InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» `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)` |
|