# Kirchoff's rules and capacitors

1. May 30, 2014

### BOAS

Hello,

I thought I had questions concerning Kirchoff's rules down to a tee, but now capacitors have entered the scene and i'm stumped.

1. The problem statement, all variables and given/known data

The circuit in the drawing (attached) shows two resistors, a capacitor, and a battery. When the capacitor is fully charged, what is the magnitude q of the charge on one of its plates?

2. Relevant equations

3. The attempt at a solution

I apologise for the terrible resolution of the image, but I think it's still readable.

From the junction rule;

IA = IB + IC

For loop A;

12V = 2IA + 4IC

My text book says a loop does not need to contain a battery, so my expression for loop B is;

4Ic = Q/C

I am confused about whether the capacitor represents a voltage drop, because to me, it seems like a fully charged capacitor should create a potential rise.

I am unsure about whether I need to make an expression for the loop created by the 'outer' sides of the circuit.

Please can you help me in this regard?

Thanks.

#### Attached Files:

• ###### kirchoff2.PNG
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2. May 30, 2014

### willem2

When the capacitor is fully charged, no more current will go through it, so you can ignore the loop B.

Even if the capacitor isn't fully charged, you can treat the resistor and the capacitor as one circuit element, with I = V/R + C (dV/dt), and you can still ignore loop B.

3. May 30, 2014

### Staff: Mentor

When the capacitor is fully charged, you can imagine removing it from the circuit and replacing it with a voltmeter, an ideal voltmeter. Whatever reading the voltmeter gives is the same voltage as on the fully charged capacitor that is/was there.

* remember, an ideal voltmeter has infinite resistance and it draws no current

4. May 31, 2014

### CWatters

Are you familiar with potential divider circuits? Work out the voltage at the junction of R1 and R2 without the capacitor present. What happens if the voltage on the capacitor reaches this voltage?

Once you figure that out you can work out the charge on the capacitor.