Question

2. Find i, in the circuit of figure 2 using superposition theorem and Verify the same using Multisim Simulation. (20 Marks) 25 F 100 cos 2000 V 3702 3 2002 50mH 4 sin 4000+ A Figure 2

Answer 1

2 Gruen cheeunt . 25 MF 70-12 17 100 cos Zoost @ Y 2001 somu som de Asenteret G orge » :. Bч бич, №stam 74. Cause when zou is Acheing, Skant van los zout aust open & Edr. tout a 200 20.. .10 2.09 29V . 350 35

case ② when 100 cas 2ooot acting is R short zou seen 4 open 4 an toont 25 prill 7 ܐܘܐ | FOR 2002 100 coscoooh2 som = Dj(2000) (509103) % = whether E- 370. (2009) 25255) 100_0 $200 |8=-2017 3702 & 280 100L00) con 100

.. Joy = -12 = 0.2 2125 Iz = tasz .55-7 2. 0 = 1 '001 = (04- (90%+ e*)s )21 aal =?I ot - (lost ot] ²1 s onog u 'I smptsang Es soll att a zase 'I of - (opstol 220 on tg 092 + (5-93) © com 0- 001 = 75 t - (!* +84)4 to ..700 a 27 ot-(!2017 102-)'s os't, leves (2-1)0£*'s ( 102-) +07001- 4- dog

whu 4 sun gooot & Actua cament into sos farm - A Son (nat +4) :. 4 sin troot 4 cos (400ot - 90°) = A cos (unt+4-90) 1 4 Li a 42-90 cas farm sin form when 4 sin (4000 t) A is acting Short on 100 cos 2000t, short 200 -101 W = 49O. 2002 Sen a = jwL =j (4000)(507:02 DO By Loop Analyse me home loop 6 4 = 4 L-90 leep 9 -10; 12 + 70 (12 -Is) +200;(12-1) =o. 1+2=2007] Y = June C4000 x 25 410) * = -10;

-200, Ir + tz( 70+2007 – 10;) - 70 Izco. 5 - 200,'*+ Izl to +1 907) - 70 I3 -0. Subst I, in above equatu - -2003 ( 41-90) + Jal 76+1707)-70 Iz to. Iz( 707190;)-70 13 = 800.-@. Loup ② 80 ( 13 -1)+ 78( I3 - Iz) 4 2001, no. , 13( 80+70 + 200) – 80.- 70 Iz zo. S 350 I3 - 80 -70 Iz zo. 9 - 7012 +350 Iz = 80 (4-6-90) > -7012+350 53= - 320 i Iz = -3201 - 350 I3 - 70. Iz = $i+513 Subst Iz in equation ④

(за: +5 г.) (+ 11) - +1 = доо. - зго і - bово + 350 Ist 4s» і т- 1 13 = 1on. ( 350 + 1 Фі -то) = 2во — згі + bов I- 715 2- 84.4 3. Tь, 2-г, = -o-166 + (- + 1 = 1:3+s Lts: 51 Ip3 - 1: tts L95 •5+ . В Xop post/km 2мм = +, + + + + Тоз = + p•2L25 + 1: Fis L25-57 1. = 1+Lt- в.] .:. Costne fanm to - 1-884 cos( 200t +96-43