GATE 2024 EE Q.28: solution and answer

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

1. Load reactance at $f=50\text{Hz}$: $X_L=2\pi fL=2\pi \times 50\times 18.37\times 10^{-3}\approx 5.77\Omega$.

2. Load impedance angle: $\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)={\tan }^{-1}\left(\frac{5.77}{10}\right)={\tan }^{-1}(0.577)$.

3. Since $\tan 30^{\circ }=1/\sqrt{3}\approx 0.577$, the load angle is $\varphi =30^{\circ }$.

4. For a single-phase AC voltage controller (triac) with an RL load, the current from one half cycle is still flowing when the next gate pulse arrives if the firing angle is $\alpha \le \varphi$. The triac then conducts continuously, the output equals the full supply voltage, and it cannot be varied.

5. Control of the output voltage exists only for $\varphi <\alpha \le 180^{\circ }$, so the minimum triggering angle is ${\alpha }_{\min }=\varphi =30^{\circ }$, which is option (B).