elec273 lecture notes set 5 circuit theoremstrueman/elec273files/elec273_5_2… · chapter 4...

ELEC273 Lecture Notes Set 5 Circuit TheoremsThe course web site is:http://users.encs.concordia.ca/~trueman/web_page_273.htm

Homework on network theorems.

Mid-Term ExamSaturday October 20, 2018 from 2:00 to 3:30.The mid-term will cover chapters 1 to 4 of Sadiku and Alexander.

The tentative final exam schedule is available.ELEC 273 is December 14, from 2:00 to 5:00.

http://users.encs.concordia.ca/%7Etrueman/web_page_273.htm

ELEC 273 Mid-Term ExamSaturday October 20, 2018

14:00 to 15:30

Section R Dr. TruemanH411 A to IH415 J to Z

Section D Dr. FayyazH620H920

Topic areas for the midterm test:

Chapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of Analysis (Node and Mesh)Chapter 4 Circuit Theorems

The test lasts 90 minutes and there will be four questions.

Circuit Theorems

• Linearity• Superposition• Thevenin’s Theorem• Norton’s Theorem• Maximum Power Transfer Theorem

Alexander and Sadiku Chapter 4 “Circuit Theorems”

Linearity

• A circuit is “linear” if multiplying the source by a factor 𝑘𝑘 also multiplies the response by the same factor 𝑘𝑘 .

• The factor 𝑘𝑘 can be positive or negative!• Are real circuits “linear”?

• Yes but over a limited range for the quantities involved. • For example if a circuit is made with resistors that can dissipate 1 watt of

power, then the circuit is linear up to the excitation that makes 𝑃𝑃 = 𝑖𝑖2𝑅𝑅 equal to 1 watt.

• If 𝑖𝑖2𝑅𝑅 exceeds 1 watt, then the resistor burns up and this behavior is not “linear”.

Alexander and Sadiku page 126 (6th edition)

𝑉𝑉𝑠𝑠+𝑣𝑣𝑜𝑜−

LinearSystem 𝑘𝑘𝑉𝑉𝑠𝑠

+𝑘𝑘𝑣𝑣𝑜𝑜−

LinearSystem

Multiply by k

Example - linearity

Demonstrate that this circuit is linear for inputs of 𝑉𝑉𝑠𝑠 = 1 volt and for 𝑉𝑉𝑠𝑠 = 2 volts. Linearity: if we double the source from 1 volt to 2 volts, then we double the response. Method: • find 𝑉𝑉21 for 𝑉𝑉𝑠𝑠 = 1 volt • find 𝑉𝑉22 for 𝑉𝑉𝑠𝑠 = 2 volts. • Show that 𝑉𝑉22 = 2𝑉𝑉21.

𝑅𝑅1 = 1 ohm, 𝑅𝑅2 = 1, 𝑅𝑅3 = 1, and 𝑅𝑅4 = 1

𝑉𝑉𝑠𝑠

𝑉𝑉2𝑉𝑉1+𝑉𝑉2−

𝑅𝑅1𝑅𝑅2

𝑅𝑅3𝑅𝑅4

3𝑉𝑉1 − 𝑉𝑉2 = 1

𝑉𝑉1 − 2𝑉𝑉2 = 0𝑉𝑉1 = 2𝑉𝑉2

So 3𝑉𝑉1 − 𝑉𝑉2 = 1 gives3(2𝑉𝑉2) − 𝑉𝑉2 = 15𝑉𝑉2 = 1

𝑉𝑉2 =15

So with 𝑉𝑉𝑠𝑠 = 1, we find 𝑉𝑉21 = 15

Solve the circuit using node analysis for 𝑉𝑉𝑠𝑠 = 1 :

1 − 𝑉𝑉11

−𝑉𝑉11−𝑉𝑉1 − 𝑉𝑉2

1= 0

𝑉𝑉1 − 𝑉𝑉21

−𝑉𝑉21

= 0

Solve the node equations:1 − 3𝑉𝑉1 + 𝑉𝑉2 = 0

For 𝑉𝑉𝑠𝑠 = 1

(A)

