GATE 2024 EE Q.63: solution and answer

Official answer: 50Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Full-bridge VSI in $180^{\circ }$ conduction: each diagonal pair conducts for half a period, so each pair has duty ratio $D=0.5$. There is no phase control here (firing angle $\alpha =0^{\circ }$). The output is a square wave $v_o=\pm V_{dc}=\pm 100\text{V}$.

2. Period $T=\frac{1}{f}=\frac{1}{50}=20\text{ms}$, so each half-cycle lasts $\frac{T}{2}=10\text{ms}$.

3. The load is purely inductive, so $v_o=L\frac{di}{dt}$. During each half-cycle $v_o$ is constant, so the current changes linearly and the load current is a triangular wave.

4. The average of $v_o$ over a cycle is zero, so in steady state the average load current is zero and the current is symmetric about zero. It swings between $-I_{pk}$ and $+I_{pk}$.

5. Change in current over one half-cycle: $$\Delta i=\frac{V_{dc}}{L}\cdot \frac{T}{2}=\frac{100}{20\times 10^{-3}}\times 10\times 10^{-3}=50\text{A}$$

6. This change takes the current from $-I_{pk}$ to $+I_{pk}$, so $$I_{pp}=\Delta i=\frac{V_{dc}}{4fL}=\frac{100}{4\times 50\times 20\times 10^{-3}}=50\text{A}$$ with $I_{pk}=25\text{A}$. The peak-to-peak load current is $50\text{A}$.