Question

In the given AC circuit below, compute the active, reactive, apparent power and the power factor of the load. Calculate the value of the capacitor which will be connected in parallel with the load to make the power factor equal to 0.98. Assume load voltage stays constant. 1. vs(t)=141.42 cos(3t+30°) V C=ly R,=2y L=y R2=2y Vs(t) LOAD a) Determine which of the diodes conduct in the circuit. Assume ideal diode model. Show your work for all possible diode states. b) Calculate I and V. D2 3y 10V 15V D1 30V m 2.

Please take y=60 for calculations.

Answer 1

O 10013.° / vs CH = 141.42.609 (37 +30) = 141.42 L300 Vs(t) = 100_309 Visme) y= 60 ; Ri= 2y = 2x60=1200 R₂ = 2y = 2x60 = 1200 IL-4 = 60H c="ly=1607 XL = WL = 3 (60) = 1801 Kes one dalle 52001 Load impedancel Zu) : (*) Jejw) + - R2 jx 1200s.jte = (Jigo) (1204-jas) 3180 - 20 (20 +j180-jo - 109.49|27.4 Ri=120 I 109-491274 -= 0.448516.936 A 120 +109-4927-4 YLE VL = 100 13109.49127-4 L 120+109-49127.4 = 49.105 | 4613399 v Apparent power (5) = (VL) (I)* =Ptja, = (49-105(44.34')(0.4485(-16.94") = 19.553 +j10.135 = 22602 27.4% YA Active power ) - 19.553 N S-22-02 VA Reactive power (Q.) = 10-135 VAR IP = 19.553 w power fuctor Coso = 19.553 = 0.8879 R = 10.135 VAR 12.02 11:f = 0.8879 11 = 100130 100130

19.553 New Power factor/Cosep.) = 0.98 → $ = 11.478° * Active power is not changes with Capacitors compensation. New Reactive power = Q4 Coso, = + 0.98 = 7 V p²+Q7 = 19.553 -19.56** = 15-169 Q = 3.91 VAR 0.98 ) Resultant Reactive power QD-Q-4,- 10-135-397 = 616 VA (Required) ei ole enam tahun ber og = 39157 Xeya waren Crep = woken 3x29115 = 0.85 x103 F Coeq = 0.65 mF ☺ a) Short ciruwt test: y=60 34.60=1604 V Izo n loy + kol at node 1: zov lt 1 154 32x60=1204 IV = 10.909 volte 1, -V Ho - 10.90941° 30.114 A Ig : V-15 10.9012 Diode Di is tum on (because (I,>0)) Diode Iz is fun off (becausen 1, 40) -0.0682 A

1802 I. - 6ou un 304 lov toi 154 Ž 1200 I= 40 30716 180 +120 = 0.1333 A 300 I= 133.33 mA V = 120x1 - 120x 0-1333 = 16V [V=16 v]