(A)

(B)

(B)

𝑉𝑉𝑠𝑠 = 1

𝑉𝑉2𝑉𝑉1+

𝑉𝑉2 =15

−

𝑅𝑅1𝑅𝑅2

𝑅𝑅3𝑅𝑅4

𝑉𝑉𝑠𝑠

3𝑉𝑉1 − 𝑉𝑉2 =2

𝑉𝑉1 − 2𝑉𝑉2 = 0𝑉𝑉1 = 2𝑉𝑉2

So 3𝑉𝑉1 − 𝑉𝑉2 =2 gives3(2𝑉𝑉2) − 𝑉𝑉2 =25𝑉𝑉2 = 2

𝑉𝑉2 =25

So with 𝑉𝑉𝑠𝑠 =2, we find 𝑉𝑉22 = 25

Solve the circuit using node analysis for 𝑉𝑉𝑠𝑠 = 2 :

2 − 𝑉𝑉11

−𝑉𝑉11−𝑉𝑉1 − 𝑉𝑉2

1= 0

𝑉𝑉1 − 𝑉𝑉21

−𝑉𝑉21

= 0

Solve the node equations:2 − 3𝑉𝑉1 + 𝑉𝑉2 = 0

For 𝑉𝑉𝑠𝑠 = 2

(A)

(A)

(B)

(B)

𝑉𝑉𝑠𝑠 =2

𝑉𝑉2𝑉𝑉1+

𝑉𝑉2 =25

−

𝑅𝑅1𝑅𝑅2

𝑅𝑅3𝑅𝑅4

𝑉𝑉𝑠𝑠

Linearity: if we double the source, then we double the response.

Is 𝑉𝑉22 = 2𝑉𝑉21?25

= 2𝑥𝑥15

Yes, linearity is verified.

So multiplying the source by 𝑘𝑘 = 2multiples the response by 𝑘𝑘 = 2and we have verified that the circuit is linear.

for 𝑉𝑉𝑠𝑠 = 2

𝑉𝑉𝑠𝑠 = 1

𝑉𝑉2𝑉𝑉1+

𝑉𝑉2 =15

−

𝑅𝑅1𝑅𝑅2

𝑅𝑅3𝑅𝑅4

𝑉𝑉𝑠𝑠 =2

𝑉𝑉2𝑉𝑉1+

𝑉𝑉2 =25

−

𝑅𝑅1𝑅𝑅2

𝑅𝑅3𝑅𝑅4

With 𝑉𝑉𝑠𝑠 = 1, we find 𝑉𝑉21 = 15.

With 𝑉𝑉𝑠𝑠 =2, we find 𝑉𝑉22 = 25.

Homework: demonstrate linearity. Use the circuit in Alexander and Sadiku in Fig. 4.75 to demonstrate linearity, as follows:

Solve the circuit with a 1 amp source to find 𝑉𝑉𝑥𝑥1.

Then solve the circuit again with a 10 amp source to find 𝑉𝑉𝑥𝑥1𝑥.

Show that 𝑉𝑉𝑥𝑥1𝑥=10𝑉𝑉𝑥𝑥1.

Superposition

The response of a circuit 𝑉𝑉𝑥 with two sources 𝑉𝑉𝑠𝑠1 and 𝑉𝑉𝑠𝑠2 acting together is equal to the sum of:• the response 𝑉𝑉𝑥1 with source 𝑉𝑉𝑠𝑠1 acting alone and 𝑉𝑉𝑠𝑠2=0 ,• plus the response 𝑉𝑉𝑥2 with source 𝑉𝑉𝑠𝑠2 acting alone and 𝑉𝑉𝑠𝑠1=0,𝑉𝑉𝑥 = 𝑉𝑉𝑥1 + 𝑉𝑉𝑥2.

Remarks:1.Superposition works for current sources as well as voltage sources.2.Superposition can be generalized to N sources.

Alexander and Sadiku page 118 6th edition

Two sources acting together. Source #1 acting alone. Source #2 acting alone.

