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The topic of this problem is nodal analysis, and

Â we're going to work with circuits with dependent sources.

Â The problem is to write the nodal equations for the circuit shown.

Â As you can see, the circuit has a dependent

Â source on the left hand side the dependent source is beta I0.

Â So we have a current controlled current source.

Â The I0, which is controlling the current controlled current source,

Â is the current down through resistor 3 in the center-right side of the circuit.

Â So we're going to write the nodal equations,

Â and what we want to remember when we have these circuits with dependent sources is

Â we're going to need an additional equation relating the controlling parameter for

Â the dependent source with our nodal voltages.

Â So we're going to find an equation which relates

Â I sub 0 with our nodal voltages as our third equation.

Â So as we have done in all our nodal analysis circuits the first thing we do

Â is we identify the nodes, that's the first step in solving the problem.

Â So, we see that we have a node at the left hand

Â side of the circuit at the top and I'll call that node one and we have a node

Â at the right hand top part of the circuit and we're going to call that node two.

Â 1:21

We also have a ground node it's at the bottom of the circuit.

Â That we know is at 0 volts reference so

Â that we have a reference point for our calculations.

Â So we're going to go through it and

Â write our nodal equations at this point after we have identified our nodes.

Â The first one is node 1 and

Â we're going to use Kirchoff's current laws to write our nodal equations.

Â That is,

Â the sum of the currents into any node at any instant in time is equal to zero.

Â So we're going to use Kirchoff's current law,

Â and we're going to sum the currents into the nodes.

Â 1:59

Again, with kirchoff's current law we can choose to sum the current into the nodes.

Â Or we can sum the currents out of the node,

Â we just need to be consistent throughout the problem.

Â So we're going to Kirchhoff's current law,

Â and we're going to sum the currents into the node, starting with node number one.

Â So for node number one, we have beta I sub 0 flowing out of that node.

Â So we're going to put in a minus Beta I0 for

Â that current we also have current flowing up through

Â R sub 1 and if we add that current is going to be

Â the current flowing up through R1 so it's going to be the bottom node.

Â Minus the top node voltage divided by the resistance, R1.

Â So that's going to be the bottom node, which is 0 volts- V sub 1,

Â which is our nodal voltage at the top for node one, divided by R1.

Â 2:59

And then we have one more current flowing into that node.

Â These are current flowing through the resistor R2 at the top of the circuit.

Â That resistance is between node 1 and node 2.

Â So if we're looking at the current flowing into node 1 through resistor R2.

Â It's V2 minus V1 divided by R2,

Â V2- V1 divided by R2.

Â And that's equal to 0, that's our last current flowing into node 1.

Â So that's our first independent equation.

Â We have a second equation that we can write around node 2.

Â And again, summing the currents into node 2.

Â We know that we have I sub A flowing into node 2,

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

Â And then we have the currents which are flowing through R3 and

Â R2 into node 2 as well.

Â So looking at R2 first, we have the current flowing left to right.

Â So it's going to be V1 minus V2 divided by R2.

Â V1 minus V2 divided by R2 and we have one other current

Â the current flowing up through R3 flowing into node 2.

Â So that's going to be our ground nodes,

Â 0 volts- the nodal voltage at that second V2, divided by R sub 3.

Â And that's all of our currents flowing into node 2.

Â So that's sum is going to be equal to 0.

Â So now we have two independent equation.

Â And as we can see, we have the two equations that we have three unknowns,

Â we have V1 and V2 for the nodal voltages and

Â we also have I sub 0 which is in there as well and

Â I sub 0 comes from our dependent source, it's beta I sub 0.

Â So again, we need that third equation which relates our controlling

Â parameter I sub 0 to our nodal voltages and so we have to think about

Â that third equation for our set of three equations and our three unknowns.

Â So how do we make that relationship?

Â We see that I sub 0 is flowing down through R3 and

Â 5:43

So, now we have this third independent equation.

Â So, we have three equations, three unknowns, and

Â we could solve for I0, V1, and V2.

Â