In the circuit in Fig. switch `S_(1)` was closed for a long time . At time `t = 0` the switch is opened.

Find the maximum potentail difference across the plates of the capacitor after the switch is opened.

Options

A. `100 V`
B. `200 V`
C. `300 V`
D. `400 V`

Answer

B

Explanation

The equaivalent inductance `L = 500 mH`. Just before opening the switch, current through the inductor is
`I_(M) = (200)/(100) = 2a`
Potential drop across the resistor is `2 xx 100 = 200 V`
hence, across capacitor potential difference is zero.
So energy stored
`U = (1)/(2) L(2)^(2) = (1)/(2) xx 500 xx 10^(-3) xx 2^(2) = 1J`
`(1)/(2) LI_(M)^(2) = (1)/(2) Cv_(max)^(2) = 1J`
`V_(max) = I_(M) sqrt((L)/(C )) = 2 xx sqrt((500 xx 10^(-3))/(50 xx 10^(-6))) = 200 V`
`omega = (1)/(sqrt(LC)) = (1)/(sqrt(500 xx 10^(-3) xx 50 xx 10^(-6))) = 200 rad s^(-1)`