So here's a circuit for our 6 volt source.

And we're looking for the contribution to V out from just the 6 volt source.

So there's a number of different ways we can solve this,

we can start perhaps by using Kirchhoff's Voltage Law and the emission analysis,

if we do that, then we can assign two different ,meshes to this problem.

I1 and I2 will be the lead currents or

mesh currents associated with those meshes.

We can then sum up the voltages around each one of these meshes,

solve for I1 and I2.

Once we have an I2, we can multiply it by 6 kilohms to find V out prime.

So starting with loop 1 we're going to sound the currents

around loop 1 in a clockwise fashion.

Starting the lower left hand corner it's going to be minus 6 volts,

this is our first equation, minus 6 volts.

Continuing, we run into the 4K resistor, with a current I1 minus

I2 which is flowing in the opposite direction through the 4K.

So it's 4K I1 minus I2 and then going down a lower part

of this loop we encounter a two kilohm resistor with current

I1 through it since there's no current in this open circuit mesh below.

So it's plus 2KI1 is equal to 0.

So we have another equation over here on the right hand side of our circuit where

we can sum the voltages around this loop.

Starting in the lower left hand corner

going up we encounter a 2k resistor with only a current I sub 2 flowing through it.

The mesh current for this second mesh, so it's 2k I 2.

Continuing to the 4 kilohm resistor, voltage drop is 4k times I2

flowing in the same direction as we're summing our voltages, minus I1,