If you look at loop 2, we can sum the voltages up

around loop 2 and we can use Kirchhoff's voltage law to do that.

So starting at the lower left hand corner of this loop or

mesh we first encountered the 12 volts source and in fact the minus or

negative polarity of the 12 volts source.

We continue and we encounter the 2-kilohm resistor.

We know the voltage drop across the 2-kilohm resistor is going to

be 2 kiloohms times I2, which is flowing the same direction as we're

summing the voltages, minus I1, which is flowing the opposite direction.

So our second voltage drop as the 2-kilohm resistor I sub 2 minus I sub 1.

Then at the top as we go through the 4 kilo ohm resistor we have another

voltage drop which is simply 4 K I 2.

Because I2 is the only loop where mesh current flowing through it.