Peak current in an LC discharge through a thyristor path A capacitor of capacitance C is initially charged to voltage V and then discharged through a series path consisting of a thyristor, a diode, and an inductor L. What is the peak current attained in this lossless resonant discharge?

Introduction / Context:
LC commutation and resonant discharge are common in thyristor circuits. The peak current determines device stress and component ratings.

Given Data / Assumptions:

• Capacitor C charged to V.
• Series inductance L, negligible resistance.
• Ideal devices; energy exchange between L and C is lossless.

Concept / Approach:
In an ideal series LC, energy initially stored in the capacitor (E = 0.5 * C * V^2) converts entirely into inductor energy at peak current (E = 0.5 * L * I_peak^2). Equating energies yields the peak current.

Step-by-Step Solution:
Initial energy: E_C0 = 0.5 * C * V^2.At current peak, capacitor voltage is momentarily zero; inductor energy: E_L = 0.5 * L * I_peak^2.Set energies equal: 0.5 * C * V^2 = 0.5 * L * I_peak^2.Solve: I_peak = V * sqrt(C / L).

Verification / Alternative check:
Angular frequency ω₀ = 1 / sqrt(LC). Current waveform i(t) = I_peak sin(ω₀ t); peak occurs at quarter cycle, consistent with energy transfer.

Why Other Options Are Wrong:
V * sqrt(L/C), V * (C/L), V/L: Dimensionally incorrect or not energy-consistent.
V / sqrt(L*C): Has units of current but corresponds to V * ω₀; peak current must include sqrt(C/L) factor.

Common Pitfalls:
Ignoring losses does not change the ideal peak expression; losses only reduce the actual peak slightly.

Final Answer:
V * sqrt(C/L)