This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.
Sample Problem: Superposition (Independ Sources) 1
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From the course by Georgia Institute of Technology
Linear Circuits 1: DC Analysis
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From the lesson
Module 4
This module introduces additional methods for analyzing circuit problems and how resistors are used in sensors.
Meet the Instructors
Dr. Bonnie H. FerriProfessor
Electrical and Computer Engineering
Dr. Joyelle HarrisDirector of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer
The topic of this problem is Superposition Analysis, and
we're working with circuits with independent sources.
The problem is to determine V sub zero in the circuit shown below.
The circuit has two independent sources it has a current source and a voltage source.
We're measuring the output voltage across
a load resistance of six kilohms in our circuit.
So we're going to use superposition to solve this problem.
We know that superposition applies to linear circuits and that the circuits
that we work with, it's involved resistors and capacitors, inductors,
and dependent sources and such and independent sources are linear circuits.
And so we can use the concept of superposition to analyze our problems.
The way we're going to use it in this problem is we're going to find
the contribution to V out from each one of the individual sources independently.
And then we'll sum the voltage from the two milliamp source and
from the six volt source and get a total V out.
So our V out will be contribution from the two milliamp source and
a contribution from the six volt source.
So the first thing that we do when we're solving a problem using
superpositions is we redraw the circuit for the individual sources.
So we're going to draw a circuit looking at the contribution of
the 6 volt source first.
So when we do that, we'll take all other current sources in this circuit and
we will open circuit them.
And all other voltage sources in the circuit and we had short circuit them.
So in this case, we only have one other source.
It's a current source and we're going to open circuit that current source.
All the other elements remain.
We have a 2K resistor and
we have a 4k resistor and another 2k resistor.
We have our 6 volt source, we have our load
resistance with is 6K, and here's our V out prime.
It's a contribution from the six volt source.
It is not the total V out because we haven't taken into account what is
happening with respect to the two milliamp source.
So here's what we have so far.
We want to take this circuit and we want to be able to find V out prime from it.
The voltage across the six K resistor.
So what we might do is we might take the circuit and
re-draw it so that it looks like something that we're familiar with.
So we can take this, and we can then combine some of
the resistors together to, hopefully to solve for
the 6k, voltage across the 6 kilometer resistor.
So here's a circuit for our 6 volt source.
And we're looking for the contribution to V out from just the 6 volt source.
So there's a number of different ways we can solve this,
we can start perhaps by using Kirchhoff's Voltage Law and the emission analysis,
if we do that, then we can assign two different ,meshes to this problem.
I1 and I2 will be the lead currents or
mesh currents associated with those meshes.
We can then sum up the voltages around each one of these meshes,
solve for I1 and I2.
Once we have an I2, we can multiply it by 6 kilohms to find V out prime.
So starting with loop 1 we're going to sound the currents
around loop 1 in a clockwise fashion.
Starting the lower left hand corner it's going to be minus 6 volts,
this is our first equation, minus 6 volts.
Continuing, we run into the 4K resistor, with a current I1 minus
I2 which is flowing in the opposite direction through the 4K.
So it's 4K I1 minus I2 and then going down a lower part
of this loop we encounter a two kilohm resistor with current
I1 through it since there's no current in this open circuit mesh below.
So it's plus 2KI1 is equal to 0.
So we have another equation over here on the right hand side of our circuit where
we can sum the voltages around this loop.
Starting in the lower left hand corner
going up we encounter a 2k resistor with only a current I sub 2 flowing through it.
The mesh current for this second mesh, so it's 2k I 2.
Continuing to the 4 kilohm resistor, voltage drop is 4k times I2
flowing in the same direction as we're summing our voltages, minus I1,
And then one more voltage drop across the 6k resistor.
So it's plus,
6k I2 is equal to 0.
So we can solve for I1 and I2 and ultimately we find out that
Vout prime is equal to 6k times I2 because using the passer
sign conversion we will assign a polarities across the 6k resistor.
Such that a positive I 2 flows into the positive polarity of
the voltage drop across the 6k resistor.
So if we find I 2 from our two equations above, plug it into this equation,
then we get a V out prime which is equal to eighteen-sevenths of a volt.
Now we're going to look at the current source and
the contribution from the current source.
So that will be our second contribution.
We'll sum the first one with whatever we get for the second one to get a v out
total because v out is equal to v out prime plus v out double prime.
So let's do that, let's look at the 2miliamp source.
First, we want to redraw the circuit for 2miliamp source, so we will short
circuit the voltage source, we leave our current source in, like we see here.
We have a 4 kilohm resistor at the top.
We have a 2 kilohms resistor below it and from left to right we have
a 2 kilohms resistor and we have a 4 kilohms resistor as well,
And 6 kilohm resistor as well.
So, this is our 6K resistor and our 4K,
our 2K and other 2K resistor and the 2 mA.
So, I simply redrawn our original circuit just looking
at the contribution from the 2 mA source and this is what we end up with.
V out double prime to measure across our 6k load resistance.
So we see that we have a circuit that has some resistors in it and a current source.
So we can redraw the circuit, if we wanted to,
to make it look a little simpler to us.
If we did that, we could put our 2
milliamp source on the left hand side of our newly drawn circuit.
And then we would have our 6k resistor on the far right hand side of the circuit.
And we're going to replace these 4 and
2k resistors in the middle with the equivalent resistance.
And that equivalent resistance would be in the case of the 4k and the 2k at the top.
The equivalent resistance is four thirds kiloohm.
That is 4k in parallel with 2k, and we still have our 2k resistor down below.
And the 6K resistor which is our load resistance,
and this is our V out double prime.
So V out double prime is going to be equal to the 6k
resistor times the current through the 6k resistor,
I'm going to call it I 6k.
And so V out is 6 kiloohms times the current I, 6k.
What is I6k?
I6k is equal to whatever amount of this 2 milliamp source,
travels through the 6 K resistor, when it's split between the two paths.
It can travel through this combination of resistors.
Or it can also travel through the 6k resistor.
So we're going to use current division of the 2 milliamp source
to find how much of it flows through the 6 kilohm resistor.
So let's do that.
So we would take the two milli-amp source and
we multiply it by the resistance of the opposite leg which
is 4 3rdsk plus 2k and we divide that by the sum of all
the resistances between the two legs 4k plus 2k.
Plus 6k.
That will give us the amount of the current which flows
through that 6 kilohm resistor.
We can then plug that into our V out double prime equation.
If we do that then we get a V out double prime which is equal to 30 7ths volts.
So V out is going to be equal to 18
7ths volts plus 30-7ths volts.
And that gives us 48-7ths volts for V Out.
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