GATE 2026 EE Q.61: solution and answer

Official answer: 1.04 to 1.14Source: Official answer key, GATE 2026 EE (IIT Guwahati)

1. The loop is the source $v_s$, the ideal diode $D$, the resistor $R=10\Omega$ and the battery $V_{dc}=50\text{V}$, all in series. The battery's + terminal is on the resistor side, so it opposes the current that the positive half-cycle of $v_s$ drives through the diode.

2. An ideal diode conducts only while $v_s>V_{dc}$. With $\theta =\omega t$, this means $100\sin \theta >50$, so $\sin \theta >\frac{1}{2}$ and $\theta$ runs from $\frac{\pi }{6}$ to $\frac{5\pi }{6}$. The conduction angle is $\frac{2\pi }{3}$.

3. During conduction, $$i=\frac{100\sin \theta -50}{10}=10\sin \theta -5\text{A}$$ Otherwise $i=0$, because the diode blocks.

4. The average over one full cycle ($2\pi$) is $$I_{avg}=\frac{1}{2\pi }{\int }_{\pi /6}^{5\pi /6}(10\sin \theta -5)d\theta$$

5. The first integral is ${\int }_{\pi /6}^{5\pi /6}10\sin \theta d\theta =10\left(\cos \frac{\pi }{6}-\cos \frac{5\pi }{6}\right)=10\sqrt{3}\approx 17.3205$. The second is ${\int }_{\pi /6}^{5\pi /6}5d\theta =5\cdot \frac{2\pi }{3}\approx 10.4720$.

6. Therefore $$I_{avg}=\frac{17.3205-10.4720}{2\pi }=\frac{6.8485}{6.2832}\approx 1.09\text{A}$$