net capacitance of 24uF ,12uF and 8uF is (1/Cx) = (1/24)+(1/12)+(1/8)
Cx = 4 uF
this Cx is in parallel with 4uF then Cy = 4+4 = 8 uF
this Cy ,5uF and 6uF are in series
then net capacitance of the circuit is 1/(C_AB) =
(1/8)+(1/5)+(1/6)
C_AB = 2.03 uF
energy stored in this circuit is U = 0.5*C*V^2 = 0.5*2.03*10^-6**120^2 = 0.014616 J = 14.61 mJ
b) potential difference across 4uF capacitor is Qnet /Cy =
Cnet*V/Cy = (2.03*10^-6*120)/(8*10^-6) = 30.45 V
then charge on 4uF is Q = Cy*V1 = 4*10^-6*30.45 = 121.8*10^-6 C = 121.8 uC
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