When K is open, circuit is balanced wheatstone ∴Req=
20×40
20+40
+10=
70
3
Ω ∴i=
55
Req
=2.36A Now when key is closed We get 8l1+12(L1−I2)−40(I′−L1)=0....... ..(i) 2I2−12(L1−I2)=−10 . . . . . . . (ii) 8I1+12(L1−I2)+10l′=55...... . . . . . . . (iii) solving(i), (ii) and(iii) IH=2A,l′=2.7A,I2=1A Hence, Charge on capacitor =4.8µc