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The topic of this problem is mesh analysis and

Â we are working with circuits with independent sources.

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

Â 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.

Â So we know that one where doing our problem and

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

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

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

Â So when we're performing Mesh Analysis,

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

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

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

Â 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

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

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

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

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

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

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

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

Â the voltage that drops across the source.

Â That adds another variable to the problem.

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

Â we also have a fourth unknown, V2 milliamps.

Â So in order to avoid that,

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

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

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

Â which is tied between two adjacent loops or

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

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

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

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

Â where we avoid that 2 milliamp source, yet

Â 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

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

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

Â 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

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

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

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

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We continue on our path and

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

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

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

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

Â So that's our first equation.

Â 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.

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

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

Â which is flowing the opposite direction of that current source.

Â 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?

Â Is equal to I sub 1 minus I sub 3.

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

Â So if we solve these three independent equations for

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

Â And if I sub 1 is equal to 1.2 milliamps,

Â then I sub 0 is equal to 5.2 milliamps.

Â