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I am solving this circuit and according to the dot convension,the j6 ohm and j8 ohm have opposite polarities. Applying kvl to both loops,
I got, 10 = j5I1 – j3I2 + j3I2
Or, I1 = -j2 A
But I am stuck in the second loop.
It seems like, I2=0 because the mutual impedance of both the opposite polarity inductors with the j5 ohm inductor is same, j3 ohm. Also no active source is present in second loop. Am I right here with dot convention? Please ensure me.
One way to think of a inductor from the circuit's point of view is that it gives inertia to current. This gives you some intuitive feel for why current builds up slowly as voltage is applied, and why the voltage goes as high as it needs to on a attempt to shut off the current abruptly. This isn't how the physics works, but this can be a very useful mental model for understanding and designing circuits with inductors.
The dot only determines the polarity of the mutual voltage, it won't affect other directions in the calculation. In figure (b), it shows the equivalent circuit take into the mutual voltage. Because the direction of \$I_{1}\$ in loop 1 is flow into coil 1, and the direction of \$I_2\$ is defined to flow out from coil 2. So the mutual voltage's polarity in loop 2 is positive at the dot side.
Best Answer
You can technically write KVL for the second loop and find that only \$I_2=0\$ satisfies it. I suppose the purpose of this task is to train you in application of the dot rule (which is simple: mutual inductance e.m.f. has the same sign as self inductance e.m.f. when both the loop current and the coupled current are flowing in or out of dotted ends; the signs are opposite otherwise); you have to apply it 4 times for your second loop.
You did it well with the first loop.
Physically you have a transformer with two secondary windings, connected in opposite directions. Voltage from the one winding cancels voltage from the other, so the second loop has no current.