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The topic of this problem is mesh analysis and

we are working with circuits with independent sources.

The problem is to determine i sub 0 in the circuit shown below.

The circuit has one independent voltage source and two independent current sources

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and the way we going to solve this problem is using Mesh Analysis.

So we know that one where doing our problem and

solving it using Mesh Analysis, that we're looking for the Mesh currents.

Hopefully, if we can find the Mesh currents, then we're able to find any

other important quantity that we're interested in in our circuit.

So when we're performing Mesh Analysis,

the first thing we do is we assign our meshes and our mesh currents.

So we're going to start with the lower left-hand corner and we're going to assign

our first mesh and we're going to assign a direction as clock wise for that mesh.

We're going to assign a second one to the top mesh, again clock wise from mesh

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The next thing we do when we're performing Mesh Analysis is we, some of the voltages

around our meshes ultimately, so that we can find the values for I sub 1,

I sub 2 and I sub 3 or mesh or loop currents.

So when we look at this first mesh and we try to do that,

we see that we have, we can certainly sum the 12 volt source and

we can certainly sum the voltage across the 2 kilo ohm resistor, but

we don't know the voltage drop across the 2 milliamp source.

Sure we can assign another variable, maybe the 2 milliamps for

the voltage that drops across the source.

That adds another variable to the problem.

And instead of having just three unknowns, I1 through I3,

we also have a fourth unknown, V2 milliamps.

So in order to avoid that,

we use this concept of supermesh in order to solve the problem.

So if we're trying to find the solutions to I1,

I2 and I3 in Mesh Analysis, and we have a current source

which is tied between two adjacent loops or

meshes, then we need to use this concept of supermesh to solve the problem.

And what supermesh allows us to do is it allows us to

avoid that current source when we're summing the voltages around the loop.

So our supermesh that we're going to find is the one in red,

where we avoid that 2 milliamp source, yet

we still create another independent equation, so we have that third equation.

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So we're going to use this as our super mesh and

we're going to first of all, sum up our voltages around the super mesh.

So starting at the lower left hand corner of the super mesh, we first

encounter the 12 volt source and the negative polarity for that 12 volt source.

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Continuing, we encounter the 1 kilo ohm resistor, and the voltage drop across

1 kilo ohm resistor is 1K, and then the current I 2 is flowing the same direction.

We're summing the voltages, so it's I 2 minus I sub 3,

which is flowing the opposite direction through the 1 kilo ohm resistor.

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We continue on our path and

we encounter the 2 kilo ohm resistor on the bottom centre of our circuit.

The voltage drop for that is 2k I sub 1 which is flowing

in the same direction as we're summing the voltages minus I sub 3 and

that's equal to 0 because that gets us back to our starting point.

So that's our first equation.

And we can see from this, we have our three unknowns, I1, I2 then I3.

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shared by two adjacent loops or meshes.

And so we know that 2 milliamp source is going to be equal to

I1 which is flowing in the same direction as it minus I2

which is flowing the opposite direction of that current source.

So now we have three independent equations, we have

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know I sub 0 which is ultimately what we're looking for, is equal to what?

Is equal to I sub 1 minus I sub 3.

So we need to find I sub 1 in order to solve for I0.

So if we solve these three independent equations for

I sub 1, we go to I sub 1 which is equal to 1.2 milliamps.

And if I sub 1 is equal to 1.2 milliamps,

then I sub 0 is equal to 5.2 milliamps.