Kirchoff's rules and capacitors

[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.

Homework Statement

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?

Homework Equations

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;

I_A = I_B + I_C

For loop A;

12V = 2I_A + 4I_C

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

4I_c = 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.

 

[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.

 

[NascentOxygen]

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

 

[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.