Hi,

I'm trying to work out the loop equations of the circuit attached ... I actually have the answer but I am have troubling understanding how it came about.

The second loop equation (the right one) is :

0 = (4 - j2 + j6 +j8 + (2xj4))I2 - (4+j8)I1 -j4I1

The bold is the part which I cannot understand ... Why is the mutually induced voltage term multiplied by 2?

Thanks in advance!!

 

How do you account for the mutual inductance in the j6 ohm coil, in relation to I2? How about in the j8 ohm coil? Which coil does I2 go through?

 

Ok I got it I didn't realise the self induced voltage term has the same polarity as the mutual induced term in the second loop!!