Question

(a) Calculate the sinusoidal steady state expression of the current io(t) using nodal analysis as shown in Figure 1(a). [10 Marks) C R L -0000 10mH 200 10(1) 50 4F + vs(t) + 10 cos 1000+ V R22022 410(1) R3=300 Figure 1(a) (b) Calculate the sinusoidal steady state expression of the current is(t) using superposition principle as shown in Figure 1(b). [15 Marks) с ist) v.1) 1012 R: 1012 0.1 F 5 cos 2t VR L 0000 4H 2 sin 4t A R3 512 Vs (4V Figure 1(b)

Answer 1

@ Angular frequency w = 1000 rad/s Capacitive reactance Xc= 100 x 50 x106 Xc=con Inductive reactance * = WL = 1000 x 10x103 ^ x = lor Frequency domain circuit: -izor -1- cor V sor mm 1) >zor 3or ve Yolór + + = 0 20 20 Applying kel at node we get V-1010 G- - 20 Ootions ) v -(0.05) 7,-0.5 -> 0 Applying kel at node 2 G-5 Ve +470 = we get + 520 30+ilo To = ; Substituting to in above equation we get 20

Ja -0. Iz - + 4 4 ( 130 - 320 30+'o (0.2 – 0. b) +(0.03 + 0. 04 ), so > 0 Solving equations 0 and @ we set V 1.544(-24.150 V TE 20 1.54442401 10 = A 0 20 % = 78.69 (-24.15° mA Time domain expression is (t)= 68.69 Cos Clooot -24.15°) mA

Short Vs. 6 Consider valt) and Open ibct) Source Vact)= 5cosat Angulan frequency w=2 rad/s Capacitive reactance . La Xe= 2xool Xc=54 Inductive reactance X = col=2x4 m X=84 -552 HI Gostov slov 회 loop jor I2 100p(2) Applying kul in loopy we set (-15+38 ) + 10 I-I) =50 (10+13 ) I' -10I, = 50 Applying kul in loop 2 we get

103, +5 I,' +10 (1, -57) -0. --107 + 25 = 0 -> Solving equations ) and (@ I)' = 0.298 |–2 boy8º A we get is استلم (0.248.2260480) - 0.298/153.52A is' (t)= 0.298 cos (2t +153.52) A Consider i(t), Short Vact) and vs Tb (t)=2sinut Angulon frequency 4 radls. Capacitive reactance Xe= COC uxo.1 Xc = 2.54 Inductive reactance X = W2=4 xur X = lor

52.5m دسته sor IL W son loop mam. slor IM O2LA 100p 2 Lia Appling kul in loop we get (-12.5) I" +10 ( Ī" - T," ) +336 ( 559 – 210 )=0 (10+ 113.5)5,5 - 10 932 3 Applying kul în loop @ we get 105 + 10 (E-I)+5 (5"-210) =0, -10 1"+25 I₂" = 0 -> Solving equations and y we get I," = 1.262 211.58 A is" =-1" -- (1-2626.1.5*° )A Ts" - 1.262 1–168.43' A

is" Ct)1.262 Sin Cut - 168.439) A Consider Vs Short Valt) and open ibct). For de, inductor is shorted and Capacitor is open Circuited lor sor Wy Jr 4 711 13" A => is = 0,16A lotiots lis" (t) = 0.16A From Superposition Principle is = (+) + (+)+1" (4) Is = 0.298 Cos (2t +153.52°) + 1.262. Sin (4t-168.43) + 0.16 A