0:14

We have a circuit that has dependent sources and independent sources in it.

So we have a 4 milliamp independent source,

we have a 12 volt independent source, and we have a voltage

source that is a current controlled voltage source in this case.

And the current that's controlling it is the current that's flowing through

the 1 kOhm resistor.

1:01

So the first thing we do when we're working with, problems where

we're performing mesh analysis is we assign the meshes and

along with that we assign mesh currents.

So we have mesh 1, we have mesh 2.

1:48

You go around it and we end at the same point that we started and

sum up the voltage drop across each element.

We see for a loop one that we have two resistors and

a current source that we're summing voltages for.

We don't know what the voltage drop is for the four milliamp source and

so if want to sum up the voltages in the matter that we normally do for

the Kirchhoff's Voltage Law then we have to add another variable.

We press we're you call it the voltage for the four milliamp source.

But instead of doing that we know that what we're looking for

when doing mesh analysis is in the end going to be the mesh currents.

2:44

So if we'd look at mesh 2, we have a similar situation.

We have two resistors and

we can find voltage drop four and we have this current source.

It's a dependent source but it's a current source where again we'll have the same

problem knowing what the voltage drop is across the current source.

We have to introduce yet another variable.

So instead of doing that, we again notice that what we're looking for

is the mesh current.

And in fact we're given the mesh current for this.

We're given that I2 is equal to V sub x divided by 2K.

3:25

Okay, so going to mesh 3 and mesh 3 is in a lower left hand corner of our circuit.

If we start to lower left hand corner of that mesh and

start assuming the voltage drops as we go around the loop back to the start again.

We first encounter the negative polarity of the current controlled voltage source.

And so we have minus 1K I sub x.

We continue on and we run into the 2 kilo-ohm Resistor.

We know the voltage drop for the 2 kilo ohm resistor is V sub x but it's also 2K

4:08

I1 which is in the same direction we're summing these drops for

I3 minus I1.

So we have 2K I sub 3 which is flowing in the same

direction as our summing minus I1.

So we continue around this loop and we encounter the 1 kilo ohm resistor and

would use at the same approach to find the voltage drop for it.

It's going to be 1K I sub 3 which is flowing the same direction as we're

summing the voltages minus I sub 4.

5:01

So if we go to loop 4, our mesh 4 and

do the same thing, we can sum the voltages up just like we did before.

So, the first voltage drop is across 1 kilo ohm resistor,

and it's 1k I4 minus I3 plus voltage

drop across 1 kilo ohm resistor in the center

of the circuit, 1k, I4 minus I sub 2.

And then we have the voltage drop from the voltage source, which is 12.

And that is equal to zero.

So they're for loop or mesh equations.

We know what I1 and I2 are, I2 in terms of another variable V sub x.

So we have one unknown or

one variable which I1, two fro I2,

three for V sub x, four for I sub 3,

five for I sub 4 and six for I sub x.

So we need six equations.

So we need to have find two more equations that relate our variables to one another.

That's means the independent of the prior four.

So we can investigate those first of all by looking first at V sub x.

What is V sub x?

It's the voltage drop across the 2 kilo ohm resistor.

So we can write an equation V sub x is

equals to 2K I sub 3 minus I sub 1).

Because I sub 3 flows into the positive polarity of the voltage drop

7:23

In the end, if we do that,

we end up with an I sub one equal to 4 milliamps as we have seen above.

We have I sub 2 equal to minus 6 milliamps.

We have I sub 3 equal to minus 2 milliamps and

we have an I sub 4 which is equal to minus 10 milliamps.