and V1 minus V2, for the voltage.

So, the current is V sub 1 minus V sub 2,

divided by 2K, and that's equal to zero.

So that's the sum of the currents into or out of node one.

Now, if we look at node two and node three,

we see that we have this voltage source between node two and node three.

And we really don't have a way of determining what the current is through

the voltage source immediately.

Unless we want to assign another variable,

maybe I12 volts with a current that flows through it.

But that would then give us a third or a fourth unknown in our equations.

Right now, we have unknown V sub one and V sub two.

As we continue to write equations, we're going to have an unknown V sub three.

So we're going to need three equations for those three unknowns.

And if we add another one for the current through the 12 volt source, and

that would be our fourth equation.

We wouldn't be able to come up with a fourth independent

equation to solve the problem.

So the next thing we notice is that, since we have this voltage source,

which is floating between node two and node three.

Then we have to use the concept of a super node to solve the problem.

And the way we do that is we kind of block off this voltage source like this.

And we'll sum all the currents out of what we call a super node,

which is the node which encompasses what we assigned as node two and

node three, and the 12-volt voltage source.

So if we sum the currents out of the super node,

it's going to be the current right to left, through the two-kilo ohm resistor.

Top to bottom through the one -kilo ohm resistor,

top to bottom through the two kilo-ohm resistor.

And then we also have the four milliamps which is flowing in.

So we have a minus four milliamp contribution flowing out from this path.

So it's going to have four different components to it.

So this is the second equation.

It's our super node equation where we have, first of all,

the current flowing right to left through the two kilohm resistor.

Which we know is V2 minus V1, over 2K, plus the current

flowing top to bottom through the one kilo ohm resistor.

It's equal to I sub zero, by the way.

And that is going to be V2 minus 0, divided by 1K,

because we know that the bottom of our circuit is a ground node.

And we have the current down through the two kilo ohm resistor

on the right-hand of our circuit, V sub 3 minus 0, over 2K.

And the current which is flowing up which is minus four milliamps, so

it's minus four milliamps.

And that's all the currents flowing out of the super node.

And so again we have an equation that has three unknowns V1, V2, and V3.

We have two independent equations at this point.

So we need one more equation relates V1, V2, and V3.

And our third equations comes from the constraining equation of our super node,

that is V sub 3 minus V sub 2, is equal to 12 volts.