# Homework Help: RC circuit-general questions

1. Oct 16, 2011

### pyroknife

Let's say you have a circuit with a resistor, capacitor and voltage source and a switch.
I got a few questions.
1. The instant the switch is closed, the voltage drop would all be across the resistor right?
2. After a long time the switch is closed, the current would approach 0 and the voltage drop would be all across the capacitor?

The above is charging a capacitor.
Below is discharging
If we have the same circuit, except that the switch was initially closed for a long time then opened.
1. The voltage the instant the switch is opened would be all across the capacitor still right?
2. After a longtime when, the charge on the capacitor has diminished, how would you calculate the current/voltage across the resistor?

2. Oct 16, 2011

### cepheid

Staff Emeritus
It sounds like you're talking about a series RC circuit, from the context. Let's say that we start with the switch open and placed in such a way that it interrupts the connection between the voltage source (battery or whatever) and the resistor and capacitor. Let's says also that the voltage source has a voltage of V0. If, at time t = 0, we close the switch, then the voltage across the capacitor vs. time is given by the function

$$v_C (t) = V_0 (1-e^{-t/RC})$$

If you plot this (e.g. look at 1-e-x on your graphing calculator or plotting computer program), you'll see that this starts at 0 and increases asymptotically to V0.

Since the voltage across the resistor and the voltage across the capacitor have to add up to the source voltage, the voltage across the resistor is given by:

$$v_R (t) = V_0 - v_c (t)$$

$$= V_0e^{-t/RC}$$

You can see that this starts at V0 and decays asymptotically to 0. So, your statements (1 and 2) are correct!

The way to derive the function for the capacitor voltage above is to start with KVL (sum of voltages around a closed loop must equal zero):

$$V_0 - v_C(t) - v_R(t) = 0$$

$$v_C(t) + v_R(t) = V_0$$

Now, since V = q/C for a capacitor, where q(t) is the charge on the capacitor (which is a function of time in this case) and also since Ohm's law says the voltage drop across a resistor is iR where i(t) is the current in the circuit (a function of time here), the equation above becomes:

$$\frac{q(t)}{C} + i(t)R = V_0$$

Noting that i = dq/dt by definition, this equation becomes:

$$\frac{dq}{dt} + \frac{q(t)}{RC} = \frac{V_0}{R}$$

If you solve this differential equation for q(t) and then note that q = CvC, you'll get the result that vC(t) is equal to what I wrote in the very first equation above.

Last edited: Oct 16, 2011
3. Oct 16, 2011

### cepheid

Staff Emeritus
Well, if you open the switch, the charge on the capacitor has nowhere to go (because it's an open circuit) and so nothing happens. However, if you then introduce a wire to bypass the voltage supply by connecting the resistor directly to the capacitor (so that the circuit now consists of only R and C in series), then the capacitor will discharge through the resistor. Starting with KVL:

$$v_R(t) = -v_C(t)$$

Or:

$$i(t)R =- \frac{q(t)}{C}$$

$$\frac{dq}{dt} = -\frac{q}{RC}$$

Solving this differential equation (which is easier than the one in my previous post, because it is separable), and assuming that the capacitor starts out with charge CV0 at t = 0, you get:

$$q(t) = CV_0e^{-t/RC}$$

The current as a function of time i(t) will just be the derivative of this.

4. Oct 16, 2011

### Staff: Mentor

Yes, providing there was no initial charge on the capacitor.

Yes.

You arrange this as a series circuit without the battery, I suppose you mean? If so, the capacitor has the full voltage across it, and the switch places all of this across the resistor. So immediately after the switch is closed, called t=0+ the resistor has a voltage across it equal to the capacitor's fully-charged voltage.

You don't have to wait a long time, you can calculate it at ANY time because it follows the exponential decay equation with a time constant = R.C seconds.

Last edited: Oct 16, 2011