Capacitance Problem

1. May 25, 2014

rajumahtora

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

In the above circuit, find the potential difference across AB.

2. Solution Online

"C34 = 4μf
C2,34 = 12μf
CEQ = 4.8μf
q = CEQ x V
the q on 1 is 48μC, thus
V1 = q/c
V1 = 6v
VPQ = 10 - 6 = 4v
By Symmetry of 3 and 4 , VAB = 2v
"

3.My question is -
(a).What is Symmetry and how is it used in this case?
(b).How are they combining C2,34 with C1 in Series?

Thanks

2. May 25, 2014

BvU

The symmetry is that if you swap C3 and C4, you get the same situation, so the charge on each must be the same.

3. May 25, 2014

rajumahtora

How is same charge possible on both C3 and C4 and potential difference between B and A is different and P and A is different?
and how is VAB = 2v using this ?

4. May 25, 2014

willem2

C3 and c4 are in series, so the current through them will always be the same. If they start out with the same charge.m they will always have the same charge across them.

Why do you think that the potential difference between P and A and the potential difference between A and B are different?

5. May 26, 2014

BvU

Charge on right plate of capacitor 3 + charge on top plate of capacitor 4 = 0
Q = C V, Q same, C same, therefore V same. Vpb = 4 Volt so Vpa = Vab = 2 V

You know about calculating equivalent capacitance of capacitors in series ?