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
Right Answer is:
(4.5 + j 30) V
SOLUTION
From the given figure
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