But again, we have to account for

the mutual inductance term, minus i1 j omega M.

Again, because of the opposite reference directions of the dots.

If this dot were down at the bottom instead of at the top,

these then would become pluses.

So, using this and some linear algebra, we can actually find and

solve for the values of I1 and I2.

So these are types of equations that we were using,

and then here's an expression of I1.

It's a very lengthy and convoluted equation, but it turns out that

it might sometimes be a little bit easier to work with an equivalent impedance.

So if I want to know the impedance that this source sees in al of this,