The capacitor in the circuit shown below is initially charged. After closing the switch, choose the correct option(s)

Energy of the capacitor becomes half of its initial value in time.

R C \frac{(l n 2)}{2}

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Charge on the capacitor becomes half of its initial value in RC(ln2).

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Current in the circuit becomes half of its initial value in time RC(ln2).

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Current in the circuit becomes half of its initial in time

R C \frac{l n 2}{2}

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Solution

The correct options are
A Energy of the capacitor becomes half of its initial value in time. $R C \frac{(l n 2)}{2}$
B Charge on the capacitor becomes half of its initial value in RC(ln2).
C Current in the circuit becomes half of its initial value in time RC(ln2).
$q = q_{0} e^{\frac{- t}{C R}}$
For (A) part $\frac{q_{0}}{\sqrt{2}} = q_{0} e^{\frac{- t}{C R}}$
Taking log on both sides
$\frac{1}{2} l o g_{e} 2 = \frac{t}{C R}$
$t = R C l o g_{e} 2$
(B) part $\frac{q_{0}}{2} = q_{0} e^{\frac{- t}{C R}}$
$t = R C l o g_{e} 2$
(C) part $\frac{i_{0}}{2} = i_{0} e^{\frac{- t}{C R}}$
$t = R C l o g_{e} 2$
(D) part $t = R C l o g_{e} 2$

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