JEE Mains 29-July-2022 Shift 2 Solved Paper

At steady state charge stored on the capacitor,
qmax=CV
=500×10−6×100
=5×10−2CEnergy stored in the capacitor,
Umax=‌

qmax2

2C

Now, when electrostatic energy of capacitor converted to magnetic field energy then all energy of capacitor is transferrd to the inductor.
∴ Maximum energy stored in the inductor
UL‌max=‌

1

2

LImax2
∴‌

1

2

LImax2=‌

qmax2

2C

⇒Imax=‌

qmax

√LC

=‌

5×10−2

√50×10−3×500×10−6

=‌

5×10−2

5×10−3

=10A