0:03

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 V0 at the circuit below.

Â You can see in the circuit that we have one current source 2 Milli amps and

Â we have 2 voltage sources a 6 volts source and a 12 volts source.

Â And V out is measured across a 1 kilo ohms load resistor and

Â it's on the right hand side of the circuit.

Â So we are going to use Mesh analysis to solve this problem.

Â So the first thing that we know is that Mesh analysis

Â is performed using Kirchhoff's voltage law.

Â So we're going to use Kirchhoff's voltage law and after we realize that our first

Â step as far as circuit analysis will be to determine and assign loops so

Â that we can sum up the voltages around the loops using Kirchhoff's voltage law.

Â So we're going to assign our loops now, we're going to start with the lower left

Â hand corner we're going to call this loop one and it'll have loop current I1.

Â We have another loop at the top of the circuit we're going to

Â call it loop 2 or mesh 2 and it has current I2 and

Â our last independent loop is the lower right hand loop and

Â we'll give it a current I3.

Â So now that we've assigned our loops and

Â our loop currents, we now want to go around and

Â write our equations using Kirchhoff's voltage law for each one of the loops.

Â So we're going to start in the lower left hand corner first

Â summing up the voltages around loop one, so

Â we'll start the lower left hand corner and we'll go around loop 1.

Â What we're looking for when we're doing mesh analysis is we're looking for

Â the loop currents.

Â If we can find the loop or mesh currents, that's really what we're interested in.

Â Once we find I1, I2, and I3, we could solve for

Â any other value that we want in our circuit.

Â So that's really what we're after.

Â And so we see with loop 1 that we have a voltage drop across a 2 milliamp source,

Â and 1 kilo Ohm resistor, and a 1 kilo Ohm resistor, and

Â then a voltage source at the bottom toward the end of our loop.

Â So we don't know the voltage across the 2 milliamp source and

Â in fact if we wanted to sum up the voltage around this loop

Â we'd have to add another variable for like the voltage across the 2 milliamp source.

Â So we immediately realize that this 2 milliamp source is I sub 1,

Â because I sub 1 is the only current flowing up through this left most

Â leg of our circuit, a branch of our circuit.

Â 2:49

Our second equation for loop 2.

Â We're going to do the same thing we're going to start in the lower left hand

Â corner and we're going to sum up the voltages as we go around loop two.

Â It's a closed path from start to finish.

Â So let's do that, first thing we encounter is a one kilo Ohm resistor at the top it

Â has I sub 2 as the current flowing through it so the voltage is going to be 1k(I2).

Â Our second voltage is the voltage for the voltage source and

Â we effect the negative polarity of that voltage source first.

Â So we're adding a -6 V.

Â We continue around our loop.

Â [COUGH] The next resistor we encounter is this 1 kilo Ohm resistor

Â 3:37

and it has a 1 kilo Ohm resistance associated with it.

Â And the current, as we're traveling counter-clockwise around loop 2,

Â would be I sub 2 which is flowing in the same direction.

Â -I3, which is also flowing through the 1 kilo-ohm resistor but

Â in the opposite direction.

Â So it's going to be +1k(I2- I3).

Â And continuing past that resistor, on to the next 1 kilo Ohm

Â resistor which is shared Between loop 1 or mesh 1 and mesh 2.

Â It's going to be 1 kilo Ohm times the current through it, which is I2,

Â flowing in the same direction as we're adding up our voltages, -I1.

Â So I'm going to add that to our Kirchhoff's voltage law equation for

Â this loop.

Â 5:09

So let's get our last equation.

Â We start at the lower left hand corner,

Â traveling upward with the negative polarity of the 12 volt source first.

Â It's -12 volts, and then we hit the 1 kilo Ohm resistor after that.

Â And so it's 1k times I3, which is flowing the same

Â direction as we're summing our voltages minus i1.

Â And as we continue around the circuit the next resistor is 1 kilo Ohm.

Â At the top of loop 3 its going to be 1 k (i3- I2) for

Â the voltage drop across that resistor.

Â 6:07

So there's our third equation, which also has I1, I2 and I3 as unknowns in it.

Â So we can solve for each one of those loop currents if we want to.

Â The problem that we have is to find V sub 0 which is the voltage drop

Â across the one kilo Ohm resistor.

Â So the only current we're really interested in is I3.

Â 6:30

Once we have I3, you can then multiply by the 1 kilo Ohm.

Â And since it's flowing in that direction,

Â we use the passive sign convention to assign polarities.

Â And so I3 times 1k is going to be equal to the voltage V sub 0.

Â So V sub 0 is 1k (I sub 3).

Â So if we solve our three equations above for

Â I sub 3, we end up with an I3 = 6mA.

Â So if that's the case,

Â then we end up with a Vout equal to 6v.

Â