Discharging Capacitor Voltage Calculation

The question asks for the voltage across a capacitor while it is discharging through a resistor, specifically at a time equal to its time constant.

We are given the following values:

• Capacitance, \(C = 100 \, \mu F = 100 \times 10^{-6} \, F\)
• Resistance, \(R = 1 \, K\Omega = 1 \times 10^3 \, \Omega\)
• Initial voltage across the capacitor, \(V_0 = 50 \, V\)
• The time at which we need to find the voltage is \(t = \tau\), where \(\tau\) is the time constant.

The formula for the voltage across a capacitor during discharging is given by:

\(V(t) = V_0 e^{-t/RC}\)

First, let's calculate the time constant (\(\tau\)) of the RC circuit. The time constant is given by the product of the resistance and the capacitance:

\(\tau = RC\)

Substituting the given values of \(R\) and \(C\):

\(\tau = (1 \times 10^3 \, \Omega) \times (100 \times 10^{-6} \, F)\)

\(\tau = 1 \times 10^3 \times 100 \times 10^{-6} \, s\)

\(\tau = 100 \times 10^{(3-6)} \, s\)

\(\tau = 100 \times 10^{-3} \, s\)

\(\tau = 0.1 \, s\)

Now, we need to find the voltage \(V(t)\) at time \(t = \tau\). We substitute \(t = \tau = RC\) into the voltage formula:

\(V(\tau) = V_0 e^{-\tau/RC}\)

Since \(\tau = RC\), the exponent becomes \(-\tau/RC = -RC/RC = -1\).

\(V(\tau) = V_0 e^{-1}\)

We know that \(V_0 = 50 \, V\) and the value of \(e^{-1}\) is approximately \(0.36788\).

\(V(\tau) = 50 \, V \times e^{-1}\)

\(V(\tau) \approx 50 \, V \times 0.36788\)

\(V(\tau) \approx 18.394 \, V\)

Comparing this calculated value with the given options:

• Option 1: 50 V (Initial voltage)
• Option 2: 12.50 V
• Option 3: 18.5 V (Closest to 18.394 V)
• Option 4: 23 V

The calculated voltage of approximately 18.394 V is closest to 18.5 V among the given options.

Therefore, the voltage across the capacitor at the time of its time constant during discharging is approximately 18.5 V.