Answer to Question #153582 in Electrical Engineering for ahmed

Question

A balanced Y load having a 15+j25 Ω load impedance in each leg is connected to a three-phase, four-wire, Y-connected generator having a line voltage of 230 V. The line impedance per phase is 0.5+ j1.5 Ω . Answer the following:

a. Draw the circuit showing the given values.

b. Calculate the line voltages of the load.

c. Calculate the phase currents of the load.

d. Calculate the power consumed by the 3-phase load.

Answer 1

The image attached shows the schematic.

b) Phase voltage of source

$=\frac{230}{\sqrt{3}} =132.79 V.$

Phase voltage on load side$=132.79 \times \frac{15+j25}{(15+j25) +(0.5+j1.5) }=126.1-j1.41=126.1\angle-0.64\degree$

Hence line voltage on load$=126.1\times \sqrt{3}= 218.41V$

C) phase current of the load $=\frac{132.79}{(15+j25) +(0.5+j1.5) } = 4.32\angle-59.68A$

d) power consumed by 3 phase load=$3\times i^2\times Rload=3\times4.32^2\times15=839.81W$