In the single-phase rectifier shown (figure), the average load current is 15 A. Determine the root-mean-square (rms) current in the transformer secondary winding.

Introduction / Context:
This problem tests understanding of current waveforms in transformer-fed rectifiers. For a typical single-phase full-wave rectifier using a center-tapped secondary, each half of the secondary conducts for half the cycle. Hence, rms current in a half-winding is not the same as the average direct current delivered to the load.

Given Data / Assumptions:

• Average (dc) load current Idc = 15 A.
• Full-wave rectification with a center-tapped secondary (each half-winding conducts for 50% of the time at the magnitude equal to the load current during its conduction interval).
• Ideal diodes/thyristors, negligible ripple for current shape reasoning.

Concept / Approach:

For a center-tapped full-wave rectifier, current in each half of the secondary is a rectangular pulse of amplitude approximately equal to load current during its half-cycle conduction window (duty = 0.5). The rms of a rectangular waveform over a full period is amplitude * √(duty).

Step-by-Step Solution:

Verification / Alternative check:

If you integrate i^2(t) over one full period with i(t) = 15 A for half the time and 0 otherwise, Irms = √( (15^2 * 0.5) ) = 10.61 A, confirming the shortcut formula.

Why Other Options Are Wrong:

• 15 A: equals average current, not rms per half-winding.
• 14.14 A: equals 15/√(0.5) misconception; wrong application.
• 7.5 A: half the average, not rms.
• 12.0 A: arbitrary and not tied to duty-cycle math.

Common Pitfalls:

• Confusing load dc current with rms transformer current.
• Forgetting each half-winding conducts only half the cycle.

Final Answer:

10.61 A