JEE Advanced 2020 Concept Recapitulation Test 3 Paper 1

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