Question

The equivalent ciscuite c eal tsans tomer isgiven belpw. A iopedances avenchms 2-25: i J 3003 4 2チ子 Lo acwith no Lod connected nd the b with a oad 10onnected cgain ind the out put vottraya C. Determine the banc4 revoltage x4ulation (ff the condi 6ons are consid ered tul Ld

Answer 1

VID load: I_2 = I_1 = 0 I^1_1 = I_0 z_eq = 400(3300)/400 + 3300 = 1200,00 angle 90 degree/500 angle 36.87 degree rightarrow 240 angle 53.13 degree rightarrow 144 + 9[194] z_in = z_eq + (2 + J2) = 146 + J(194) rightarrow 242.8 angle 53.04 degree I^1_1 = I_0 = 227 angle 0 degree/242.8 angle 53.04 degree = 1.14 angle - 53 .04 degree E_1 = z_eq Vertical-bar I_0 Vertical-bar = Vertical-bar (2400) angle 53.13 degree Vertical-barVertical-bar 1.14 angle - 53.04 degree Vertical-bar rightarrow 273.8 angle 0.09 degree almostequalto 273.8 angle 0 degree E_2 = N_2/N_1 E_1 - 1/2.25 (273.8 angle 0 degree) = 121.7 angle angle 0 degree \r\n load = 10 Ohm z_0 = 10 + (0.2) + J(0.5) rightarrow 10.2 + J(0.5) = 10.212 angle 2.80 degree I_2 = E_2/z_0 = 121.7 < 0 degree/10.212 angle 2.80 degree = 11.92 angle - 2.80 degree q_0 = I_2 R_L Vertical-bar 10 Vertical-barVertical-bar 11.92 Vertical-bar angle - 2.80 degree rightarrow 119.2 angle - 2.80 degree q. R = Vertical-bar E_2 Vertical-bar - Vertical-bar q_0 Vertical-bar / Vertical-bar q_0 Vertical-bar = 121.7 - 119 Ohm/119.2 times 100 %