In the circuit shown in the figure, `E = 50.0 V, R = 250 Omega` and `C = 0.500 muF.` The switch `S` is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of `150 V`. What is the inductance `L`?

Answer

B

Explanation

In steady state when switch was closed,
`i_0=E//R=(1//5)A=0.2A`
After switch is opened it becomes `L-C` circuit in which peak value current is `0.2 A`.
`:. 1/2 Li_0^2=1/2CV_0^2`
or `L=V_0^2/i_0^2.C`
`=((150)^2)/((0.2)^2)xx0.5xx10^-6`
`=0.28H`