The two systems are shown in Fig. Let R be the resistance per conductor.
P1 = 2V I1 cos φ; Cu loss, W1 = 2I12 R
Percentage Cu loss
= (W1/P1) × 100
= (2I2 R/2VI1 cos φ) × 100
= I1R × 100/V cos φ
P2 = 3VI2 cos φ
W2 = 3I2 2 R
% line loss = (W2/P2) × 100
= (3I22 R/3VI2 cos φ) × 100
= I2R × 100/V cos φ
Since percentage line losses are the same in both cases
∴ I1R × 100/V cos φ = I2R × 100/V cos φ
∴ I1 = I2
Now, P1 = 2VI1 cos φ = 20 MW ;
∴ VI1 cos φ = 10 MW
P2 = 3VI2 cos φ = 3VI1 cos φ = 3 × 10 = 30 MW