Find the Norton Equivalent as seen by resistor R 7. Problems With Assistance Module 4 – Problem 3...

Dave ShattuckUniversity of Houston

© Brooks/Cole Publishing Co.

Find the Norton Equivalent as seen by resistor R7.

R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

Problems With AssistanceModule 4 – Problem 3

Filename: PWA_Mod04_Prob03.ppt

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Go straight to the Problem Statement

Go straight to the First Step



Overview of this Problem

In this problem, we will use the following concepts:

• Equivalent Circuits

• Norton’s Theorem

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Go straight to the Problem Statement

Go straight to the First Step



Textbook Coverage

The material for this problem is covered in your textbook in the following sections:

• Circuits by Carlson: Sections #.#• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections

#.#• Basic Engineering Circuit Analysis 6th Ed. by Irwin and

Wu: Section #.#• Fundamentals of Electric Circuits by Alexander and

Sadiku: Sections #.#• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections

#-#

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Coverage in this Module

The material for this problem is covered in this module in the following presentation:

• DPKC_Mod04_Part03

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© Brooks/Cole Publishing Co. Problem Statement

Next slide


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]


© Brooks/Cole Publishing Co. Solution – First Step – Where to Start?

How should we start this problem? What is the first step?

Next slide


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]



Problem Solution – First Step

How should we start this problem? What is the first step?

a) Define the open-circuit voltage.

b) Label the terminals of resistor R7 and remove it.

c) Define the short-circuit current.

d) Combine resistors R2 and R3 in parallel.

e) Combine resistors R6 and R7 in parallel.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]


© Brooks/Cole Publishing Co. Your choice for First Step – Define the open-circuit voltage

This is not a good choice for the first step.

The key here is that we are finding an equivalent seen by something; when we do this, the first step should always be to get rid of that something. Here, we need to get rid of R7 before doing anything else.

Remember that in circuits, a component or device does not “see” itself. We suggest that you go back and try again.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]



Your choice for First Step – Label the terminals of resistor R7 and remove it

This is the best choice for this problem.

The key in these kinds of problems where we have been asked for an equivalent “as seen by” something, is to remove the something as the first step.

The something is assumed to not “see itself”, and thus it needs to be removed. Its terminals become the place where the equivalent is found. Failure to do this can lead to errors. Let’s go ahead and remove R7.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]


© Brooks/Cole Publishing Co. Your choice for First Step – Define the short-circuit current

This is could be the first step, since R7 will not affect the short circuit current. However, it is a dangerous choice. The key is get rid of R7 right away, and then we don’t have to worry about whether it has an effect or not.

So, let’s go back and try again.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]


© Brooks/Cole Publishing Co. Your choice for First Step was – Combine resistors R2 and R3 in parallel

This is a valid step, but is not a good choice for the first step.

The key here is that we are finding the equivalent seen by R7. Thus, we need to handle R7 right at the beginning. Therefore, although R2 and R3 are in parallel, we recommend that you go back and try again.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]


© Brooks/Cole Publishing Co. Your choice for First Step was – Combine resistors R6 and R7 in parallel

This is not a good choice.

The problem here is that we are finding the equivalent as seen by R7. If we combine R6 and R7 in parallel, then R7 will be gone, and we will not be able to complete the problem correctly.

Please go back and try again.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]



Label the terminals of resistor R7 and remove it

We have named the terminals of R7, and then removed resistor R7 from the circuit. We named the terminals A and B.

Let’s go to the next slide and consider what to solve for first.

Next slide


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B



What Should We Solve for First?

We need to consider what to solve for first. Click on your choice from the choices below.

The open-circuit voltage.

The short-circuit current.

The equivalent resistance.

The total power dissipated.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B



You Chose: The open-circuit voltage

You said that the first thing to solve for would be the open-circuit voltage. This solution is not the easiest available to us. For example, this would have five essential nodes, and would require four simultaneous equations. There are better choices. Please go back and try again.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B



You Chose: The short-circuit current

You said that the first thing to solve for would be the short-circuit current. This is a good choice. Once we apply the short circuit, the R6 resistor could be neglected, we would have four essential nodes, and three equations, one of which would be very simple. Let’s find the short-circuit current.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B



You Chose: The equivalent resistance

You said that the first thing to solve for would be the equivalent resistance. This is a good choice. However, while this will be the next thing we will choose, there is another thing that must be found as well; it will be either the open-circuit voltage, or the short-circuit current. Go back, and determine which you think would be better.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B



You Chose: The total power dissipated

You said that the first thing to solve for would be the total power dissipated. This is not a good choice. The total power dissipated is not useful to us at this point. Go back and try again.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B



Finding the Short-Circuit Current

Let’s find the short-circuit current. The first step is to define the polarity. That is done in the circuit shown here. We have also defined some voltages to help us get this value.


R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B

iSCvC

+

-

vD

+

-

Next slide



R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B

iSCvC

+

-

vD

+

-

Writing the KCL’s (The Node-Voltage Method)

We wish to find vC and vD first, and then use those values to get iSC. We write the equations here:


12[V]5[A] 0, and

22[ ] 60[ ]

6[V] 12[V]5[A] 0.

