Question

Question 1 30 pts 220 V 212 -j522 2290° A 3412 > j3 Ω 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 = 2 А iz = А iz = A

Answer 1

For loop 220v KVL 210-35(1,-13) +2(9,-1) = 0 15 HH (2-35) i, -212 +53 iz--210 j3.0 mm 2296 422 KVL for loop - 2 Flis & Here iz = 2290 aj KUL for loop-3 uliz-iz) -;5(iz-i,) +j3i z=0 j5i, - uia +(4-22) iz=0 → Substitute - Eqn 2 = 2jt in equation and ② (2-j5) i, -ujo + 5j iz = -2 (2-55) i, + 5jiz = -2+uj ig - (0.6 + o uj) + (0928 +30.069), & Eqn@ j5i, - 4125) + (4-12) iz = 0 Substitute is value in above equation 35, -85 +(u-j2) [60-8+0.43) +(0.828+30.069)) - (3.448 + 3.621j) i + 14-8i) = 0 i = 1.79(10.16 A Add is = (68+043) + (0-828 +30.069) (179/70-16) (: from Egn 3 = 2.18 L57. 16 A we already have iz = 2L904)