In the given network V1 =100∠0°V, V2 =100∠120°V,

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In the given network V₁ =100∠0°V, V₂ =100∠120°V, V₃ = 100∠ +120° V. The phasor current i (in Ampere) is

1. 173.2∠ – 60°
2. 173.2∠ – 120°
3. 100.0∠ – 60°
4. 100.0∠ – 120°

Correct Option: A

From the given network,

-i =	(V₁ - V₂)	+	(V₂ - V₃)
	-j		j

-i =	100 ∠0° - 100 ∠120°	+	100 ∠- 120° - 100 ∠120°
	1 ∠- 90°		1 ∠ 90°

i = 173.2 ∠- 60°