Electrical Machinery and Apparatus
Expect transformer equivalent circuits, motor slip and torque, efficiency, and starting methods. The exam favours numerical questions on losses, regulation, and parallel operation.
Core rules
- Transformer: V1/V2 = N1/N2 = I2/I1. Referred impedance scales by the square of the turns ratio.
- Efficiency is highest when copper loss equals iron loss; at part load x, copper loss = x² × full-load copper loss, while iron loss stays constant.
- Voltage regulation = (V_no-load − V_full-load)/V_full-load × 100%, and can be negative for leading power factor.
- Parallel operation requires the same voltage ratio, polarity, phase sequence, and similar percentage impedance and X/R ratio.
- Induction motor slip s = (n_s − n)/n_s; rotor frequency = s × supply frequency; torque is proportional to V² near rated conditions.
Worked example
A 100 kVA transformer has 1 kW iron loss and 2 kW full-load copper loss. At 75% load and unity power factor, output = 75 kW. Copper loss = 0.75² × 2 = 1.125 kW. Total loss = 1 + 1.125 = 2.125 kW. Efficiency = 75 / (75 + 2.125) = 97.2%. The copper loss falls with the square of load while iron loss remains fixed, which is why part-load efficiency can exceed full-load efficiency.
Traps
Applying full-load copper loss at part load, or confusing kW and kVA in the efficiency denominator, are frequent slips. In slip questions, calculate synchronous speed from frequency and pole number (n_s = 120f/P). For regulation, check the sign convention for leading loads.