𝑉𝑉𝑠𝑠1 𝑉𝑉𝑠𝑠2

+𝑉𝑉𝑜𝑜 −

Linear Circuit 𝑉𝑉𝑠𝑠2

+𝑉𝑉𝑜𝑜2 −

Linear Circuit𝑉𝑉𝑠𝑠1

+𝑉𝑉𝑜𝑜1 −

Linear Circuit

Demonstrate the Superposition TheoremTwo sources acting together:

1 − 𝑉𝑉𝑥2

−𝑉𝑉𝑥3−𝑉𝑉𝑥 − 2

1= 0

3 1 − 𝑉𝑉𝑥 − 2𝑉𝑉𝑥 − 6(𝑉𝑉𝑥 − 2) = 0

3 − 3𝑉𝑉𝑥 − 2𝑉𝑉𝑥 − 6𝑉𝑉𝑥 + 12 = 0

11𝑉𝑉𝑥 = 15

𝑉𝑉𝑥 =1511

With both sources acting together, the response is 𝑉𝑉𝑥 = 15

11volts.

1 v 2 v2 Ω

3 Ω

1 Ω

𝑉𝑉𝑥

+𝑉𝑉𝑥−

Source #1 acting alone:Source #1 acting alone, so source #2 is set equal to zero:

1 − 𝑉𝑉𝑥12

−𝑉𝑉𝑥13−𝑉𝑉𝑥1 − 0

1= 0

3 1 − 𝑉𝑉𝑥1 − 2𝑉𝑉𝑥1 − 6(𝑉𝑉𝑥1 − 0) = 0

3 − 3𝑉𝑉𝑥1 − 2𝑉𝑉𝑥1 − 6𝑉𝑉𝑥1 + 0 = 0

11𝑉𝑉𝑥1 = 3

𝑉𝑉𝑥1 =3

11

With source #1 acting alone, the response is 𝑉𝑉𝑥1 = 3

11volts.

1 v

2 Ω

3 Ω

1 Ω

𝑉𝑉𝑥1

+𝑉𝑉𝑥1−

Set source #2 to zero.

Source #2 acting alone, so source #1 is set equal to zero:

0 − 𝑉𝑉𝑥22

−𝑉𝑉𝑥23−𝑉𝑉𝑥2 − 2

1= 0

3 0 − 𝑉𝑉𝑥2 − 2𝑉𝑉𝑥2 − 6(𝑉𝑉𝑥2 − 2) = 0

−3𝑉𝑉𝑥2 − 2𝑉𝑉𝑥2 − 6𝑉𝑉𝑥2 + 12 = 0

11𝑉𝑉𝑥2 = 12

𝑉𝑉𝑥2 =1211

With source #2 acting alone, the response is 𝑉𝑉𝑥2 = 12

11volts.

2 v2 Ω

3 Ω

1 Ω

𝑉𝑉𝑥2

+𝑉𝑉𝑥2−

Source #2 acting alone:

Set source #1to zero.

Source #2 is active!

Add the responses with the sources acting alone:

𝑉𝑉𝑥1 =3

11

Superposition: add the responses.

𝑉𝑉𝑥1 =3

11

𝑉𝑉𝑥2 =1211

So the overall response is 𝑉𝑉𝑥 = 𝑉𝑉𝑥1 + 𝑉𝑉𝑥2.

𝑉𝑉𝑥 =3

11 +1211 =

1511

This agrees with what we calculated when we solved the circuit with the two sources acting together.

1 v

2 Ω

3 Ω

1 Ω

𝑉𝑉𝑥

+𝑉𝑉𝑥−

2 v2 Ω

3 Ω

1 Ω

𝑉𝑉𝑥2

+𝑉𝑉𝑥2−

𝑉𝑉𝑥2 =1211

Superposition with Dependent Sources

Find 𝑣𝑣𝑎𝑎 using the Superposition Theorem.

Dependent sources do not “act alone” and are never set to zero.

