2:07

and V1 minus V2, for the voltage.

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

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

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

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

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

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

Â the voltage source immediately.

Â Unless we want to assign another variable,

Â maybe I12 volts with a current that flows through it.

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

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

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

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

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

Â that would be our fourth equation.

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

Â equation to solve the problem.

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

Â which is floating between node two and node three.

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

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

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

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

Â node three, and the 12-volt voltage source.

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

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

Â Top to bottom through the one -kilo ohm resistor,

Â top to bottom through the two kilo-ohm resistor.

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

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

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

Â So this is the second equation.

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

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

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

Â flowing top to bottom through the one kilo ohm resistor.

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

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

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

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

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

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

Â it's minus four milliamps.

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

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

Â We have two independent equations at this point.

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

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

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

Â 5:39

Okay, so now we have the three equations and three unknowns.

Â We know that we get from nodal analysis, we get nodal voltages.

Â So the result of our analysis is going to give us V1, V2 and V3.

Â If we want to find something like I sub zero,

Â we an find it easily once we know what V sub two is.

Â If we know the voltage at this node,

Â I sub zero simply that voltage divided by one kilo ohm to get that current.

Â So, first thing we notice is that if we solve these equations,

Â we get a V sub 2 equal to minus 16 over 3 volts.

Â And we know that I sub zero

Â is going to be equal to minus

Â 5.33 milliamps.

Â So that's our nodal analysis for this problem.

Â What we're going to do is we're going to also solve this using mesh analysis.

Â If we use mesh analysis, we'll see that we get the same answer for

Â I sub zero is going to be equal to minus 5.33 milliamps.

Â 7:43

So if we look at mesh one,

Â we see that we can determine sub one directly from our mesh.

Â That is, I sub 1 is equal to minus 2 milliamps.

Â Our second equation for our second loop

Â gives us a similar way to find I sub two.

Â I sub two in this case is the only current that's flowing through this four-milliamp

Â source at the top of our circuit.

Â So we know I sub two is equal to four milliamps.

Â So that's two out of our three mesh currents.

Â The third one that we need and in fact, we need it to find I sub zero,

Â is we're going to need to find I sub three.

Â So we can sum up the voltages around this third loop to find I sub three.

Â So if we start the lower left hand corner of this loop and go up clockwise,

Â we first encounter the one-kilo ohm resistor.

Â The voltage drop across the one-kilo ohm resistor is 1K times I sub 3,

Â which is flowing in the same direction that we're summing our voltages minus I1,

Â which is flowing the opposite direction.

Â We continue up around our mesh or loop, we encounter the 12-volt source.

Â In fact, we encounter the negative polarity of that 12-volt source.

Â So it's minus 12.

Â We continue and we encountered the two-kilo ohm resistor on the right-hand

Â side of our circuit as we complete our mesh.

Â And the voltage drop for that element is going to

Â be 2K times I sub 3, and that's equal to zero.

Â So we have a third equation which is independent of our first two and

Â it tells us that, if we know I1 and I2, we can find I sub three.

Â And I sub three in this case,

Â comes out to be 3.33 milliamps for I sub three.

Â Now, if we want to find I sub zero,

Â I sub zero is going to be equal to I1 minus I sub 3.

Â This is I sub one.

Â And so that's equal to I sub one,

Â which was minus two milliamps, minus I sub three,

Â which was 3.33 milliamps.

Â So we get the same answer for I sub zero.

Â We have an I sub zero equal to minus 5.33 milliamps,

Â just as we did for nodal analysis.

Â