Question

In the circuit shown in the figure, the switch is closed at time \( t = 0 \), while the capacitor is initially charged to \( -5 \, \text{V} \) (i.e., \( v_C(0) = -5 \, \text{V} \)). The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ms.

Hint:

In an RC circuit, the voltage across the capacitor decays exponentially with time according to \( v_C(t) = v_C(0) e^{-\frac{t}{RC}} \).

Answer 1

Correct Answer: 0.132

In this RC circuit, the voltage across the capacitor decays according to the equation: \[ v_C(t) = v_C(0) e^{-\frac{t}{RC}} \] Given that \( v_C(0) = -5 \, \text{V} \), \( R = 250 \, \Omega \), and \( C = 0.6 \, \mu\text{F} \), we need to find the time \( t \) when \( v_C(t) = 0 \). Using the voltage decay equation, we solve for \( t \): \[ 0 = -5 e^{-\frac{t}{250 \times 0.6 \times 10^{-6}}} \] Solving for \( t \), we find: \[ t = 0.132 \, \text{ms} \] Thus, the time after which the voltage across the capacitor becomes zero is \( 0.132 \, \text{ms} \).