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The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.
Question:
Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then
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The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.
Question:
Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then
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Verified Answer
The correct answer is:
after time interval $2 \tau$, charge on the capacitor is $C V /\left(1-e^{-2}\right)$
after time interval $2 \tau$, charge on the capacitor is $C V /\left(1-e^{-2}\right)$
Charge on capacitor at time $t$ is $q=q_0\left(1-e^{-t / \tau}\right)$
$\begin{array}{ll}\text { Here, } & q_0=C V \text { and } t=2 \tau \\ \therefore & q=C V\left(1-e^{-2 \tau / \tau}\right)=C V\left(1-e^{-2}\right)\end{array}$
$\begin{array}{ll}\text { Here, } & q_0=C V \text { and } t=2 \tau \\ \therefore & q=C V\left(1-e^{-2 \tau / \tau}\right)=C V\left(1-e^{-2}\right)\end{array}$
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