Initially, the capacitor is charged to a potential of `5 V` and then connected to position 1 with the shown polarity for `1 s`. After `1s` it is connected across the inductor at position 2

(a) Find the potential across the capacitor after `1 s` of its connection to position 1.
(b) Find the maximum current flowing in the `L- C` circuit when capacitor is connected across the inductor. Also, find the frequency of `LC` oscillations.

Answer

A

Explanation

In the capacitor,
`q_i=CV_i=(10xx10^-3)(5)=0.05C`
`q_f=CV_f=(10xx10^-3)(10)=0.1C`
`:.` Charge in capacitor will increase from `0.05C` to `0.1C` exponentialy.
Time constant for this increase would be
`tau_L=CR=1s`
`:.` Charge at time `t` will be
`q=0.05+(0.1-0.05)(1-e^(-t//tau_C))`
`=0.1-0.05e^(-t//tau_C)`
b. At `t=1s, q=0.1-0.05e^-1=0.0816C`
`V=q/C=0.0816/(10xx10^-3)=8.16V`
b. This charge `0.0816C` is also the maximum charge `q_0` of `L-C` oscillations.
From energy conservation equation,
`1/2q_0^2/C=1/2Li_0^2` we have,
`i_0=q_0/sqrtLC=0.0816/sqrt(25xx10^-3xx10xx10^-3)`
`=5.16A`
Further, `omega=1/sqrt(LC)`or `f=omega/(2pi)=1/(2pisqrtLC)`
`=1/(2pisqrt(25xx10^-3xx10xx10^-3))`
`=10Hz`