Question

Determine the complete power information of a series circuit composed of R=17.32 Ω and having pf=0.866 lag, for an applied voltage v(t)=100*sqrt(2)*cos(1000t + 30)

Answer 1

ceries RL - cuit * you know, Generally anductor can absorb the Logging Reactive power series RL- circuit alts as logging Recitive power circuit. It can absorb the Reactive power from the SOURCO. in indulter current lays the vertege, a 90° * Appavent power s= VI* s=ptio power factor= cos lagging o = Tant (WL) - tanto Larver factor angle PA Active power series.Re-circuit Á + Reactive power Generally capacitor can deliver the lagging Reartive power. In capacitor cunnet leads the voltage by an angle 96 Capautor in a source of Recutive power power farter = cm leading Ø = tant (WER) Vi ☺ Ls parver factor angle

In this problem Here that power factor =colo = 0.866 lagging the power factor is lagging, it means. the circuit is series RL-circuit Se=17032.01. XL W2 VI = 100 12 607(10004 +30°) V; W = 1000Rad/sec Mat) - 10062 130°-V power fouter = .009% =0.866 las Ø2005! (0.866) Ø = 30° RL- circuit power factor angle 8=300 0 = tan! (WL For tam (30°) = wl. = 17.3 2 = con

1000xL 17.32 17.32 x 103 13 [2 = 10mt through direit Cevyen vel. I = 1002 130 RAIN2 Rtite I - 1000 L30 17.32.7110 I = 7.071200 I =7.071 000(1000 €) Lo make mum value of current 76al Apparent power delivered by source S = voltage x(current) S = 1 VI

S = -x look (30 87.07120)* s = f x 10050 130 X 7.07120 s = 433.01 + j2u9.99.7 S = 433.01 +1250 s = ptig L Realtive power delivered by source Getive power delivered by searce # Active power delivered by source for delivered by Resistor R P = 433.01 . & Reartive power absribed by the delivered by source (ay) inductor. 250 var. (OR) Alive power delivered by SOUVE P = Ilms R

you know already Resistor can deliver the artive power * R.m-s value of current for cosine tunition IRMS = I maximum - = (7071 8 x 17:32 Tp = 433 w. * Reartive power absorbed by inductor. o = I XL = 17.07%)** 10 = 249.999 6 = 250 var