GATE 2026 EE Q.47: solution and answer

Official answer: BSource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. The period is $T=20\mu \text{s}$. S1 and S2 conduct for $0<t<10\mu \text{s}$, so the duty ratio is $D=0.5$. There is no phase control, so the firing angle is $\alpha =0^{\circ }$. The load voltage is $+V_{dc}$ in this interval and $-V_{dc}$ in the next 10 µs.

2. From the plot, $i_L$ is a symmetric triangle between $-5\text{A}$ and $+5\text{A}$. It rises linearly from $-5\text{A}$ to $+5\text{A}$ while $v_o=+V_{dc}$, that is during $0<t<10\mu \text{s}$. It falls linearly back to $-5\text{A}$ during the next 10 µs.

3. While S1 and S2 are on, the left-leg midpoint is tied to the positive rail through S1, so the whole load current passes through S1. For $i_L>0$ it flows in the MOSFET channel. For $i_L<0$ it flows in the reverse direction through the channel or the body diode. In both cases $i_{S1}=i_L$ for $0<t<10\mu \text{s}$. For $10<t<20\mu \text{s}$, S1 is off and its leg current flows through S4 or its diode, so $i_{S1}=0$.

4. On the conducting interval, $i_L$ is linear from $-I_p$ to $+I_p$ with $I_p=5\text{A}$. Write $i_L=I_p\left(\frac{2t}{t_1}-1\right)$ with $t_1=10\mu \text{s}$. Then $$\frac{1}{t_1}{\int }_0^{t_1}{i}_L^{2}dt=I_p^2{\int }_0^1(2x-1{)}^{2}dx=\frac{I_p^2}{3}=\frac{25}{3}{\text{A}}^2.$$

5. The rms value over the full period is $$I_{S1,rms}=\sqrt{\frac{1}{T}{\int }_0^{T/2}{i}_L^{2}dt}=\sqrt{D\cdot \frac{I_p^2}{3}}.$$

6. Substituting $D=0.5$ and $I_p=5\text{A}$ gives $$I_{S1,rms}=\sqrt{0.5\times \frac{25}{3}}=\sqrt{4.167}\approx 2.04\text{A}.$$ This is option (B).