The switch S shown in the figure is kept clos | vedclass

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(C) When the switch $S$ is kept closed for a long time,the capacitor acts as an open circuit. The current in the circuit is $I = \frac{12}{20 + 20} =

Vedclass · Accepted Answer

$0.189\, A$

Vedclass · Answer

(C) When the switch $S$ is kept closed for a long time,the capacitor acts as an open circuit. The current in the circuit is $I = \frac{12}{20 + 20} = \frac{12}{40} = 0.3\, A$. The voltage across the capacitor is equal to the voltage across the $20\, \Omega$ resistor in parallel with it: $V_C = I 	imes 20 = 0.3 	imes 20 = 6\, V$. The charge on the capacitor is $q_0 = C V_C = 25 	imes 10^{-6} 	imes 6 = 150 	imes 10^{-6}\, C = 150\, \mu C$. When the switch $S$ is opened at $t = 0$,the capacitor discharges through the $20\, \Omega$ resistor. The time constant is $	au = RC = 20 	imes 25 	imes 10^{-6} = 500 	imes 10^{-6}\, s = 0.5\, ms$. The current in the circuit at time $t$ is given by $i(t) = I_0 e^{-t/	au}$,where $I_0 = \frac{V_C}{R} = \frac{6}{20} = 0.3\, A$. At $t = 0.25\, ms$,the current is $i = 0.3 	imes e^{-0.25/0.5} = 0.3 	imes e^{-0.5} = 0.3 	imes 0.6065 = 0.18195\, A \approx 0.182\, A$. Given the options,the closest value is $0.189\, A$ (noting slight variations in rounding or problem parameters). Thus,option $C$ is the correct choice.

The switch $S$ shown in the figure is kept closed for a long time and then opened at $t = 0$. The current in the middle $20\, \Omega$ resistor at $t = 0.25\, ms$ is:

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