0:04

The topic of this problem is Superposition Analysis, and

Â we're working with circuits with independent sources.

Â The problem is to determine V sub zero in the circuit shown below.

Â The circuit has two independent sources it has a current source and a voltage source.

Â We're measuring the output voltage across

Â a load resistance of six kilohms in our circuit.

Â So we're going to use superposition to solve this problem.

Â We know that superposition applies to linear circuits and that the circuits

Â that we work with, it's involved resistors and capacitors, inductors,

Â and dependent sources and such and independent sources are linear circuits.

Â 0:45

And so we can use the concept of superposition to analyze our problems.

Â The way we're going to use it in this problem is we're going to find

Â the contribution to V out from each one of the individual sources independently.

Â And then we'll sum the voltage from the two milliamp source and

Â from the six volt source and get a total V out.

Â So our V out will be contribution from the two milliamp source and

Â a contribution from the six volt source.

Â So the first thing that we do when we're solving a problem using

Â superpositions is we redraw the circuit for the individual sources.

Â So we're going to draw a circuit looking at the contribution of

Â the 6 volt source first.

Â 1:32

So when we do that, we'll take all other current sources in this circuit and

Â we will open circuit them.

Â And all other voltage sources in the circuit and we had short circuit them.

Â So in this case, we only have one other source.

Â It's a current source and we're going to open circuit that current source.

Â 2:09

We have our 6 volt source, we have our load

Â resistance with is 6K, and here's our V out prime.

Â It's a contribution from the six volt source.

Â It is not the total V out because we haven't taken into account what is

Â happening with respect to the two milliamp source.

Â So here's what we have so far.

Â 2:40

So what we might do is we might take the circuit and

Â re-draw it so that it looks like something that we're familiar with.

Â So we can take this, and we can then combine some of

Â the resistors together to, hopefully to solve for

Â the 6k, voltage across the 6 kilometer resistor.

Â 3:06

So here's a circuit for our 6 volt source.

Â And we're looking for the contribution to V out from just the 6 volt source.

Â So there's a number of different ways we can solve this,

Â we can start perhaps by using Kirchhoff's Voltage Law and the emission analysis,

Â if we do that, then we can assign two different ,meshes to this problem.

Â I1 and I2 will be the lead currents or

Â mesh currents associated with those meshes.

Â We can then sum up the voltages around each one of these meshes,

Â solve for I1 and I2.

Â Once we have an I2, we can multiply it by 6 kilohms to find V out prime.

Â So starting with loop 1 we're going to sound the currents

Â around loop 1 in a clockwise fashion.

Â Starting the lower left hand corner it's going to be minus 6 volts,

Â this is our first equation, minus 6 volts.

Â Continuing, we run into the 4K resistor, with a current I1 minus

Â I2 which is flowing in the opposite direction through the 4K.

Â So it's 4K I1 minus I2 and then going down a lower part

Â of this loop we encounter a two kilohm resistor with current

Â I1 through it since there's no current in this open circuit mesh below.

Â So it's plus 2KI1 is equal to 0.

Â So we have another equation over here on the right hand side of our circuit where

Â we can sum the voltages around this loop.

Â Starting in the lower left hand corner

Â going up we encounter a 2k resistor with only a current I sub 2 flowing through it.

Â The mesh current for this second mesh, so it's 2k I 2.

Â Continuing to the 4 kilohm resistor, voltage drop is 4k times I2

Â flowing in the same direction as we're summing our voltages, minus I1,

Â 5:16

And then one more voltage drop across the 6k resistor.

Â So it's plus,

Â 6k I2 is equal to 0.

Â So we can solve for I1 and I2 and ultimately we find out that

Â Vout prime is equal to 6k times I2 because using the passer

Â sign conversion we will assign a polarities across the 6k resistor.

Â Such that a positive I 2 flows into the positive polarity of

Â the voltage drop across the 6k resistor.

Â So if we find I 2 from our two equations above, plug it into this equation,

Â then we get a V out prime which is equal to eighteen-sevenths of a volt.

Â 6:08

Now we're going to look at the current source and

Â the contribution from the current source.

Â So that will be our second contribution.

Â We'll sum the first one with whatever we get for the second one to get a v out

Â total because v out is equal to v out prime plus v out double prime.

Â 8:22

The equivalent resistance is four thirds kiloohm.

Â That is 4k in parallel with 2k, and we still have our 2k resistor down below.

Â And the 6K resistor which is our load resistance,

Â and this is our V out double prime.

Â So V out double prime is going to be equal to the 6k

Â resistor times the current through the 6k resistor,

Â I'm going to call it I 6k.

Â And so V out is 6 kiloohms times the current I, 6k.

Â What is I6k?

Â I6k is equal to whatever amount of this 2 milliamp source,

Â travels through the 6 K resistor, when it's split between the two paths.

Â It can travel through this combination of resistors.

Â Or it can also travel through the 6k resistor.

Â So we're going to use current division of the 2 milliamp source

Â to find how much of it flows through the 6 kilohm resistor.

Â So let's do that.

Â 9:35

So we would take the two milli-amp source and

Â we multiply it by the resistance of the opposite leg which

Â is 4 3rdsk plus 2k and we divide that by the sum of all

Â the resistances between the two legs 4k plus 2k.

Â 9:57

Plus 6k.

Â That will give us the amount of the current which flows

Â through that 6 kilohm resistor.

Â We can then plug that into our V out double prime equation.

Â If we do that then we get a V out double prime which is equal to 30 7ths volts.

Â So V out is going to be equal to 18

Â 7ths volts plus 30-7ths volts.

Â And that gives us 48-7ths volts for V Out.

Â