How it works
In a balanced star (Y) connection, the neutral carries zero current, so the wire is often omitted in industrial loads. Line current equals phase current in star, but in delta the line current is √3 times the phase current — IL = √3 × Iph. Total three-phase power is P = √3 × VL × IL × cosφ; for a balanced 10 kW load at 0.8 pf, the line current at 415 V works out to about 17.4 A. The two-wattmeter method measures total power with only two wattmeters regardless of load balance, and the reading difference gives reactive power Q = √3(W1 − W2).
Key points to remember
Phase sequence RYB is positive (clockwise phasor rotation); RBY is negative. For balanced loads the neutral current is exactly zero — this is a favourite proof question. Star-delta starter reduces starting voltage to VL/√3, cutting starting torque to one-third and starting current to one-third compared to direct-on-line. Power factor from two-wattmeter readings: tanφ = √3(W1 − W2)/(W1 + W2). At unity power factor both wattmeters read equal values; at pf = 0.5 one wattmeter reads zero.
Exam tip
Every Anna University paper has a question on the two-wattmeter method, so practise deriving both wattmeter expressions W1 = VL IL cos(30° + φ) and W2 = VL IL cos(30° − φ) from the phasor diagram.