In: Physics
Given,
C1 = 4F , C2 = 12 F and C3 = 1 F
The equivalent cap of the circuit will be :
C12 = C1 + C2 = 4 + 12 = 16 F
Ceq = C12 x C3 / C12 + C3 = 16 x 3 / (16 + 3) = 2.53 F
Charge in the circuit:
Q = CV = 2.53 x 12 = 30.36 F
We know that the same charge flows through the capacitors connected in series, so
C12 = C3 = 30.36 F
We know that, V1 = V2, since they are in parallel
Q1/C1 = Q2/C2
Q1/4 = Q2/12 =>Q2 = 3Q1
Also, Q1 + Q2 = 30.36
Q1 + 3Q1 = 30.36 => Q1 = 7.59 C
Q2 = 3 x 7.59 = 22.7 C
V1 = V2 = Q1/C1 = 7.59/4 = 1.9 V
V3 = V - V1(or V2) = 12 - 1.9 = 10.1 V
Hence,
a) C1 = 7.59 C ; C2 = 22.7 C and C3 = 33.6 C
b) V1 = V2 = 1.9 V and V3 = 10.1 V
c) Energy is given by:
E = 1/2 C V2
E1 = 0.5 x 4 x 10-6 x 1.9 x 1.9 = 7.22 J
E2 = 0.5 x 12 x 10-6 x 1.9 x 1.9 = 21.7 J
E3 = 0.5 x 1 x 10-6 x 10.1 x 10.1 = 51 J