GATE 2020 EE Q.11: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. Very large $L$: the dc-side current is essentially constant, $I_o$, with negligible ripple, because the inductor will not let the load current change appreciably within a cycle.

2. The bridge is built from uncontrolled diodes, so the firing angle is $\alpha =0^{\circ }$ (no commutation overlap is assumed, since the source inductance is not given). The source current is $i_s=+I_o$ while D1, D2 conduct (positive half-cycle of $v_s$) and $i_s=-I_o$ while D3, D4 conduct (negative half-cycle). So $i_s$ is a square wave of amplitude $I_o$ at the supply frequency, in phase with $v_s$.

3. A square wave with half-wave symmetry, $i_s(t+T/2)=-i_s(t)$, has zero average value and no even harmonics. Only odd harmonics are present.

4. Fourier series: $$i_s(t)=\frac{4I_o}{\pi }\left[\sin \omega t+\frac{1}{3}\sin 3\omega t+\frac{1}{5}\sin 5\omega t+\cdots \right],\omega =2\pi \times 50\text{rad/s}$$ so the peak of the $n$th harmonic is $I_n=\frac{4I_o}{n\pi }$ for odd $n$.

5. The largest component is the fundamental, $n=1$, at $50\text{Hz}$. The next largest is $n=3$, at $3\times 50=150\text{Hz}$, with amplitude $\frac{1}{3}$ of the fundamental. The $n=5$ component (250 Hz) is only $\frac{1}{5}$ of the fundamental, which is smaller.

6. 0 Hz is absent because the source current is alternating with zero average, and 100 Hz is an even harmonic and is absent. Hence the two most dominant components are $50\text{Hz}$ and $150\text{Hz}$: option (C).