# Calculating an R-RC circuit

This is not homework. I have this circuit, and I want to calculate V2. I know it is equal to V1 at t=0, and equal to $V1 \cdot \frac{R2}{R1+R2}$ at t=$\infty$, but I don't know how to calculate the charging of the capacitor.

All I find on Google is charging of an RC, without the parallel resistor.

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en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem Thevenin says R1 and R2 are in parallel and the voltage is relative to the resistor divider. Is that hint enough to get you going? –  jippie Jul 8 '12 at 14:41

The answer is in Thévenin, like Alfred and jippie also suggested. Thévenin claims that any 1-port network consisting of voltage sources and resistors can be replaced by a voltage source and a series resistor across that port, and who am I not to believe him?

Let's consider your circuit without the capacitor and assign its connections as the circuit's port.

First we look for $V_{th}$, which we do by leaving the output open-circuit, so that $R_{th}$ can't cause a voltage drop. Then R1 and R2 form a voltage divider with $V_{AB}$ = V1 $\times$ R1/(R1 + R2) = 3 V. (I'm using actual values for voltage and resistors to make it more graphic.) That's $V_{th}$. Fine.

Next we have to find $R_{th}$. You can do that by shorting all voltage sources and measure the resistance between A and B. But let's do it the alternative way: short-circuit A to B, and measure the current through that point. That should be $V_{th}/R_{th}$. Both methods give the same result, and it depends on the kind of circuit which way is best.

So shorting A-B we get I = V1/R2 = 12 V/ 12 Ω = 1 A. (What a coincidence! :-)) Then $R_{th}$ = 3 V/ 1 A = 3 Ω. If we now reconnect our load we have the typical RC circuit where C1 is charged via a series resistor (let's say C1 is 1 F):

$V_C(t) = V_\infty + (V_0 - V_\infty) e^{\dfrac{-t}{RC}}$

$V_\infty$ is $V_{th}$ because after C1 is charged there won't be a voltage drop across $R_{th}$. And $V_0$ is 0, we start with an uncharged capacitor. Then

$V_C(t) = 3 V + (0 V - 3 V) e^{\dfrac{-t}{3 s}} =3 V (1 - e^{\dfrac{-t}{3 s}})$

And that's the well-known charging equation.

The blue curve is the voltage between A and B, the purple curve is the voltage at B with respect to ground.

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Haha! Nicely explained and was really funny. I got a little lost when you have posted both the thevenin circuit and the original circuit without the capacitor, but then I recovered quickly. –  abdullah kahraman Jul 9 '12 at 21:40
@abdullah - Funny??? Not the exponential, I hope :-) –  stevenvh Jul 10 '12 at 4:15
lol. what happened to the exponential ? :) –  abdullah kahraman Jul 10 '12 at 6:44

There's always more than one approach to solving a circuit problem but the approach I generally find most useful in this type of problem is to find the Thevenin equivalent resistance $R_{TH}$ "seen" by the capacitor. This will allow you to find the time constant, $\tau = R_{TH}C$.

To find the Thevenin resistance, remove the capacitor and zero the voltage source (replace with wire). Now, find the resistance between the terminals where the capacitor connects; that resistance is $R_{TH}$

If you've already found the voltage across the capacitor at t = 0 and t = $\infty$, just "connect them together" with the exponential function:

$v_C(t) = [v_C(\infty) - v_C(0)](1 - e^{t/\tau}) + v_C(0)$

For $v_C(0) = 0$, this simplifies to:

$v_C(t) = v_C(\infty)(1 - e^{t/\tau})$

Now that you have $v_C(t)$, you have $v_2(t) = V_1 - v_C(t)$

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