# RC Circuit Lab

1. Apr 1, 2013

### Bashyboy

Hello,

At the moment, I am writing the theory portion of my lab report. In the RC circuit, will the electric potential across the capacitor be at the same electric potential as the battery? Or will it depend on the voltage drop across the intermediate resistor?

2. Apr 1, 2013

### Staff: Mentor

Electric potential is a property of individual point, not a difference. I think you mean "potential difference" or "voltage". Do you know Kirchhoff's circuit laws? They give a direct answer to your question. Otherwise, the concept of a potential itself is sufficient to see it.

3. Apr 1, 2013

### Bashyboy

Let me repose my question: Suppose we have a capacitor connected to a battery that has a voltage V. If the capacitor is initially uncharged, then current will flow out of the to the capacitor, until the capacitor is at the voltage as the battery. Now, if I were to supplement this circuit with a resistor, whose position would be such that the current first passes through it before hitting the capacitor, would the capacitor reach the same voltage that the battery has, or would it somehow be less because of the voltage drop across the resistor?

4. Apr 1, 2013

### Staff: Mentor

Current is nothing which can "pass something before it hits something else". Current flows in the whole circuit.
If you wait long enough.
Keep in mind that a voltage drop in a resistor requires a current.

5. Apr 1, 2013

### Staff: Mentor

The resistor would add resistance to the circuit and increase the amount of time it takes for the capacitor to charge fully, but the voltage would still be the same once charged.

Once fully charged the capacitor is acting like an "open", and as such the resistor would have zero voltage drop across it, as all of the voltage drop would be across the capacitor.

6. Apr 1, 2013

### Bashyboy

From my understanding, current, or charges, cannot pass through the region between the capacitor plates. So, electrons build up on the one plate, repelling the electrons on the other plate. Is this not correct?

7. Apr 1, 2013

### Bashyboy

So, the resistor just increases the time it takes for the capacitor to acquire the same voltage as the battery? Also, what exactly do you mean by the capacitor acting like and "open"ing? I don't quite understand how there isn't a voltage drop across the resistor, nor do I understand how the voltage drop is across the capacitor.

8. Apr 2, 2013

### Staff: Mentor

In a DC circuit, once the capacitor is charged it isn't possible for current to flow through the circuit anymore. This situation is exactly what would happen if you simply cut the conductor and made an open.

Now, the voltage drop across each component in series must add up to equal the applied voltage. If you used a multimeter to measure the voltage across each component you would find that the resistor has zero volts across it. You could add a hundred resistors before or after the capacitor and the voltage across each one would still be zero. This is because there is no difference in electric potential across the resistor. All of it is across the capacitor.

Think about it. If your power source is applying voltage to the circuit, and your capacitor is applying the exact opposite voltage, then the two cancel out. Again, if you simply replaced the capacitor with a cut piece of wire this is exactly what would happen. The cut piece of wire would act like a very tiny capacitor, with charges building up on each side that oppose the applied voltage.

9. Apr 2, 2013

### Bashyboy

How does the capacitor apply an exact opposite voltage to that of the battery's? If the capacitor is initially uncharged, the capacitor has no voltage across the plates; once the charging begins, however, doesn't it obtain the exact same voltage as the battery?

10. Apr 2, 2013

### Staff: Mentor

With a resistor in the circuit, it needs some time to charge, as the resistor limits the current (to some charge-state-dependent value).

11. Apr 2, 2013

### Staff: Mentor

Yes, exactly. Once charged the voltage across the two plates will be equal to the applied voltage. But with infinite resistance no current can flow. Perhaps my choice of words was poor. I shouldn't have said the capacitor has opposite voltage to the battery. I merely meant that the charges that build up on each plate oppose the buildup of more charges because the voltage is the same as the source. Which just means that once the voltage of the capacitor reaches the applied voltage, no more current flows.

12. Apr 3, 2013

### Basic_Physics

The additional resistor just slows the charging up of the capacitor down, and yes it will still charge up to the voltage of the battery. At that point the flow of charge, or current, in the circuit stops since the two voltages opposes each other and are the same.