For that voltage drop across the source.

We then come around and encounter the four kilo ohm resistor with just I1 flowing

through it so the voltage drop there as we just had talked about is

four kilo ohms times I sub 1.

And then we have one more resistor in this loop, it's the 2 kilo ohm resistor, and

we know the voltage drop across it.

If we're adding up the voltages in the clock-wise fashion,

it will be plus to minus 4 positive current flowing into the two kilo ohm

resistor using the pass of sign convention.

So, in this case,

this voltage drop across two kilo ohm resistor is going to be 2K I1,

which is flowing the same direction as we're summing, minus I2.

So it's going to be 2K, I sub 1 minus I sub 2,

and that's equal to zero.

That's our first equation for loop one or mesh one.

Now, if we go down the mesh two.

And we look at mesh two, we see that if we start at the lower left-hand corner,

we're going around this loop.

We first encounter this two-milliamp source and we are adding voltage drops.

And we don't know what the voltage drop is across this source.

And in fact, we can write it in our equation as maybe a V2 milliamp, but

that would add another unknown to our equations.

And we only have two loops, we'll come up with two independent equations for

those loops.

So if we added this third unknown,

then we wouldn't have enough equations to solve for our unknowns.

So instead of doing that,

we recognize that this loop has a current source which is just I2.