And that's equal to 0.

So that's our second independent equation, and

upon inspection it has two unknowns as well, I2 and I3.

Now our third equation for the right-hand side of the circuit, so

the right-hand lower loop has three voltage drops.

It has first of all starting at the lower left-hand corner,

voltage drop across the 3 kilometer resistor, then the 6, and

then the 12 volt on the right-hand side of the circuit.

So starting with the 3K voltage drop,

we see that we are in the negative

polarity of that voltage drop,

so it's -3K (I2- I1).

Now I'm going up to the 6 kiloohm resistor in the negative

polarity first of that voltage drop, it was 6K (I1- I3).

That was that voltage drop.

We see that from equation 1.

And then we have voltage drop associated with the 12K resistor.

And it's going to be 12K times I sub 3,

and that's equal to 0.

So we have these three equations and three unknowns, so

we would solve these equations for I1, I2, and I3.

And then we could find any other value we wanted for

voltages or currents in the circuit, as well as power.