Answer to Question #161334 in Electric Circuits for Param Dhingana

Question

Three impedances each of (3-j4) ohm are connected its delta to a 230V, 3-phase, 50 Hz balanced supply. Calculate the live and phase currents in delta connected load and the power delivered to the load.

Answer 1

Given, 230 V(line), 3-phase, 50 Hz balanced supply

$V_{RY}=230\angle0ºV_{YB} = 230 \angle-120ºV_{BR}=230 \angle-240º\; V \\ Z_P = Z_{RY}=Z_{YB}=Z_{BR}=(3-j4)Ω = 5 \angle -53.13º \;Ω$

As this is delta connected load, the phase voltages are equal to the corresponding line voltages.

The phase currents are given by:

$I_{RY} = \frac{V_{RY}}{Z_{RY}} = \frac{230 \angle0}{5 \angle -53.13} = 46 \angle53.13º \;A \\ I_{YB}=46 \angle -66.87º \;A \\ I_{BR}=46 \angle -186.87º \;A$

Vph = 230 V

Iph = 46 A

The line currents are 1.732 times the magnitude of phase current, lag behind the reference phase current by 30 degrees and are given by:

$I_R = \sqrt{3} \times 46 \angle (53.13º -30º)A = 79.67 \angle23.13º \\ I_Y = \sqrt{3} \times 46 \angle (-66.87º -30º)A = 79.67 \angle -96.87ºA \\ I_B = \sqrt{3} \times 46 \angle (-186.87º -30º)A = 79.67 \angle -216.87ºA \\ \phi_p = \angle (Phase \; voltage - Phase \; current) = 53.13º$

The total active power and reactive power delivered to the delta connected load:

$P = 3 \times V_{ph} \times I_{ph} \times cos \phi_p = 3 \times 230 \times 46 \times cos(53.13º) = 19.044 \;kW \\ Q = 3 \times V_{ph} \times I_{ph} \times sin \phi_p = 3 \times 230 \times 46 \times sin(53.13º) = 25.39 \;kVAR$