# RC circuits

1. Jul 17, 2005

### naissa

I am doing a lab on RC circuits and I do not understand anything about it.
In the lab, I charged an discharged a capacitor. The voltage decreased with the equation V=Voe^(-t/RC)

Can you explain me this lab?

Thanks

2. Jul 17, 2005

### Curious3141

Have you studied radioactivity yet ? Exponential growth and decay ? These are highly analogous situations.

Do you know how to set up and solve first order linear differential equations ? You'll need to know that in order to understand the derivation of that equation.

3. Jul 17, 2005

### naissa

yes I do

but I do not understand the cocnepts of the circuit.

4. Jul 17, 2005

### HallsofIvy

Staff Emeritus
It's hard to explain a lab when you haven't told us what you did! You give an exponential function but we have no idea what that has to do with the lab.

5. Jul 17, 2005

### Jelfish

You have a capacitor and resistor in series. The capacitor starts out with a charge V0 and starts to discharge onto the resistor. Now, because of the way capacitors work, the voltage they suply is proportional to how much charge is on the capacitor. Therefore, it makes sense that as time passes, not only does the voltage drop, but the change in voltage drops as well.

First let's draw a picture of the circuit:

Code (Text):
C
+-----| |------+
|              |
|              |
+-----~~~------+
R
Now, according to Kirchoff's Voltage Law, the sum of the voltage drops from those two components should equal zero, i.e.
$$IR+\frac{Q}{C}=0$$​
Since the voltage we're interested in is the voltage across the resistor, we can substitute IR for V:
$$V+\frac{Q}{C}=0$$
$$V=-\frac{Q}{C}$$​
We want to know how these things change with time, so we take the derivative with respect to time:
$$\dot{V}=-\frac{\dot{Q}}{C}$$​
You should know that the current is the time-derivative of charge:
$$\dot{V}=-\frac{I}{C}$$​
You also know that the current will be proportional to the voltage (Ohm's Law):
$$\dot{V}=-\frac{V}{RC}$$​
So now, doing a little rewritting and rearranging, we get:
$$\frac{dV}{V}=-\frac{dt}{RC}$$​
From here, you should be able to solve the differential equation. Hope that helps!

6. Jul 18, 2005

### BobG

Current will only flow through a capacitor until it reaches the same voltage as the source. If you apply a DC voltage to the RC circuit, current flows only until the capacitor charges up (you're measuring your voltage drop across the resistor?). Then the circuit reaches an equilibrium where the capacitor becomes an open circuit. Your lab is measuring how long this takes?

If so, you should be substituting in different capacitors and/or resistors and should discover the time required to charge a capacitor in an RC circuit is very predictable. Obviously, the fact that the value of the resistor and capacitor are in the equation implies that, but you're trying to find the precise relationship.

You could also find the precise relationship mathematically.

If t=0, what's the voltage?

If t=RC, what's the voltage? (this is particularly important)

How long until the voltage is practically zero; in other words, how long to charge the capacitor? (this is a little arbitrary - when t is 5 times larger than RC, the result is considered close enough to zero to be ignored).

This is important even for AC circuits (the type where capacitors and inductors are most useful), since the capacitor is going to take time to reach equilbrium with the rest of the circuit. When you first turn on your AC circuit, there will be a very short period of time where the performance of the circuit doesn't match what you want for the long term performance. That short time to charge up is called the transient response and is what you're calculating in your lab. The long term performance is called the steady state.

The others gave a good explanation of how the equation you're using was derived - in other words, why the equation does such a good job explaining what you should see in your lab.

Last edited: Jul 18, 2005
7. Jul 21, 2005

### Antiphon

The lower the voltage on the capacitor, the more slowly the current will come out.

As the current comes out of the capacitor, the voltage drops.

Imagine a water tank with a small hole in the bottom. The voltage
is like the pressure. As the water level (charge) comes down the
pressure (voltage) decreases and so the water coming out the hole
(current) slows down (or is less).

At the end you have barely a trickle when at first you had a strong
stream.

Last edited: Jul 21, 2005