Question

120 Loo El For The circuit shown Zlor - a) Find S, Q,P and VEMS ( draws power triangle ३२२m 60Hz - 6) Find power factor of angle, and reactance needed for power factor correction.

Answer 1

| += 60 mg Buon M = 2xf 12010* A = 376.99 rad/s 60Hz mH Inductive reactance, X = w2 = 376.99 X 22 X10-3 . 29 So, total impedance = 8.2912 of the circuit Z= R+ jx2 = (40+; 8.29) =100)*+8:29j* _tan" 3:29) (Complen to poler form, (atib)=820, V= 1a +62 and 0 = tan'la)) z = 10 +j 8.29 = 12.99_39.6622 Current, 'I = Ž = 12.99239.66* = 9.24_-39.66' A Now, apparent power is givers by * s=NIR 1* conjugate of I)

= '12040 (9.244-39.66") ( Use conjugate of 8L0 = 820) : = 120X9.24_39.66 5 = 1108.8 239.66 VA Convert to complex forms using g LQ = rl Coso + i Sino) S = 1108.8l Cos 39.66+j Sin39.66) ..= (853.60 +j 707.67) VA Here, the real part is the (P) active power and imaginary port is the reactive power. Therefore, S=1853.60 +3707.67) VA P = 853.60 W and Q = 707.67 VAT and the power triangle is given by, 170 (S=P+jo) (1108.8 ray 1 (707.67 O is the argument " Var) (angle) of s. 0 = 39.66 (853.cow) (S = r20 - angle)

(6) Power factor is the ratio of active power to the apparent power. ( part of apparent power, 5 used as active power, P) So, power factor = Cos @ = E . (p.f) 1853.60 1208.87 le pof. = Coso = 0.77 pf legg I lage because, Current Calculated in bartla) { 9.24L-39.66} is I logging the voltage) where, o is the pit angle and gif angle, o= Cost ( 8 ) or 0 = 39.66 - see power triangle) . Now, for pot correction and better pf (Coso=1), capacitive To reactance is required to absorb the reactire power to make s=p, as Shown aside. (Coso = 1 ise o=0)

Suppose this Capacitive reactance R is ac So to compensate lj x 21 = 1-jxl se Xc=XL from part (a), XL = L = 8.2912 Therefore, X = 8.292 Which is the required C reactance. ( Capacitive reactance).