GATE 2023 EE Q.55: solution and answer

Official answer: 7.1 to 7.5Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Before firing, T is OFF and no steady direct current flows through the series branch $C$-$L$ or the $4\Omega$ resistor. The capacitor is therefore charged to the source voltage: $v_C(0)=100\text{V}$ (upper plate positive) and $i_L(0)=0$.

2. At $t=0$ the thyristor is fired. Taking it as an ideal switch, it connects the top wire to the lower node. The source then drives a constant current through $T$ and the $4\Omega$ resistor: $$I_R=\frac{100\text{V}}{4\Omega }=25\text{A}$$

3. The capacitor discharges through the loop $C\to T\to L\to C$. Its positive plate feeds current through $T$ in the forward direction (anode to cathode). The loop is series resonant: $${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{4\times 10^{-6}\times 1\times 10^{-6}}}=5\times 10^5\text{rad/s}$$

4. The characteristic impedance is $Z_0=\sqrt{L/C}=\sqrt{4}=2\Omega$. With $v_C(0)=100\text{V}$ and $i_L(0)=0$, the resonant current is $$i_{LC}(t)=\frac{V_C(0)}{Z_0}\sin ({\omega }_{0}t)=50\sin ({\omega }_{0}t)\text{A}$$

5. The thyristor current is the sum of the two currents: $$i_T(t)=25+50\sin ({\omega }_{0}t)\text{A}$$

6. With latching and holding currents zero, $T$ turns off when $i_T$ first returns to zero. This gives $\sin ({\omega }_{0}t)=-\frac{1}{2}$. The first such instant after $t=0$ is ${\omega }_{0}t=\frac{7\pi }{6}$, because $\sin$ is positive until $\pi$ and $i_T$ stays above zero until then.

7. $$t=\frac{7\pi /6}{5\times 10^5}=\frac{3.6652}{5\times 10^5}\text{s}=7.33\mu \text{s}$$