Difficulty: Medium
Correct Answer: 90%
Explanation:
Introduction / Context:
Transformer efficiency quantifies how effectively input power is converted to useful output power. Real transformers exhibit copper losses (I^2R), core losses (hysteresis and eddy currents), and stray losses. A quick efficiency check validates design sizing and reveals abnormal loading or faults.
Given Data / Assumptions:
Concept / Approach:
Compute input power P_in = V_p * I_p (power factor assumed near unity for simplicity; many small transformers under resistive loading are close). Compute output power P_out = V_s^2 / R_L. Efficiency η = P_out / P_in expressed as a percentage.
Step-by-Step Solution:
Verification / Alternative check:
Secondary current I_s = V_s / R_L = 18 / 10 = 1.8 A. If the turns ratio were measured, P_out should still be near 32.4 W; any large discrepancy from 36 W input indicates realistic losses yielding efficiency below 100%.
Why Other Options Are Wrong:
Common Pitfalls:
Multiplying V and I on the secondary without considering the load, or forgetting to square V when using P = V^2 / R; also, mixing rms and peak values leads to errors.
Final Answer:
90%
Discussion & Comments