this very easily using Laplace Transform methods, but it may be above your pay grade right now. Nov 13 '20 at 0:15
• Awesome, thank you so much! It really helped me to understand it. Nov 14 '20 at 3:47
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