Question

802 jXL -j502 Given: The phasor circuit shown above. Required: Calculate the value of inductive reactance, XL, such that the complex impedance, Zab, at terminal pair, a-b, is real; i.e., the imaginary part of Zab is zero. Also, calculate the value of Zab at that value of X Solution: Zab =

Answer 1

Combination is Zu jx, is in parallel with (80-150) , jxz C80-550) Sjx2 +80-350 or and Zx are in Series, a Combination is Zas= 6 + 2x Zab Zab=b+ 5x (80-550) ot jx 780-550 tanaht Lab - b + j 80X2+sox 10 80 +(x+50) divide with 80-3(*2-0) as shown Multiply and Zas = 6+ 180XL +50x180-3(x2-10) 7 80+}(41-50) ( 80-3(x2-50) ) 2 - ( 36400 + 8ox(x-)+ 4000x - Đx (x-7) 802+ (x2-5053

Imaginary Part of Zab must equal zero 6400 Xl - 50x2(x2-50) 802 + (x2-50)2 =o 6400 x - 50x² +2500x zo 8900 XV = 50 x? X = 8900 so = 1784 [X,= 178r Substitute X in equation we set - 980(178) +50 (178) Zab= 6+ 80 + (1+8-) TZab= 117.25e