In the circuit shows in Fig the capacitor is initially uncharged and the two - way switch is connected in the position `BC`. Find the current through the resistence `R` as a function of time `t`. After time `t = 4` ms, the switch is connected in the position `AC`. Find the frequency of oscillation of the capacitor of the circuit in the position, and the maximum charge on the capacitor `C`. At what time will the energy stored in the capacitor be one-half of the total energy stored in the circuit? It is given `L = 2 xx 10^(-4)H, C = 5 mF, R = (In 2)/(10) Omega` and emf of the battery `= 1 V`.

Answer

`(10)/(In 2) (1 - ((1)/(2))^((t)/(2xx 10^(-3)))) ; 10^(3) rad s^(-1) ; (7.5 xx 10^(-3))/(In 2) C`;
`4 ms + (n pi + (-1)^(n) pi//4)/(omega)`

Explanation

Current in `R` or `L` at any time is given by:
`I = (varepsilon)/(R )(1 - e^(-Rt//L)) rArr I = (1)/((In 2)//10) [1 - e^(-(t In 2)/(10xx2xx10^(-4)))]`
`rArr I = (10)/(In2) [1-e^(In 2^-((t)/(2xx10^(-3))))] rArr I = (10)/(In 2) [1-((1)/(2))^ ((t)/(2xx10^(-3)))]`
Now at `t = 4 ms`
`I = (10)/(In2)[1-((1)/(2))^((4xx10^(-3))/(2xx10^(-3)))] = (10)/(In 2)[ 1- (1)/(4)] = (15)/(2 In 2) A`
Now the switch is shifted between `A` and `C`. Just after shifting the switch, current in inductor will be as calculated above. It can't change at once. Now due to this current, the capacitor will start charging. Frequency of oscillation is given by :
`omega = (1)/(sqrt(LC)) = (1)/(sqrt(2 xx 10^(-4) xx 5 xx 10^(-3)))= 10^(3) rad//s`
To find maximum charge : `(Q^(2))/(2C) = (1)/(2) LI^(2)`
`rArr Q = (sqrt(LC))I = (10^(-3) xx 15)/(2 In 2) rArr Q = (7.5 xx 10^(-3))/(In 2) C`
Energy in the capacitor is half of total energy when charge on
capacitor is `q = (Q)/(sqrt(2))`
Charge on capacitor at any time
`q = Q sin omegat [ :' at t = 0, q = 0]`
`rArr (Q)/(sqrt(2)) = Q sin omega t rArr omegat = npi + (-1)^(n) (pi)/(4)`
(1) `rArr t = (n pi + (-1)^(n) pi//4)/(omega)`
Total time, `T = 4 ms + (n pi + (-1)^(n) pi//4)/(omega)`