0:02

The topic of this problem is mesh analysis.

Â And we're going to be working with circuits with independent voltage and

Â current sources.

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

Â Circuit has three independent sources, two current sources and a voltage source.

Â We're measuring the output voltage, V sub 0, across the load resister,

Â a 4 kilo ohm load resister, level resistor in our problem.

Â It's on the right most side of our circuit.

Â 0:46

So the first thing we do when we have a problem and

Â we want to solve it using mesh analysis is we assign loops or meshes to the circuit.

Â And we also assign loop or mesh currents as well.

Â So letâ€™s do that first.

Â First loop Iâ€™m going to assign is the upper left hand loop and

Â I'm going to give it mesh current I1.

Â 1:26

So now we've assigned our meshes and

Â we've assigned our mesh currents that's the first step in mesh analysis.

Â Then the second step is to go to each one of these meshes and to add up or

Â sum up the voltages around the loop that close path, which we have.

Â Starting at the, typically I start at the lower left hand corner and

Â going around the loop, back to that same corner.

Â So we'll do that for each one of our loops.

Â We'll go all the way around it from start to finish,

Â ending in the same point, summing up the voltages.

Â What we're going to find in this problem is that we have a couple of

Â issues with using this straightforward approach and doing the problem.

Â So we have to use some of our additional rules of mesh analysis and

Â Kirchhoff's voltage law.

Â So let's start with Loop 1 which is the upper left hand corner.

Â We need to sum up the voltages around this loop.

Â So if we start in the lower left hand corner,

Â we start up, we see that we run into a 3 milliamp source.

Â We don't know what the voltage drop is across this, so we'd have to assign

Â another variable D sub 3 milliamps for the voltage drop across it.

Â So we'd add another variable in addition to our Mesh current.

Â Our four mesh currents would have a fifth variable in that case.

Â And we could go on around and we can find the voltage drop

Â across the two kilo ohm with respect to the loop current I1 and I sub 2.

Â And the 4k we could do the same using I1 and

Â I4 and the current as, to find the current through the 4k.

Â But we have this problem with the 3 milliamp resistor or

Â 3 milliamp source on the left hand side of the circuit.

Â 3:11

Once we have those, we can find any other value that we want in the circuit,

Â any voltage drop, any power whatever else might be of interest to us.

Â So that's really what we're after.

Â If we look at this loop one and we inspect it closely we see that,

Â I sub 1 is equal to 3 milliamps.

Â So the problem itself gives us I1 and we don't need to,

Â 3:49

Let's go to the second loop and see if we can come up with an equation for it.

Â So starting in the lower left hand corner,

Â we encounter first the voltage drop across the 2 kilo ohm resistor.

Â That voltage drop is going to be 2K.

Â 4:01

And since we're going clockwise,

Â the current which is going the same direction as I2, it's going to be I2- I1,

Â which is also flowing through the 2 kilo ohm resistor, I2- I1.

Â Because it's flowing through in the opposite direction of I2.

Â 4:21

And we continue on around the loop and we

Â encounter the four kilo ohm resistor which just has a current I2 flowing through it.

Â So the voltage drop is 4k(I2) and

Â continuing down through the 12 kilo ohm resistor at the bottom of

Â loop two the voltage drop is 12 kilo ohms times I2 minus I3.

Â 4:49

And that's our last voltage drop before we get back to our starting point in our loop

Â that we had for loop two, back to the lower left-hand corner of it.

Â So let's go down and look at loop three.

Â If you look at loop three,

Â we encounter really the same type of problem that we did up with loop one.

Â But this time we see that when we hit this

Â current source that we don't know the voltage drop across.

Â We don't have this current source assigned to just one loop,

Â as shared between loop three and loop four.

Â So we have an equation that we can write for that and

Â it will be that 1 milliamp = I3, which is following

Â the same direction as the 1 milliamp source,- I4.

Â So we can at least write that.

Â 5:40

So, that gives us another equation, but we need four equations to solve

Â this because we have four unknowns I1, I2, I3 and I4.

Â So, we need one more independent mesh,

Â 6:05

In order to get around this 2 milliamp source, I'm sorry the 1 milliamp source,

Â in the center of the these two loops I'm going to assign a mesh which

Â we can sum the voltages around just like we could any other mesh.

Â But is independent of mesh one and

Â mesh two above, so we get our, another independent equation.

Â 6:36

And we can again sum up our voltages,

Â starting in the lower left hand corner of this mesh.

Â Going around this mesh, we have the ability to write an equation for

Â all the voltage drops from start to finish.

Â And this is, again, the supermesh.

Â We use this concept when we have a current source in a mesh analysis

Â problem that is shared between two adjacent loops.

Â Whenever we have that the case we have to rely on mesh analysis and

Â supermesh analysis in order to solve the problem.

Â So let's write our last equation for the lower two loops.

Â So starting in the lower left hand corner we run into the negative polarity of

Â the 6v source first.

Â It's a -6v for that voltage drop.

Â And then we encounter the 4 kilo ohm resistor.

Â Since we're going in the clockwise fashion, I4 is also travelling

Â the same direction through that loop, it's going to be 4k I4- I1.

Â 8:11

And that's our last voltage drop before we get back to the starting point.

Â So that's equal to zero, so now we have four equations and four unknowns.

Â What we're really after in this problem is we're after V sub out.

Â Which is the voltage drop across the four kilo ohm load resistor.

Â We know V out = 4K times I sub 2,

Â pass of sign convention tells us the positive I2 would flow

Â into the positive of the voltage drop across the 4K resister.

Â So it's 4K( I sub 2) for V out.

Â So really the only loop, or mesh, current that we're interested in is I sub 2.

Â So that's the one we're going to solve for.

Â If we solve for I sub 2 from this equations above,

Â we get a I2 = 31 15ths milliamps.

Â