Sample Problem: Norton (Independ Sources) 2

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From the course by Georgia Institute of Technology

Linear Circuits 1: DC Analysis

168 ratings

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

168 ratings

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.

From the lesson

Module 4

This module introduces additional methods for analyzing circuit problems and how resistors are used in sensors.

Sample Problem: Max Power(Thevenin)6:26

Sample Problem: Thevenin (Independ Sources) 111:38

Sample Problem: Thevenin (Independ Sources) 28:29

Sample Problem: Thevenin (Independ Sources) 38:12

Sample Problem: Thevenin (Depend Sources)19:52

Sample Problem: Norton (Independ Sources) 114:07

Sample Problem: Norton (Independ Sources) 27:55

Sample Problem: Superposition (Independ Sources) 110:41

Sample Problem: Superposition (Independ Sources) 26:03

Meet the Instructors

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Norton's analysis.

And we're going to work with circuits with independent sources.

The problem is to use Norton's equivalent circuit to determine V sub zero in this

problem, where V sub zero is the output voltage measured across our load resistor,

which is a 6 kilo ohm resistance.

So, when we are using Norton's equivalent circuit to solve for problems,

we first of all have to realize what that circuit looks like and

what we have to, ultimately,

get to in order to analyze this problem using Norton's equivalent circuit.

And so, I'm drawing Norton's equivalent circuit for us, a general form of it.

And what it involves is a current source I short circuit,

and a resistance, R thevenin,

that replace all of our circuit except for the load, our R sub L.

So in our case, our load is 6K, and it would represent the right hand side

of our circuit in our Norton's equivalent circuit.

What we have to come up with is the two R,

the two elements on the left hand side of the circuit.

We have to come up with the current source I short circuit, and

we have to come up with R Thevenin's.

These two are equivalent representations.

This is an equivalent representation for the circuit that we have originally.

We take the original circuit, and we reduce everything except for the load.

We've taken the load out.

We reduce everything except for the load to this circuit.

If we do that, then we can solve the circuit very easily using current division

to find what the current is through the load resistance, or

ultimately what the voltage is across the load resistance.

So, we need to come up with I short circuit and R thevenins.

In order to do that, we redraw our circuit for the short circuit condition.

We take the load resistor on the right hand side of the circuit, and

we replace it by a short circuit.

We'll draw everything out to that point first.

So, we have our sources, we have our 1 and

2K resistor on the left hand side of the circuit.

We have our three volt source at the top of the circuit.

We've taken our 6K resistor out as our load, and

we're replacing it with a short circuit like this.

And the current is I short circuit.

That's the current that we're looking for.

So we can analyze this circuit to find I short circuit.

Perhaps the easiest way to do that is to just look at the other loop of the circuit

and use Kirchhoff's voltage law, and some of the voltages around this outside loop.

So, let's do that.

If we sum the voltages around the outside loop We would

get a voltage across one kilo ohm resistor, and

we want to assign, since we're using mesh analysis,

we're want to assign mesh one a current, mesh one.

And we have mesh 2, here.

And we have a current, I sub 2, associated with mesh 2.

So the voltage drop across a 1 kilo ohm resistor,

because we're going to sum up the voltages around our loop,

is going to be 1K times I1.

And it's going to be 2K

Times I1 Minus 3 volts.

And this is a short circuit going back to the beginning, is equal to zero.

So we have an equation which is a function of I sub 1.

And so, we find that I1 is equal to 1 milliamp.

So, if I1 is one milliamp, is flowing into this node

I2 is also flowing into this node as 2 milliamps.

It's not I2, it's the 2 milliamp source.

That results in 3 milliamps flowing from left to right

through this second loop, the outer part of the second loop.

So, what that tells us is that I2, which is equal to I short circuit,

Is equal to 3 milliamps.

So that's the first of our elements that we need for

our Norton's euivalent circuit.

A second element is R thevenins.

The way we determine R thevenins is we take our original circuit,

again taking the load out, and we find the equivalent resistance for

the remaining part of our circuit.

And so, we redraw it with that in mind.

We have a 1 kilo ohm resistor, and a 2 kilo ohm resistor.

Like so.

When we're finding R thevenins, we open circuit all the current sources.

So, we have an open circuit for the current source.

And we short circuit all the voltage sources.

So this is our equivalent circuit.

We were looking for R thevenins.

So this is R thevenins equivalent resistance for our network with

the current sources open circuited, the voltage sources short circuited.

So that's pretty easy to determine.

It's 1K plus, should be 2K at the top.

And that gives us a 3 kilo ohm equivalent resistance for our network.

So, we can redraw our Norton's equivalent circuit above, and

put in our values for I short circuit, which is 3 milliamps.

We can put in our equivalent R thevinins resistance for our network.

And that is 3K.

And we can put our load back in.

In our case, our load resistance was 6 kilo ohms.

And we're looking for the voltage drop across that 6 kilo ohm resistor.

So, we can use current division to find the current through the 6K.

And we can multiply it by the resistance and get V out.

V out is equal to 6K times the current.

Let's call it I 6K.

We know that I 6K Is equal to what?

It's equal to 3 milliamps times

3K divided by 3K plus 6K.

So, if we solve for V out in this problem,

we get a V out which is equal to 6 volts.

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