0:14

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

Â So we have a 4 milliamp independent source,

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

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

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

Â the 1 kOhm resistor.

Â 1:01

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

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

Â along with that we assign mesh currents.

Â So we have mesh 1, we have mesh 2.

Â 1:48

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

Â sum up the voltage drop across each element.

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

Â a current source that we're summing voltages for.

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

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

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

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

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

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

Â 2:44

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

Â We have two resistors and

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

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

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

Â We have to introduce yet another variable.

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

Â is the mesh current.

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

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

Â 3:25

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

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

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

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

Â And so we have minus 1K I sub x.

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

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

Â 4:08

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

Â I3 minus I1.

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

Â direction as our summing minus I1.

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

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

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

Â summing the voltages minus I sub 4.

Â 5:01

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

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

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

Â and it's 1k I4 minus I3 plus voltage

Â drop across 1 kilo ohm resistor in the center

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

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

Â And that is equal to zero.

Â So they're for loop or mesh equations.

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

Â So we have one unknown or

Â one variable which I1, two fro I2,

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

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

Â So we need six equations.

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

Â That's means the independent of the prior four.

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

Â What is V sub x?

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

Â So we can write an equation V sub x is

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

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

Â 7:23

In the end, if we do that,

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

Â We have I sub 2 equal to minus 6 milliamps.

Â We have I sub 3 equal to minus 2 milliamps and

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

Â