Question

ind the Thevenin equivalent circuit for the portion of the circuit external to the component etween point a and b. Also, Find the Zt of the network for Maximum Power transfer and find the value of the power (25 Points). 2-3 + 6

Answer 1

From Voltage division rule v th = (20 angle 0 degree) (-j 3600/2000 - j 3000) = 20 (-3j)/(2 - 3j) v th = 16.64 angle -33.69 while finding z_th (v - source - short circuited I - source - Open circuited) Z_th = (2000// - j 3000) + j 6000 = 1664.1 angle -33.69 + j 6000 = 1384.6 - j923.07 + j6000 = 5262.3 angle 74.7 z_th = (1384.6 + j 5076.9) Ohm \r\n N.w Equivalent Circuit for getting Maximum power transfer to z_L Z_L = Z_s* (or) Z*_ht R_L + j x_L = (1384.6 + j 5076.9)* R_L + j x_L = (1384.6 - j 5076.9) Ohm I_ = 16.64/2(1384.6) 1-33.69 = 6 times 10^-3 Amps P_max [max power transfer] = Vertical-bar T_max Vertical-bar ^2 R = (6 times 10^-3)^2 times (1384.6) P_max = 0.04999 = 50m watts (mill watts)