Electromagnetic Induction And Alternating Current Ques 48
- 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
$(2006,6$ M)
(a) after time interval $\tau$ charge on the capacitor is $C V / 2$
(b) after time interval $2 \tau$ charge on the capacitor is $C V\left(1-e^{-2}\right)$
(c) the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged
(d) after time interval $2 \tau$, charge on the capacitor is $C V\left(1-e^{-1}\right)$
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Answer:
Correct Answer: 48.(b)
Solution:
Formula:
Growth Of Current in Series R-L Circuit:
- Charge on capacitor at time $t$ is
$$ \begin{array}{rlrl} & & q & =q _0\left(1-e^{-t / \tau}\right) \\ \text { Here, } \quad q _0 & =C V \text { and } t=2 \tau \\ \therefore \quad & q & =C V\left(1-e^{-2 \tau / \tau}\right)=C V\left(1-e^{-2}\right) \end{array} $$
BETA
