Page 379 - Basic Electrical Engineering
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Voltage drop across the parallel branch
V AC = I Z 2
2
= 10 ∠0° × 10.63 ∠ − 48.8°
= 106.3∠ − 48.8°V
= (70.02 − j79.98)V
Current in inductive branch
Circuit current,
I = I + I = 10 + j0 − j2.13 − j13.44 = 7.87 − j13.44
1
2
= 15.57 ∠ −59.56° A
Voltage drop across series branch,
V CB = IZ 3
= 15.57 ∠ − 59.65° × 12.8 ∠ 51.34°
= 199.4 ∠ − 8.31° V
Voltage applied across terminals AB,
V AB = V AC + V CB
= (267.33 − j 108.8) V
= 288.62 ∠22.15°V
