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In the circuit shows in Fig the capacito...

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`.

Text Solution

Verified by Experts

The correct Answer is:
`l_t=l_"max"(1-e^(-Rt//L)), 10^3 s^(-1), 3.6xx10^(-3) C` , 4 ms(initial time) + `((2n+1)pi)/(4omega)`
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