GATE EE 2026 Set 1 — Question 43

The circuit consists of a $10 \text{ V}$ voltage source, a $1 \text{ u}(t) \text{ mA}$ current source, a $25 \text{ k}\Omega$ resistor, a $10 \text{ }\mu\text{F}$ capacitor, and a $100 \text{ k}\Omega$ resistor.1. Find $V_C(0^+)$:
At $t=0^-$, the current source is $0$ (since $u(t)=0$ for $t<0$). The capacitor acts as an open circuit. The circuit consists of the $10 \text{ V}$ source, $25 \text{ k}\Omega$ resistor, and $100 \text{ k}\Omega$ resistor. The capacitor voltage $V_C(0^-)$ is the voltage across the $100 \text{ k}\Omega$ resistor.
$V_C(0^-) = 10 \text{ V} \times \frac{100 \text{ k}\Omega}{25 \text{ k}\Omega + 100 \text{ k}\Omega} = 10 \times \frac{100}{125} = 10 \times \frac{4}{5} = 8 \text{ V}$.
Since the voltage across a capacitor cannot change instantaneously, $V_C(0^+) = V_C(0^-) = 8 \text{ V}$.2. Find $V_C(\infty)$:
At $t=\infty$, the current source is $1 \text{ mA}$ (since $u(t)=1$ for $t>0$). The capacitor acts as an open circuit (steady state for DC). We can use superposition to find $V_C(\infty)$.

• Due to $10 \text{ V}$ source (current source open):

$V_{C1}(\infty) = 10 \text{ V} \times \frac{100 \text{ k}\Omega}{25 \text{ k}\Omega + 100 \text{ k}\Omega} = 8 \text{ V}$.

• Due to $1 \text{ mA}$ source (voltage source short):

The $25 \text{ k}\Omega$ and $100 \text{ k}\Omega$ resistors are in parallel. The equivalent resistance is:
$R_{eq} = 25 \text{ k}\Omega \parallel 100 \text{ k}\Omega = \frac{25 \times 100}{25+100} = \frac{2500}{125} = 20 \text{ k}\Omega$.
The voltage across this parallel combination is $V_{C2}(\infty) = 1 \text{ mA} \times 20 \text{ k}\Omega = (1 \times 10^{-3} \text{ A}) \times (20 \times 10^3 \Omega) = 20 \text{ V}$.

• Total voltage:

$V_C(\infty) = V_{C1}(\infty) + V_{C2}(\infty) = 8 \text{ V} + 20 \text{ V} = 28 \text{ V}$.3. Find $V_C(0.50 \text{ s})$:
For $t \ge 0$, the capacitor voltage follows the transient response: $V_C(t) = V_C(\infty) + (V_C(0^+) - V_C(\infty))e^{-t/\tau}$.
First, calculate the Thevenin resistance $R_{Th}$ seen by the capacitor. Turn off all independent sources (voltage source short, current source open).
$R_{Th} = 25 \text{ k}\Omega \parallel 100 \text{ k}\Omega = 20 \text{ k}\Omega$.
The time constant $\tau = R_{Th} C = (20 \times 10^3 \Omega) \times (10 \times 10^{-6} \text{ F}) = 0.2 \text{ s}$.
Now, substitute the values into the transient response equation:
$V_C(t) = 28 + (8 - 28)e^{-t/0.2} = 28 - 20e^{-5t}$.
At $t=0.50 \text{ s}$:
$V_C(0.5) = 28 - 20e^{-5 \times 0.5} = 28 - 20e^{-2.5}$.
Using $e^{-2.5} \approx 0.082085$,
$V_C(0.5) = 28 - 20 \times 0.082085 = 28 - 1.6417 = 26.3583 \text{ V} \approx 26.36 \text{ V}$.Thus, the voltages are $8.00 \text{ V}$, $26.36 \text{ V}$, and $28.00 \text{ V}$.The final answer is $\boxed{\text{B}}$