Ideal transformer power transfer (self-contained repair): Assume the transformer discussed earlier delivers 100 W of input power at its primary winding. If the transformer is considered 100% efficient (no copper loss, no core loss, and no leakage), what is the power available at the secondary winding?

Introduction / Context:
An ideal transformer is a theoretical model used in basic electrical engineering. It assumes perfect magnetic coupling, zero winding resistance, and no core losses. Under these assumptions, all input power at the primary is transferred to the secondary. This question applies that ideal-energy-conservation concept directly to determine the secondary power from a known primary power.

Given Data / Assumptions:

• Primary input power = 100 W (explicitly stated to repair the incomplete cross-reference).
• Efficiency = 100% (ideal transformer: no losses).
• No need for turns ratio, voltages, or currents since we compare total power.

Concept / Approach:
For an ideal transformer, power_in = power_out. While the voltage and current values change according to the turns ratio (V scales with turns ratio, I scales inversely), their product (P = V * I) remains the same when losses are zero. Therefore, secondary power equals primary power.

Step-by-Step Solution:
Define efficiency: η = P_out / P_in.Set η = 1.0 (100%).Compute P_out = η * P_in = 1.0 * 100 W.Therefore, P_out = 100 W at the secondary.

Verification / Alternative check:
If a turns ratio changed voltage by factor a and current by 1/a, then P_out = (a * V_in) * (I_in / a) = V_in * I_in = P_in, confirming conservation in the ideal case.

Why Other Options Are Wrong:
200 W, 1000 W: imply power gain, impossible without an external energy source.20 W: implies losses or regulation to only 20%—contradicts the 100% efficiency premise.None of the above: incorrect because 100 W is valid.

Common Pitfalls:
Confusing voltage/current transformation with power amplification; forgetting that real transformers are less than 100% efficient (typically 95–99+%), but the ideal model uses 100% for teaching and quick estimation.

Final Answer:
100 W