Single-phase full-wave AC regulator – relation between device RMS current and load RMS current In a single-phase full-wave AC regulator using two antiparallel thyristors and a resistive load, if the RMS current in each thyristor (measured over a full cycle) is Ir, what is the RMS load current?

Introduction / Context:
Device current ratings are specified in RMS terms averaged over the full mains cycle. In an antiparallel thyristor regulator with a resistive load, each thyristor conducts during alternate portions of the cycle. Relating the device RMS current to the overall load RMS current helps translate device ratings into load capability.

Given Data / Assumptions:

• Two identical thyristors in antiparallel, resistive load.
• Each thyristor conducts in disjoint intervals (complementary half-cycles or phase-controlled segments).
• Ir is the full-cycle RMS current of each thyristor.

Concept / Approach:

Because the conduction intervals of the two devices do not overlap in time for a resistive load, the mean-square load current over a cycle is the sum of the mean squares of the two device currents. Thus, I_load_rms^2 = Ir^2 + Ir^2 = 2 Ir^2 → I_load_rms = √2 * Ir.

Step-by-Step Solution:

Verification / Alternative check:

Special case at full conduction (no phase delay) also satisfies the same relationship, confirming the factor √2 holds for symmetrical sharing.

Why Other Options Are Wrong:

• Ir: ignores contribution from the other device.
• 2 * Ir: double-counts RMS magnitudes linearly; RMS components add in power (square) sense.
• Ir/√2: would be correct in the opposite inference (device RMS from load RMS).

Common Pitfalls:

Adding RMS values arithmetically instead of in quadrature; forgetting that the two device conduction intervals do not overlap for a resistive load.

Final Answer:

√2 * Ir