C1 = 5μf V1 = 24V
q1 = C1V1 = 5 × 24 = 120μc and C2 = 6μf
V2 = R
q2 = C2V2 = 6 × 12 = 72
∴ Energy stored on first capacitor
(c) U1 = (1/2) C1V2
U2 = (1/2) C2V2
Uf = (1/2)V2 (C1 + C2) = (1/2) (4.36)2 (5 + 6) = 104.5 × 10–6J = 0.1045mJ
But Ui = 1.44 + 0.433 = 1.873
∴ The loss in KE = 1.873 – 0.1045 = 1.7687 = 1.77mJ