30[ ] 27[ ] 50[ ]

C C

D D D

v v

v v v

W W

W W W

Solving for vC and vD we get

62.1[mS] 5[A] 0.2[A], and

90.4[mS] 5[A] 0.12[A].

C

D

v

v

This simplifies to yield

77.3[V], and

56.6[V].C

D

v

v

Next slide



Finding iSC

Using the values for vC and vD that are shown here, we can find iSC. We write a KCL for the red dashed closed surface:


Solving for iSC we get

77.3[V], and

56.6[V].C

D

v

v

2[A] 0, or22[ ] 27[ ] 30[ ]

77.3[ ] 56.6[V] 56.6[V]2[A] .

22[ ] 27[ ] 30[ ]

C D DSC

SC

v v vi

Vi

W W W

W W W

1.53[A].SCi

R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B

iSCvC

+

-

vD

+

-

Next slide



Next Step

We have found one of the three possible quantities. The best choice for the next quantity to find is probably the equivalent resistance. To find this, we set the independent sources equal to zero, which will simplify the circuit a great deal. Let’s do this in the next slide.

1.53[A].SCi Find the Norton Equivalent as seen by resistor R7.

R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

BNext slide



Finding the Equivalent Resistance, Step 1

To find the equivalent resistance, we set the independent sources equal to zero. With this we have the circuit below. We note that R4 and R5 are in series, and R2 and R3 are in parallel. Let’s combine these and redraw.

1.53[A].SCi Find the Norton Equivalent as seen by resistor R7.

R6=40[W]

R3=27[W]

R4=60[W]

R5=22[W]

R1=50[W]

R2=30[W]

A

B

next slide

Next slide



Finding the Equivalent Resistance, Step 2

Having combined these resistors, it should be clear that the series combination of R1 and R9 is in parallel with R8, which is in parallel with R6. Note that each is connected between terminals A and B. Thus we have the equivalent resistance, as seen by A and B, as

1.53[A].SCi


R6=40[W]

R8=82[W]

R1=50[W]

R9=14.2[W]

A

B

1 9 8 6( ) || ||

(50[ ] 14.2[ ]) || 82[ ] || 40[ ]

19.0[ ].

EQ

EQ

EQ

R R R R R

R

R

W W W W

W

Next slide



The Norton EquivalentThe Norton equivalent is a current source equal to the short-circuit

current, in parallel with the equivalent resistance. Thus, we have as the answer the circuit drawn below.

1.53[A].SCi

Find the Norton Equivalent as seen by resistor R7.19.0[ ].EQR W

R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B

R7=10[W]

iN=1.53[A]

RN=19[W]

A

B

Check this solution

Go back to Problem Statement



Checking the SolutionLet’s check our answer to see if it works. Let’s find the current

through the R7 resistor for the Norton equivalent. We will call this current iX. We have

19[ ]1.53[A]

19[ ] 10[ ]

1.00[A].

X

X

i

i

W

W W

R7=10[W]

iN=1.53[A]

RN=19[W]

A

B

iX

Next slide



R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B

iXvE

+

-

vF

+

-

vG

+

-

Checking the Solution, Original CircuitLet’s check our answer to see if it works. Let’s find the current

through the R7 resistor for the original circuit. Again, we will call this current iX. We will also define node voltages, to allow us to solve. We have the circuit below.

Next slide



R6=40[W]+

-vS2=

12[V]

R3=27[W]

R4=60[W]

R5=22[W]

R7=10[W]

iS2=2[A]

iS1=5[A]

+

-

vS1=6[V]

R1=50[W]

R2=30[W]

A

B

iXvE

+

-

vF

+

-

vG

+

-

Checking the Solution, Original Circuit Equations

The node-voltage equations are:

12[V]5[A] 0;

60[ ] 22[ ]

2[A] 0; and40[ ] 10[ ] 22[ ] 27[ ] 30[ ]

6[V] 12[V]5[A] 0.

50[ ] 30[ ] 27[ ]

E E F

F G F GF F F E

G G F G F

v v v

v v v vv v v v

v v v v v

W W

W W W W W

W W W

The solutions are:

84.6[V];

10.0[V]; and

48.8[V].

E

F

G

v

v

v

Using this, we can find iX, which is

10.0[V]

10[ ] 10[ ]

1.00[A].

FX

X

vi

i

W W

This is the same answer as before.

Go to Comments Slide


© Brooks/Cole Publishing Co. Was This Worth It?• This is a good question. Again, the best answer is, “It

depends.”• We have gone through a fair amount of work, but by doing so

we have a simpler circuit. Whether it was worth the work depends on what we were going to use the circuit for.

• For example, if we were to connect the circuit to 12 different resistors, or to 12 different current sources, it would be much easier to solve the simpler circuit each time, and in the end it would be worth it. For one resistor, it was probably not a good use of our time.

• Note, though, that Norton’s Theorem also has benefits as a way of thinking about a circuit. This will pay off in many areas, among them when we are designing circuits.

Go back to Overview

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Find the Norton Equivalent as seen by resistor R 7. Problems With Assistance Module 4 – Problem 3...

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