Transformer Turns, Regulation and Primary Current
A student builds a step-down transformer for a 12 V, 2.0 A lamp. The primary has 1100 turns and is connected to a 228 V rms, 50 Hz supply. The secondary is wound with 55 turns. On open circuit the secondary reads 11.4 V rms. When the lamp is connected and drawing 2.0 A, the secondary terminal voltage drops to 10.3 V. The primary current under this load is 0.13 A. The student claims the transformer is ideal because the turns ratio predicts about 11.4 V. The examiner asks for the turns needed for a nominal 12 V secondary at 228 V, the percentage regulation at full load, and whether the primary current is consistent with an ideal transformer. Assume the lamp is the only load and the supply frequency is constant.
1. Calculate the number of secondary turns needed for a nominal 12 V secondary at 228 V. Why is the existing 55-turn winding not sufficient?
Read the worked answer
For an ideal transformer, N_s / N_p = V_s / V_p.
Thus N_s = N_p × V_s / V_p = 1100 × 12 / 228 = 57.9 turns.
A practical winding should use about 58 turns. The existing 55 turns gives V_s = 228 × 55 / 1100 = 11.4 V, which matches the open-circuit measurement but is below the 12 V nominal target. The discrepancy is caused by the turns choice, not by an ideal turns-ratio failure.
2. Find the percentage regulation at full load and identify likely causes of the drop.
Read the worked answer
Percentage regulation = (V_no-load - V_full-load) / V_no-load × 100.= (11.4 - 10.3) / 11.4 × 100 = 9.65% ≈ 9.6%.
A significant drop means the transformer is not ideal. Likely causes are resistance of the copper windings, leakage flux that does not link both windings, and core losses. The open-circuit voltage alone cannot prove ideal behaviour because it is measured at nearly zero current.
3. Check whether the measured primary current is consistent with an ideal transformer. Show the comparison.
Read the worked answer
Using the actual secondary load, the output power is approximately P_s = 10.3 × 2.0 = 20.6 W. An ideal transformer would draw about 20.6 W from the primary, givingI_p,ideal = 20.6 / 228 = 0.090 A.
If the nominal 12 V output is used instead, the ideal primary current is about 12 × 2.0 / 228 = 0.105 A. The measured 0.13 A is larger than both estimates, so the transformer is not ideal; the extra primary current supplies magnetising current and losses.