Answer to Question #171917 in Electric Circuits for Ayush

Question

A three phase 400 V, 50 Hz supply is

connected to a delta connected load

which has impedance of (15 + j12) S2 in

each phase. Find : (1) line current (2)

phase current (3) active power (4)

apparent power

Answer 1

1) Phase current:

$I_P=U/|Z|$

$|Z|=\sqrt{15^2+12^2}=19.21\ \Omega$

$I_P=400/19.21=20.82\ A$

2) Line current:

$I_L=I_P\sqrt{3}=20.82\sqrt{3}=36.07\ A$

3) Apparent power:

$P=3UI_P=3\cdot400\cdot20.82=24.984\ kW$

4) Active power:

$P_A=Pcos\varphi$

$cos\varphi=R/|Z|=15/19.21=0.7808$

$P_A=24.984\cdot0.7808=19.509\ kW$