Question

Chapter 8, Problem 8.097 (Circuit Solution) Find V, in the network in the figure below. 11.12 + 0 12% v 320 22 1112 16% A (12/0° A (a) Find the real part of V.. (b) Find the imaginary part of Vo DO (a) (b)

Answer 1

13té en 72 168VÓ Jim Using Superposition the em, bet is find out vo V, with voltage source 1260°. V, nullifying other vollege & Cervent Sources : jis mo to die mm Tiin Yo = 1210_* = 210422 = 11.981-2.6 V 22431 – 22.02 12.60*** Vi = 11.981-26 = 11.96-j0.543 V V with voltage source 160 V and bullifying other voltage and current source: co 160V 221V=V2 N, = -168 x V2 = -15.981-36 22+11 := -1616 * 20126 =1548124 = 15.88|199.4 = -15.96 +30.125 V

Van with current source 26 and nullifying other voltage sources . On applying the nodal, analysis, a At ve At Var 218 +420 + Mua o te thote to : V = V/+YA+Yg = 11.96 - jó.543 + (-15.96 +jo.725) +0 W: = (f1.76 -15.90)+(-jo.54b +36-425) = -4+10.182 Real part of V 24 V CENA Imaginary part of Yo = 1+jo-182 4 1