Question

8.22 In the phasor-domain shown in Fig. P8.22, V = 120/0° V, I = 0.3/30° A, a = 1000 rad/s, R1 = 200 92, R2 = 200 S2, R3 = 1.2 ks2, L= 0.2 H, and C = 10 uF. Determine the complex power for each passive element, and verify that conservation of energy is satisfied. in hinde Figure P8.22: Circuit for Problem 8.22.

Answer 1

Frequency w=1000 rad/s Capacitive reactance. Xc = Xe=loor c = 1000 x 10x10 Inductive reactance X, = col= 1000 x 0.2 Xq=2005 2oon 2002 j 2004 www. Esioon In V, §1.2km 90.3 235 A 120Lovy түтігу м 3 Joop D se+ 2007. – 3100 (I-Iz ) = 120 10 (200-5100) I, +(0100 )= 1200 Appiting kul in loop @ we set (200 +1200) [ ] J -5100 (I7-11).; 71200112+ 0.3 L30 J.-o. (1100) 7 +(1400+1003), =-3.1.77-1805 Solving equations ① and 2 we set. Is = 0.483 237.86A Ina 0.255 1-146.37"A

Complex Power of voltage Source is Sw=10206")* Sus- ļ (12010) (0.183_-37.86°) Sv=-22.88 7.17.786; VA Complex Power of his Seitli51]'R, +30 Se 3.0.48334200 +50 = 23.329 +50 VA Complex Power of Capacitor is Sc=0-31 | I - Iz ²x100 Jy -= 0.483 (37.86°– 0.255 L-146.34° I-I = 0.38 136.4°A 1.5, -52 1 = 0.738 :: Se-0-1-1 (0.738)? xlou . Sc= 0-5 27.232 VA Complex Power of R₂ is Sezta (1 [2] J R + 50 Serat (0.255.}*200 tjo Sp2= 6.502 +50-YA

200 Complexe pain of induction is Syzja (111]x200 +0 S = otj Ž (0.255)²x200 XA S = 0+ 36.503 VA Complex power of R3 is : SR3 = 1 / 2 1 I2 + 0.3 (30) ²x 1200 tio . Iq +0.3 (36 = 0.255 %-1463+* +0.3130

152 +0.3133 ) = 0.0483 A Spg=(0.0483)*x 1200 +jo Spa = 1.4 tsova Voltage across Current Source is v = 1200 ( I +0.3 130°) Vi = 1200 (0.2552346.37° +0.3230) V; = 57. qu8 210.47°r. Complex Power of Current source S;= -(5;) (0.31300)* S; :-}(57.9u8 10.47') *0.3 1-30 +2.9063 VA = -8.192 Sum of Complex Powers is Esa Sat Sen + Se + spz + Se + S&3 +S; Es= -22.88 +17.7863 + 23.329+jo +0-3 270232 +6.502tjo tot ; b. 503 tloutjo - 8.19272.9063 Hence conservation of energy is Sarified