SATHEE: Electromagnetic Induction and Alternating Current 4 Question 12

Electromagnetic Induction and Alternating Current 4 Question 12

12. Initially, the capacitor was uncharged. Now, switch $S_{1}$ is closed and $S_{2}$ is kept open. If time constant of this circuit is $τ$ , then

$(2006, 6$ M)

(a) after time interval $τ$ charge on the capacitor is $C V / 2$

(b) after time interval $2 τ$ charge on the capacitor is $C V (1 - e^{- 2})$

(d) after time interval $2 τ$ , charge on the capacitor is $C V (1 - e^{- 1})$

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Answer:

Correct Answer: 12. (b)

Solution:

Charge on capacitor at time $t$ is

$\begin{aligned} q & = q_{0} (1 - e^{- t / τ}) \\ Here, q_{0} & = C V and t = 2 τ \\ ∴ & q & = C V (1 - e^{- 2 τ / τ}) = C V (1 - e^{- 2}) \end{aligned}$

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Electromagnetic Induction and Alternating Current 4 Question 12

12. Initially, the capacitor was uncharged. Now, switch S1 is closed and S2 is kept open. If time constant of this circuit is τ, then

Answer:

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12. Initially, the capacitor was uncharged. Now, switch $S_{1}$ is closed and $S_{2}$ is kept open. If time constant of this circuit is $τ$ , then