GATE 2024 EE Q.62: solution and answer

Official answer: 0.940 to 0.970Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. The load current is ripple free, so $i_o=I_0$ (constant). In a single-phase half-controlled bridge the freewheeling path makes the source current a quasi-square wave: $+I_0$ for a width $\pi -\alpha$, then $0$ for a width $\alpha$, then $-I_0$ for $\pi -\alpha$, then $0$ for $\alpha$. The wave is symmetric about the centre of each pulse, so it has no cosine term when the origin is moved there. This only shifts the fundamental in phase by $\alpha /2$ and does not change its magnitude.

2. The firing angle is $\alpha =60^{\circ }=\pi /3$. Each pulse therefore has width $\beta =\pi -\alpha =2\pi /3$ (120°).

3. Total rms input current: the square of the current equals $I_0^2$ for a fraction $\frac{2\beta }{2\pi }$ of the cycle. $$I_s=I_0\sqrt{\frac{2\beta }{2\pi }}=I_0\sqrt{\frac{\pi -\alpha }{\pi }}=I_0\sqrt{\frac{2}{3}}=0.8165I_0$$

4. Fundamental component: for a quasi-square wave of height $I_0$ and pulse width $\beta$, the peak of the fundamental is $\frac{4I_0}{\pi }\sin \frac{\beta }{2}$. With $\beta =\pi -\alpha$, $\sin \frac{\beta }{2}=\cos \frac{\alpha }{2}$. $$I_{s1}=\frac{1}{\sqrt{2}}\cdot \frac{4I_0}{\pi }\cos \frac{\alpha }{2}=\frac{2\sqrt{2}}{\pi }{I}_0\cos \frac{\alpha }{2}$$

5. At $\alpha =60^{\circ }$: $\frac{2\sqrt{2}}{\pi }=0.9003$ and $\cos 30^{\circ }=0.8660$, so $I_{s1}=0.9003\times 0.8660I_0=0.7797I_0$.

6. Ratio of the rms fundamental to the total rms input current: $$\frac{I_{s1}}{I_s}=\frac{(2\sqrt{2}/\pi )\cos (\alpha /2)}{\sqrt{(\pi -\alpha )/\pi }}=\frac{0.7797}{0.8165}=0.9549\approx 0.955$$