0:03

The topic of this problem is Mesh Analysis.

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

current sources.

The problem is to determine V0 at the circuit below.

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

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

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

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

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

So the first thing that we know is that Mesh analysis

is performed using Kirchhoff's voltage law.

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

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

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

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

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

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

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

our last independent loop is the lower right hand loop and

we'll give it a current I3.

So now that we've assigned our loops and

our loop currents, we now want to go around and

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

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

summing up the voltages around loop one, so

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

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

the loop currents.

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

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

any other value that we want in our circuit.

So that's really what we're after.

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

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

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

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

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

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

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

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

leg of our circuit, a branch of our circuit.

2:49

Our second equation for loop 2.

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

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

It's a closed path from start to finish.

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

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

Our second voltage is the voltage for the voltage source and

we effect the negative polarity of that voltage source first.

So we're adding a -6 V.

We continue around our loop.

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

3:37

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

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

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

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

in the opposite direction.

So it's going to be +1k(I2- I3).

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

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

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

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

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

this loop.

5:09

So let's get our last equation.

We start at the lower left hand corner,

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

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

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

direction as we're summing our voltages minus i1.

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

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

the voltage drop across that resistor.

6:07

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

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

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

across the one kilo Ohm resistor.

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

6:30

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

And since it's flowing in that direction,

we use the passive sign convention to assign polarities.

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

So V sub 0 is 1k (I sub 3).

So if we solve our three equations above for

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

So if that's the case,

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