Sol. The equivalent inductance L = 500mH Just before opening the switch, current th4ough the inductors is

I M = =2A

Potential drop across the resistor is 2 × 100 = 200 V

Hence, across capacitor potential difference is zero.

∴ Energy stored u = L(2) 2 = ×500×10 –3 ×2 2 = 1 J

∴ L = C = 1 J

⇒ = I M

= 2 = 200 V

ω =

= = 200 rad/s

The impedance of the LC circuit at ω = 100 rad/s is

Z = = (j 2 = – 1)

With the given value of ω , ω 2 × LC =

∴ Z = j × 100 × ×0.5 = j Ω

The peak current = = 3A

and the current lags w.r.t source emf by a phase of ( π /2).