A single-phase AC distribution line supplies two single-phase loads as shown in the figure below. The impedances of line segments A-B and B-C are j0.25 Ω and j0.35 Ω respectively. The voltage drop from A to C is

A single-phase AC distribution line supplies two single-phase loads as shown in the figure below. The impedances of line segments A-B and B-C are j0.25 Ω and j0.35 Ω respectively. The voltage drop from A to C is A single-phase AC distribution line supplies two single-phase loads as shown in the figure below.

Right Answer is:

(4.5 + j 30) V

SOLUTION

From the given figure

A single-phase AC distribution line supplies two single-phase loads as shown in the figure below.

VA = VC − (40 × j0.35) − (40 + 30∠−cos−1(0.8)) × j0.25

IB = 30∠−cos−10.8

= 30∠−36.87° A

IC = 40∠0° A

IAB = IB + IC

VC – A = − (40 × 0.35∠90°) − (40 + 30∠−36.86) × 0.25∠90°

VC – A = 16.6207∠74.29° + 14∠90°

VC – A = 30.34∠81.46°

or

VC – A = (4.5 + j30) V

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