GATE 2018 EE Q.23: solution and answer

Official answer: 16.90 to 17.70Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Read the waveform: the current is $+I_0$ for $30^{\circ }<\omega t<180^{\circ }$, zero from $180^{\circ }$ to $210^{\circ }$, $-I_0$ for $210^{\circ }<\omega t<360^{\circ }$, and zero from $0^{\circ }$ to $30^{\circ }$. The firing angle is $\alpha =30^{\circ }$, so conduction lasts $180^{\circ }-\alpha =150^{\circ }$ in each half-cycle, a duty ratio of $D=\frac{150^{\circ }}{180^{\circ }}=\frac{5}{6}$ of each half-cycle (each pulse is $\frac{150}{360}=\frac{5}{12}$ of the full period). The average value over a full cycle is zero.

2. The wave has half-wave symmetry, $i(\theta +\pi )=-i(\theta )$, so only odd harmonics exist and the fundamental coefficients are $$a_1=\frac{2}{\pi }{\int }_{\alpha }^{\pi }{I}_0\cos \theta d\theta ,b_1=\frac{2}{\pi }{\int }_{\alpha }^{\pi }{I}_0\sin \theta d\theta .$$

3. $$a_1=\frac{2I_0}{\pi }[\sin \theta {]}_{30^{\circ }}^{180^{\circ }}=\frac{2I_0}{\pi }\left(0-\frac{1}{2}\right)=-\frac{I_0}{\pi }=-6.366\text{A}$$
4. $$b_1=\frac{2I_0}{\pi }[-\cos \theta {]}_{30^{\circ }}^{180^{\circ }}=\frac{2I_0}{\pi }\left(1+\cos 30^{\circ }\right)=\frac{40}{\pi }(1.8660)=23.76\text{A}$$

5. Peak of the fundamental: $$I_{1m}=\sqrt{a_1^2+b_1^2}=\sqrt{6.366^2+23.76^2}\approx 24.60\text{A}.$$ Check with the closed form: $a_1^2+b_1^2={\left(\frac{2I_0}{\pi }\right)}^2\left[(1+\cos \alpha {)}^2+{\sin }^2\alpha \right]={\left(\frac{2I_0}{\pi }\right)}^2\cdot 4{\cos }^2\frac{\alpha }{2}$, so $I_{1m}=\frac{4I_0}{\pi }\cos \frac{\alpha }{2}=25.465\cos 15^{\circ }=24.60\text{A}$.

6. RMS value of the fundamental: $$I_1=\frac{I_{1m}}{\sqrt{2}}=\frac{24.60}{1.4142}\approx 17.39\text{A}.$$