A two-winding transformer feeds a single-phase half-wave controlled rectifier with a purely resistive load. If the transformer secondary rms current is Is and the load rms current is Irms, what is their relationship?

Introduction / Context:
In single-phase half-wave rectifiers with resistive loads, the secondary winding current and the load current flow through the same path when the diode or thyristor conducts. Understanding whether their rms values match helps with transformer current rating and heating calculations.

Given Data / Assumptions:

• Half-wave controlled rectifier (one controlled device), resistive load.
• Same current flows through the secondary and the load during conduction intervals.
• During non-conduction, current is zero in both secondary and load.
• We consider rms over a full cycle.

Concept / Approach:

For a resistive load, instantaneous current i(t) = v_sec(t)/R during conduction, and 0 otherwise. The transformer secondary current and the load current waveforms are identical in time because the secondary is in series with the load during conduction. Therefore, their rms values over a cycle are the same, regardless of firing angle that merely shifts the conduction start time and duration.

Step-by-Step Reasoning:

Verification / Alternative check:

Compute rms by integrating i^2(t) over one period for either waveform; since the waveforms coincide sample by sample, the rms values must be equal for any firing angle.

Why Other Options Are Wrong:

• 0.5 * Irms or conditional equality at 30°: not supported by waveform identity.
• √2 factor: relates sinusoidal peak to rms, not secondary versus load rms equality here.

Common Pitfalls:

• Confusing average current equality with rms; here both are equal because the waveforms are identical.
• Assuming the transformer sees a different duty than the load; it does not in this half-wave, resistive case.

Final Answer:

Is = Irms irrespective of the firing angle