Question

In a balanced three-phase wye–delta system, the source has an abc phase sequence and Van = 120∠40° V rms. The line and load impedances are 0.5 + j0.4 Ω and 24 + j18 Ω, respectively. Find the delta currents in the load.

Answer 1

Answer:

6.51∠-26.9° Amp - rms

Steps:

1. This three-phase circuit will be equivalent to three single phase circuits in Y connection and we'll calculate current for ONE phase by dividing the load impedance by 3.
2. By applying Ohm's law to the circuit, current will be calculated.
3. Then the equivalent Delta current will be calculated by dividing it to sqrt(3)  and subtracting 30°

Explanations:

Source = 120∠40 V

Z-line=0.5+j0.4Ω

Z-load=24+j18Ω

Converting Delta to Star

Z-load-1Ф=Z-load-3Ф/3

$=\frac{24+j18}{3}\\=8+j3$

Calculating Star Current by Ohm's Law:

$\frac{V}{Z-line+Z-load-1}\\ =\frac{120∠40}{(8+j6)+(0.5+j0.4)}$

=11.27 A-rms

Converting Star current to Delta Current

$I-delta=\frac{I-star}{\sqrt{3} } \leq -30$

I-delta = 6.51∠-26.9°