A capacitor C = 100 , mu F is connected to th | vedclass

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(D) When the switch $S$ is opened,the battery and the resistor in series with it are removed from the circuit. The capacitor $C$ is left in a loop wit

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$50$

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(D) When the switch $S$ is opened,the battery and the resistor in series with it are removed from the circuit. The capacitor $C$ is left in a loop with two resistors of $1 \, k\Omega$ each,which are connected in parallel to each other.The equivalent resistance $R_{eq}$ of these two resistors in parallel is given by:$R_{eq} = \frac{1 \, k\Omega 	imes 1 \, k\Omega}{1 \, k\Omega + 1 \, k\Omega} = 0.5 \, k\Omega = 500 \, \Omega$.The time constant $	au$ for the discharging circuit is given by:$	au = R_{eq} C$$	au = (500 \, \Omega) 	imes (100 	imes 10^{-6} \, F)$$	au = 50000 	imes 10^{-6} \, s = 0.05 \, s$Converting the time constant to milliseconds $(ms)$:$	au = 0.05 	imes 1000 \, ms = 50 \, ms$.

$A$ capacitor $C = 100 \, \mu F$ is connected to three resistors,each of resistance $1 \, k\Omega$,and a battery of $emf$ $9 \, V$. The switch $S$ has been closed for a long time to charge the capacitor. When switch $S$ is opened,the capacitor discharges with a time constant of ...... $ms$.

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