Question

The circuit shown in the figure has reached steady state with thyristor \(T\) in OFF condition. Assume that the latching and holding currents of the thyristor are zero. The thyristor is turned ON at \(t=0 \, \text{sec}\). The duration in microseconds for which the thyristor would conduct, before it turns off, is ................. (Round off to 2 decimal places). \begin{center} \includegraphics[width=0.45\textwidth]{27.jpeg} \end{center}

Hint:

In resonant commutation, conduction period of thyristor equals half the LC resonant cycle, \(t = \pi \sqrt{LC}\).

Answer 1

Step 1: Identify circuit type.
The circuit is a series RLC discharge circuit (resonant commutation). Once thyristor is ON, capacitor discharges through \(L\) and \(R\).

Step 2: Resonant frequency.
\[ f = \frac{1}{2\pi \sqrt{LC}} \] Given: \(L = 4 \, \mu H = 4 \times 10^{-6}\,H, C = 1 \, \mu F = 1 \times 10^{-6}\,F\). \[ LC = 4 \times 10^{-12}, \sqrt{LC} = 2 \times 10^{-6} \] \[ f = \frac{1}{2\pi \times 2 \times 10^{-6}} \approx 79.6 \, kHz \] \[ \omega = 2\pi f \approx 5 \times 10^5 \, rad/s \]

Step 3: Conduction period.
Thyristor conducts for half a resonant cycle: \[ t_{cond} = \frac{\pi}{\omega} = \frac{\pi}{5 \times 10^5} \approx 6.28 \times 10^{-6} \, s \]

Final Answer:
\[ \boxed{6.28 \, \mu s} \]