Find Potential Difference Across AB: Capacitance Problem Solved

[rajumahtora]

Homework Statement

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


2. Solution Online


"C₃₄ = 4μf
C_2,34 = 12μf
C_EQ = 4.8μf
q = C_EQ x V
the q on 1 is 48μC, thus
V₁ = q/c
V₁ = 6v
V_PQ = 10 - 6 = 4v
By Symmetry of 3 and 4 , V_AB = 2v"

3.My question is -
(a).What is Symmetry and how is it used in this case?
(b).How are they combining C_2,34 with C₁ in Series?

Thanks

 

[BvU]

How come you ask about C_{2, 34} in series with C₁ but you do not ask about C₃₄ ?

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

 

[rajumahtora]

How come you ask about C_{2, 34} in series with C₁ but you do not ask about C₃₄ ?

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

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 V_AB = 2v using this ?

 

[willem2]

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 V_AB = 2v using this ?

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?

 

[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 ?