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Transformer under load

Curricula: GCSE, Sek I/II

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Aim

Compare the currents and powers in the primary and secondary circuits of a transformer and estimate its efficiency.

The bench

An AC voltage source (sine wave, 12 V amplitude, RMS value 8.5 V), 50 Hz, is connected through an ammeter to the primary winding of the transformer (500 turns). The secondary circuit (250 turns) contains an ammeter and a 3.5 V / 1 W lamp. The voltmeters above the transformer show the voltage on the primary and on the secondary winding; all of them are in AC mode. The wattmeter above the ammeter of the primary circuit shows the power consumed by the primary winding.

Procedure

  1. Record the readings of the four instruments: \(U_1\), \(I_1\), \(U_2\), \(I_2\).
  2. Calculate the ratios \(U_2 / U_1\) and \(I_1 / I_2\) and compare them with the turns ratio \(N_2 / N_1\).
  3. Record the power of the primary circuit \(P_1\) from the wattmeter and calculate the power in the secondary circuit \(P_2 = U_2 \cdot I_2\) (the lamp is a resistive load). Find the efficiency \(\eta = P_2 / P_1\). Compare \(P_1\) with the product \(U_1 \cdot I_1\).
  4. Set Secondary turns to 150. How did the brightness of the lamp and the current \(I_1\) change?
Expected result

The readings are about \(U_1\) = 8.38 V, \(I_1\) = 146 mA, \(U_2\) = 3.55 V, \(I_2\) = 286 mA, \(P_1\) = 1.12 W. The current in the primary circuit is smaller than the current in the secondary by about the same factor as the voltage on the primary winding is greater: \(I_1 / I_2\) ≈ 0.51 ≈ \(N_2 / N_1\). The power in the secondary circuit is about 1.01 W and the efficiency about 90%; the losses are heating of the windings. The product \(U_1 \cdot I_1\) ≈ 1.23 V·A is greater than \(P_1\): this is the apparent power, and part of the primary current is the magnetising current, which does not consume energy but only pumps it back and forth. Under load \(U_2 / U_1\) = 0.42, noticeably less than without load: part of the voltage is lost in the resistance of the windings and because of leakage of the magnetic flux. With fewer turns on the secondary winding the lamp glows more dimly, and the current of the primary circuit also decreases.

Questions

  1. Why is the current of the primary winding smaller if the transformer is step-down?
  2. Where does the power consumed by the lamp come from?
  3. Why is the efficiency of a transformer less than 100%?
Answers
  1. The transformer passes energy from the primary circuit to the secondary with small losses, so the power in both circuits is about the same. At equal power the side with the higher voltage carries the smaller current: 146 mA in the primary against 286 mA in the secondary, a ratio of about 0.51, close to \(N_2 / N_1\) = 0.5.
  2. From the AC source. The primary current creates an alternating magnetic flux in the core, and this flux induces the voltage that drives the current through the lamp, so the energy passes to the secondary through the magnetic field. Of the 1.12 W that the primary winding takes from the source, the lamp receives about 1.01 W.
  3. Part of the energy heats the windings. The simulator gives them a resistance of 2.5 Ω (500 turns) and 0.625 Ω (250 turns); \(I^2 \cdot R\) gives about 0.05 W in each, which accounts for the 0.11 W difference between \(P_1\) and \(P_2\). A real transformer also loses energy in its steel core to eddy currents and hysteresis; the simulator's core has no such losses.