Question

A circuit containing capacitors `C_(1) and C_(2)` shown in Fig. is in the steady state with key `K_(1)` closed. At the instant t = 0, `K_(1)` is opened and `K_(2)` is closed. The angular frequency of oscillation of the circuit and charge across `C_(1) at t = 125 (mu)s` are respectively

A. `25 xx 10^(3) rad s^(-1), 32 (mu)C`
B. `5 xx 10^(4) rad s^(-1), 16 (mu)C`
C. `5 xx 10^(4) rad s^(-1), -16 (mu)C`
D. `25 xx 10^(3) rad s^(-1), -32 (mu)C`

Answer 1

Correct Answer - D
Angular frequency of oscillation
`omega =(1)/(sqrt(LC)) = (1)/(sqrt(0.2 xx10^(-3)xx8xx10^(-6)))`
`=2.5 xx 10^(4)rad//s=25 xx 10^(3)rad//s`
`(Q)/(C_1)-L(dI)/(dt)=0 implies (Q)/(C_1)+L(d^(2)Q)/(dt^(2))=0`
The solution of this equation is `Q=(Q_0) sin (omega t+phi)`
where at `t=0, Q=Q_(0)`
`Q_(0)=C_(eq)V=(C_(1)C_(2))/(C_(1)+C_(2))V`
`=(8xx2xx10^(-12)xx20)/(10xx10^(-6))=32 xx10^(-6)C`
`phi=(pi)/(2)`
`Q=(32 mu C) cos (2.5 xx 10^(4) xx 125 xx 10^(-6))~=-32(mu)C`.