... [(200+1200i) ,-3 60i ;-3 60i,(150+581.1i)] Z= 1.0e+003 * 0.2000 + 1.2000i - 0.3600i - 0.3600i 0.1500 + 0.5811i >> V = [800; (-9 5.2 6-5 5i)] V= 1.0e+002 * 8.0000 -0 .9526 - 0.5500i >> I = inv(Z)*V I= 0.1390 - 0.7242i... connected in series-aiding fashion have a total inductance of 250 mH When connected in a series-opposing configuration, the coils have a total inductance of 150 mH If the inductance of one coil (L1)... Manual, you are using it without permission Chapter 13, Solution 16 To find IN, we short-circuit a-b 8Ω jΩ -j2 Ω a • • j4 Ω + 80∠0 V - j6 Ω I2 IN I1 o b − 80 + (8 − j + j 4) I − jI = ⎯ ⎯→ j