Efficiency and energy losses

A 100%-efficient transformer is not available in practice.

efficiency η = (output power) / (input power) = (V_s I_s)/(V_p I_p) × 100%

An efficiency of 80% means that 20% of the input energy is lost; if the lost energy is 10% of the original energy the efficiency is 90%. For a real transformer, V_s I_s = η V_p I_p, so the current relation becomes I_s/I_p = η V_p / V_s.

Causes of loss and remedies:

1. Heating of the copper wires (I²R) → use thick wires of low resistance.
2. Eddy currents in the core → laminated core of thin insulated sheets of silicon steel (high resistivity).
3. Hysteresis (energy used in repeatedly magnetizing and demagnetizing the core, turned into vibration of the magnetic molecules) → use soft iron, whose molecules are easily magnetized and demagnetized.
4. Leakage of flux (not all the flux links both coils) → use a closed core.

Example 7. A transformer has 400 primary turns and 100 secondary turns; V_p = 220 V, and the secondary current is 4.0 A. Assuming 100% efficiency: V_s = 220 × 100/400 = 55 V; I_p = I_s N_s/N_p = 4 × 100/400 = 1.0 A.

Example 8. A transformer of 90% efficiency is connected to a 120 V supply and delivers 12 V at 3 A. P_out = 36 W; P_in = 36/0.9 = 40 W; I_p = 40/120 = 0.33 A.

Example 8a — step-down, ideal and real. A transformer converts 2200 V to 220 V. The primary has 4000 turns, so N_s = 4000 × 220/2200 = 400 turns. The secondary supplies 10 A, so the output power is 2200 W. If the transformer is ideal, I_p = 2200/2200 = 1 A. If its efficiency is 88%, P_in = 2200/0.88 = 2500 W and I_p = 2500/2200 = 1.14 A.

Example 8b — step-up. A 12 V supply feeds a step-up transformer with 50 primary turns and a secondary of 1000 turns. V_s = 12 × 1000/50 = 240 V. If the secondary current is 0.5 A, the output power is 120 W. For an efficiency of 80% the input power is 150 W and the primary current is 150/12 = 12.5 A.

Example 8c — finding the efficiency. A transformer takes 0.5 A from a 200 V source and delivers 0.9 A at 90 V. P_in = 200 × 0.5 = 100 W; P_out = 90 × 0.9 = 81 W; η = 81/100 = 81%. The lost power is 19 W.