Transformer Efficiency Under Load
A step-down transformer has 1200 turns on the primary and 60 turns on the secondary. It is connected to a 240 V RMS AC mains supply and delivers power to a 12 Ω resistive load. Measurements show the primary current is 0.85 A RMS while the secondary current is 15 A RMS. A technician claims the transformer is ideal because the turns ratio matches the voltage ratio, and that any power discrepancy is measurement error. The core is warm to the touch after several minutes of operation.
1. Calculate the ideal secondary voltage and the actual power delivered to the load, then determine the input power and efficiency.
Read the worked answer
Ideal secondary voltage: V_s = V_p × (N_s/N_p) = 240 × (60/1200) = 12 V.
Power in load: P_out = I_s² R = 15² × 12 = 2700 W. (Check: V_s ≈ I_s R = 180 V, which is inconsistent with 12 V - so the load cannot be 12 Ω at 15 A with a 12 V secondary. Resolve by using measured values: P_out = V_s I_s = 12 × 15 = 180 W.)
Input power: P_in = V_p I_p = 240 × 0.85 = 204 W.
Efficiency: η = 180/204 ≈ 88%.
The 24 W difference appears as heat in the core and windings, consistent with the warm core. The transformer is not ideal.
2. Explain why the technician's 'turns ratio matches voltage ratio therefore ideal' reasoning is invalid, and identify what the warm core indicates.
Read the worked answer
The turns ratio fixes the voltage ratio only in the ideal case; it says nothing about power transfer. An ideal transformer would have P_in = P_out exactly. Here P_in = 204 W and P_out = 180 W, so about 24 W is lost - roughly 12% - which is not measurement error at this scale.
The warm core indicates energy dissipated as heat, from eddy currents and hysteresis in the core plus resistive heating (I²R) in the windings. These are real loss mechanisms absent from the ideal model. Efficiency, not the voltage ratio, is the test of ideality. The technician confused a necessary feature of an ideal transformer (correct turns ratio) with a sufficient one.