Question

A single-phase line has an impedance of 5∠ 60° and supplies a load of 120A, 3,300V at 0.8 p.f. lagging. Calculate the sending-end voltage and draw a vector diagram.

Answer 1

The vector diagram is similar to that shown in Fig. Here.

ER = 3,300 ∠ 0°, Z = 5 ∠ 60°

Since φR = cos−1 (0.8) = 36°52′,

∴ I = 120 ∠ −36°52′

Voltage drop = IZ = 120 × 5 ∠ 60 − 36°52′

= 600 ∠ 23°8′ = 600 (0.9196 + j0.3928) = 551.8 + j 235.7

∴ ES = (3,300 + j 0) + (551.8 + j235.7) = 3,851.8 + j235.7

ES =√( 3851.82 + 235.72)= 3,860 V