GATE EC 2025 Set 1 — Question 61

The input voltage is given by $V_I = 10 \sin(100\pi t)$.
The period of the input voltage is $T = \frac{2\pi}{\omega} = \frac{2\pi}{100\pi} = \frac{1}{50} = 0.02$ seconds, which is $20$ ms.From the circuit diagram, the anode of the diode is connected to the 5V DC source, and the cathode is connected to the input voltage $V_I$ through a resistor $R$. For an ideal diode to be forward biased, the voltage at the anode ($V_A$) must be greater than or equal to the voltage at the cathode ($V_K$).So, $V_A \ge V_K$.
Here, $V_A = 5V$.
When the diode is forward biased and conducting, it acts as a short circuit, so $V_K = V_A = 5V$. The current through the resistor $R$ would be $I_D = \frac{V_A - V_I}{R} = \frac{5 - V_I}{R}$.
For the diode to be forward biased and conduct, the current $I_D$ must be positive, i.e., $I_D > 0$.
Therefore, $5 - V_I > 0 \implies V_I < 5V$.We need to find the time duration for which $10 \sin(100\pi t) < 5$.
This simplifies to $\sin(100\pi t) < 0.5$.Let $\theta = 100\pi t$. We need to find the range of $t$ for which $\sin(\theta) < 0.5$ within one period ($0 \le t < 0.02$ s or $0 \le \theta < 2\pi$).In the range $0 \le \theta < 2\pi$, $\sin(\theta) = 0.5$ at $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
So, $\sin(\theta) < 0.5$ for $0 \le \theta < \frac{\pi}{6}$ and $\frac{5\pi}{6} < \theta < 2\pi$.Converting back to time $t$:The total time duration for which the diode is forward biased in one period is:
$\Delta t = \left(\frac{1}{600} - 0\right) + \left(\frac{1}{50} - \frac{5}{600}\right)$ s
$\Delta t = \frac{1}{600} + \left(\frac{12}{600} - \frac{5}{600}\right)$ s
$\Delta t = \frac{1}{600} + \frac{7}{600} = \frac{8}{600} = \frac{1}{75}$ sTo convert this to milliseconds (ms), multiply by $1000$:
$\Delta t_{ms} = \frac{1}{75} \times 1000 = \frac{1000}{75} = \frac{40}{3} \approx 13.333$ ms.Rounding off to two decimal places, the time duration is $13.33$ ms.The final answer is $\boxed{13.33}$