Transformer efficiency calculation: A 120 V rms primary draws 300 mA rms. The secondary supplies 18 V across a 10 Ω load. Compute the efficiency η = P_out / P_in.

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:

  • Primary voltage V_p = 120 V rms.
  • Primary current I_p = 0.300 A rms.
  • Secondary voltage V_s = 18 V rms across R_L = 10 Ω.
  • Assume sinusoidal steady state and that meter readings are true rms.


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:

P_in = 120 * 0.300 = 36.0 W.P_out = V_s^2 / R_L = 18^2 / 10 = 324 / 10 = 32.4 W.η = P_out / P_in = 32.4 / 36.0 = 0.9 = 90%.


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:

  • 88% / 92% / 95%: Close but inconsistent with the direct calculation; only 90% matches P_out and P_in.


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%

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