Each phase contains three loads in parallel: -j100, 100 and 50+50j ohm.b Assume positive phase sequence with Vab = 400 /__0 V Find Van and IaA and total power drawn by load.

I found line voltage as $$Vline=Vphasor∗sqrt3=231volts.$$ But I can't get the correct value of Ian and power. I tried

by putting

Van= 231/__-30°

and calculating

Zphasor= sum of all impedances Zphasor=-j50+150= 158/-18.4 but I cant get correct answer for Ian. Correct answers are: 4.62/-30° and 3200 W.


1 Answer 1


You got the phase voltage right.

The equivalent parallel of the three impedances is:

\$Z_{eq} = \frac{1}{\frac{1}{-100j}+\frac{1}{100}+\frac{1}{50+50j}} = 50 \Omega\$.

The current can be calculated using:

\$I=\frac{V}{Z}=\frac{230.94/-30°}{50}=4.6188/-30° A\$

And total three-phase power (three times single-phase power):

\$S = 3 V_{ph}I^{*} = 3 (230.94/-30°)(4.6188/+30°) = 3200 W\$


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