# RC Circuit Lab

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?

mfb
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.

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?

mfb
Mentor
Current is nothing which can "pass something before it hits something else". Current flows in the whole circuit.
would the capacitor reach the same voltage that the battery has
If you wait long enough.
or would it somehow be less because of the voltage drop across the resistor?
Keep in mind that a voltage drop in a resistor requires a current.

Drakkith
Staff Emeritus
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.

Current is nothing which can "pass something before it hits something else". Current flows in the whole circuit..

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?

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.

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.

Drakkith
Staff Emeritus
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.

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?

mfb
Mentor
once the charging begins, however, doesn't it obtain the exact same voltage as the battery?
With a resistor in the circuit, it needs some time to charge, as the resistor limits the current (to some charge-state-dependent value).

Drakkith
Staff Emeritus