In the given circuit the capacitor (C) m...

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L).

At the start, the capicitor was uncharged. when switch `(S_1)` is closed and `(S_2)` is kept open, the time constant of this circuit is `tau`. which of the following is correct

after time interval `tau`, charge on the capacitor is CV/2

after time interval `2tau`, charge on the capacitor is `CV(1–e^(2))`

the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.

after time interval 2`tau`, charge on the capacitor is `CV(1-e^(-1))`

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The correct Answer is:

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch (S_1) . Also when (S_1) is opend and (S_2) is closed the capacitor is connected in series with inductor (L). When the capacitor gets charged compleely, (= (S_1) is opened and (S_2) is closed, Then,

at t=0 energy strored in the circuit is purely in the form of magnetic energy

at any time tgt0, current in the circuit is in the same direction

at tgt0, thre is no exchange of energy between the inductor and capacitor

at any time tgt0, instantaneous current in the circuit may be `Vsqrt((C)/(V))`

`Q=Q_(0)cos((pi)/(2)+(t)/(sqrt(LC)))`

`Q=Q_(0)cos((pi)/(2)-(t)/(sqrt(LC)))`

`Q=-LC(d^(2)Q)/(dt^(2))`

`Q=-(1)/(sqrt(LC))(d^(2)Q)/(dt^(2))`

after time interval `tau,` Charge on the capacitor is CV/2

after time interval `2 tau`, charge on the capacitor is CV`(1 - e^(-2))`

the work done by the voltae source will be half of the heat dissipated when the capacitor is fully charged.

after time interval `2 tau`, charge on the capacitor is CV`(1 - e^(-1))`