Question

Question 1 22 - 90° A 220 V 1290° V j32 322 -j52 Find the SAE's for the assigned loop currents. Solve for the loop currents. Round answer to 2 decimal places, add O's as needed. Express magnitude as positive number and angle between 180° i = A 2 .A OA Question 2 -152 HE 20 23 2290° A 220V 3412 Find the SAE's for the assigned nodal voltages. Solve for the nodal voltages. Round answer to 2 decimal places, add O's as needed. Express magnitude as positive number and angle between 180° OV 02 = L V Us V

Answer 1

Page e 13 24-909A. dolna 2lov TO 1 Loov jer & 12 SAR JOT isr for Loop (nots in Loop 2, Ia cunant files I = 22-90° AT in hoof 2112 current in Loop 3, Iz current flow). for Loop@ -26°+2.12-13) +j312=0 (2+j3)12-2133 2000 for Loop 1/90 +12/1D+2/13 12) =0 13(2-js) - 2I2-1/90°° blom Cx280) (241 (6 +60/12 - 4I3 = 4 - (4+I2 +(13-01 - 3 - 2 (15-49)Z = 7-21 Iz = 0.468957-1-013 73.5 0-47/-1.01] A E from 2+2 7+1.07 2-94/-0-323. 12-.818-56.63 A . 0.472-1.01. (2 +13) Iz =

um Qov sun 2 2980 at node Nox 220° v.) at node 1 VL-V3 V-2 2/90°=0. -js 2V-2V3 +63(2-2) = 20 V7(2-js) -2 V3 = 20 – 10j o at nade @ V3 - Ve + V3 - 2 + 30. 12;(V3-W1) + (-20)(3-2) + 15 V3 = 0 ♡ (15-8j) – 12jV1 = -403 @ from 0,2; & (@x (2-js). 1604 941)V - 24V3 = 120 +24019 V3 - - 200 - dog f10-1153)3 = -80+160j | V3 = 155/-148-46

-T 148-U X (18) + Too = 12 V1 Vt = 3-88/34-39E