GATE EE 2020 Set 1 — Question 22

The circuit shown is a single-phase full-wave AC voltage controller (back-to-back thyristors) with an R-L load. The source voltage is $V_s = 200 \text{ V}$ (RMS) at $50 \text{ Hz}$.
The load impedance is given as $10Z - 60^\circ \Omega$. This notation is ambiguous. However, given that the maximum power is $2 \text{ kW}$, we can infer the load characteristics. For an R-L load, the impedance is $Z_L = |Z_L| \angle \phi$, where $\phi$ is the load phase angle.The maximum power delivered to the load occurs when the firing angle $\alpha$ is at its minimum possible value, which for an R-L load is equal to the load phase angle $\phi$. In this case, the controller acts like a short circuit, and the full source voltage is applied to the load.Maximum power $P_{max} = \frac{V_{s,rms}^2}{|Z_L|} \cos\phi$.
Given $P_{max} = 2 \text{ kW} = 2000 \text{ W}$.
$2000 = \frac{200^2}{10} \cos\phi$
$2000 = \frac{40000}{10} \cos\phi$
$2000 = 4000 \cos\phi$
$\cos\phi = \frac{2000}{4000} = 0.5$
$\phi = \cos^{-1}(0.5) = 60^\circ$.Since the load is inductive (as implied by the positive phase angle), the minimum firing angle $\alpha_{min}$ for continuous conduction is equal to the load phase angle $\phi$.
So, $\alpha_{min} = 60^\circ$.To achieve $0 \text{ kW}$ power, the thyristors must be completely blocked, meaning no current flows through the load. This occurs when the firing angle $\alpha$ is $180^\circ$. At $\alpha = 180^\circ$, the thyristors are never triggered within the positive or negative half-cycles, resulting in zero output voltage and thus zero power.
So, $\alpha_{max} = 180^\circ$.Therefore, the minimum range of variation in $\alpha$ to control the load power from $0$ to $2 \text{ kW}$ is $60^\circ$ to $180^\circ$.The final answer is $\boxed{\text{D}}$