GATE 2021 EC – Question 48
In the circuit shown in the figure, the switch is closed at time $t=0$, while the capacitor is initially charged to $-5$ V (i.e., $v_c(0)=-5$ V).
The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ________ ms.
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Correct answer: 0.13 to 0.15
Explanation
Let $v$ be the voltage at the capacitor node. The series $250\ \Omega$ carries $(5-v)/250$ and $V_R=5-v$. KCL at the node gives $\frac{5-v}{250}=\frac{5-v}{500}+\frac{v}{250}+i_C$, so $i_C=\frac{5-3v}{500}$. Then $C\frac{dv}{dt}=\frac{5-3v}{500}$, which has time constant $\tau=\frac{500C}{3}=0.1$ ms and final value $\frac53$ V. So $v(t)=\frac53-\frac{20}{3}e^{-t/\tau}$. Setting $v=0$ gives $e^{-t/\tau}=\frac14$, $t=\tau\ln4=0.1386$ ms $\approx0.139$ ms.