Question

(t) = A, cos(500+B) Find the complex power S, = +by received by the source Find the complex power S, 4+by received by the resistor Re- Find the complex power S, = s + byl received by the resistors Find the complex power = 14 + he received by the inductor w R Given Variables: A1: 6V B1:45 degrees R1 : 4 ohm R2: 4 ohm L1 : 8 mH Determine the following: al (W) b1 (VAR) a2 (W): b2 (VAR) a3 (W): b3 (VAR):
please also find a4(W) and b4(VAR) (got cut off in the pic), thanks!

Answer 1

VS - Al cos ( 500t+ B ) VS By 6 (45° 14 us= 6Cos( 500t+U5) = re B4 L=8mn Wh = 600 x 8 x 10 = un Input Impedance zin= 4+ (ca/lui uxui tui 6+21 Sin G27 = 6.3245 218.43 SE VS 6 cos 500 t + u5) Pin 6129 IS = 0-9487COS (5000+ 26.57) A IS = 0.9487/12 (26.57 complex poleer Received by Source Sf Vs Is' L26 +57) 9- 245°) ( 99484 ra S, - (2.8641 (18-43*) Si c-R 712 +0.905 i) = pt jQ aarthi bia-0.905 VAR Power delivered by source = 20712tjagad 9,--2.712 W V2 corrent Througth Resister RIPING Is = 0.9487 <26.57 voltage acrop Resper R, EISR, = 3.7948226:57 VE complex pocter Received By Rehan R, =SR, E VR, (JR)* 52 = SR,= (3.7948 226-57) (09487 (2697)* S2- b2 = OVAR A2 = 1.8w of SRI = 1.80 to i Q2 + baj leHH GREIGHT UP

Current througth Denme Тc x 14 60.448 9 (26.53 ) o som vi 12 SR 5047444133' 0.094)_9157 A complex Pocer Received By Resitter RL = мо, «(то)" Sa = 10:4742676.57 )(1.8798 091 557)* Sg 20.899 + 10 = az A ļ b3 bga o VAR a=0.899 w Incluar L I= ISX 4 utui Corrent Througth 9487 V 22657 ) CF4Lau5') IL Fou742 (-18:43' a voltage across Irleder = (GR) (IL) VL = 1.8968 (7642 complex power Received by Inductor L = Su = VL (ILP su =(1.896827142) (0.44422-18-43) su = 0 + 0.908 =autiby auso WAH by a 0.90G VAR bu= 0905 VAR au= Power balance SIE SL A Syt sy 2.712+ io905 - Si S2 153 +54 = (1.8+50) (0.899+10 )+0+10.905) satisfied -2.712t j 0.905 - SI rocurer delivered by source = Power A osvobou by axbferent passive Elements