Solve the circuit for 𝑣𝑣𝑎𝑎:1) With the 2A source “acting alone” and the 6 v source set

to zero. 2) With the 6 volt source “acting alone” and the 2 A source

set to zero.3) Add 𝑣𝑣𝑎𝑎 from part 1 to 𝑣𝑣𝑎𝑎 from part 2 to find 𝑣𝑣𝑎𝑎 with both

sources acting together.

This old exam question is challenging and requires excellent skills with node analysis!

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

2 amps 6 volts

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1

+𝑣𝑣𝑎𝑎 −

How should we choose the node voltages?This is the most obvious choice: one unknown node voltage at each junction.

There are two “constraint equations”:𝑉𝑉3 = 𝑉𝑉2 + 6𝑉𝑉4 = 𝑉𝑉1 + 3𝑖𝑖1

But there are FOUR unknowns!!!!

Can’t we solve this “smarter” with fewer unknowns?

Answer: yes.Build the “constraint equations” into the node voltages and use only two unknowns.

𝑉𝑉1 𝑉𝑉2 𝑉𝑉3 𝑉𝑉4

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

2 amps 6 volts

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1


What is an efficient way to solve this circuit?

We can solve the circuit with only 2 unknown node voltages by building the “constraint equations” into the node voltages.

Use two node voltages 𝑉𝑉1 and 𝑉𝑉2.

For the 6 volt source, the voltage at the “-” terminal is 𝑉𝑉2 and at the “+” terminal is 𝑉𝑉2 + 6.For the 3𝑖𝑖1 dependent voltage source, the voltage at the “-” terminal is 𝑉𝑉1 and at the “+” terminal is 𝑉𝑉1 + 3𝑖𝑖1.

Build the constraint equations into the node voltages.

𝑉𝑉1 𝑉𝑉2 𝑉𝑉2 + 6 𝑉𝑉1 + 3𝑖𝑖1

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

2 amps 6 volts

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1


Solve the circuit with the current source acting alone:−𝑉𝑉15− 2 +

𝑉𝑉2 − 𝑉𝑉1 + 3𝑖𝑖12

− 2𝑣𝑣𝑎𝑎 = 0

2 −𝑉𝑉22−𝑉𝑉22−𝑉𝑉2 − 𝑉𝑉1 + 3𝑖𝑖1

2= 0

where

𝑖𝑖1 = −𝑉𝑉22

𝑣𝑣𝑎𝑎 = 𝑉𝑉2 − (𝑉𝑉1 + 3𝑖𝑖1) = 𝑉𝑉2 − (𝑉𝑉1 + 3−𝑉𝑉2

2 ) =52𝑉𝑉2 − 𝑉𝑉1

Solve the equations to find 𝑉𝑉1 = 11.44𝑉𝑉2 = 3.43

Hence

𝑣𝑣𝑎𝑎 =52𝑉𝑉2 − 𝑉𝑉1 = −2.87

Voltage source set to zero.

𝑉𝑉1 𝑉𝑉2 𝑉𝑉2 𝑉𝑉1 + 3𝑖𝑖1

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

2 amps

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1


Solve the circuit with the voltage source acting alone:−𝑉𝑉15

+(𝑉𝑉2 + 6) − 𝑉𝑉1 + 3𝑖𝑖1

2− 2𝑣𝑣𝑎𝑎 = 0

−𝑉𝑉22−𝑉𝑉2 + 6

2−

(𝑉𝑉2 + 6) − 𝑉𝑉1 + 3𝑖𝑖12

= 0Where

𝑖𝑖1 = −𝑉𝑉22

𝑣𝑣𝑎𝑎 = (𝑉𝑉2+6) − 𝑉𝑉1 + 3𝑖𝑖1 = 𝑉𝑉2 + 6 − 𝑉𝑉1 + 3−𝑉𝑉2

2Hence

𝑣𝑣𝑎𝑎 =52𝑉𝑉2 − 𝑉𝑉1 + 6

Solve the equations to find 𝑉𝑉1 = −2.13𝑉𝑉2 = −3.14

Hence

𝑣𝑣𝑎𝑎 =52𝑉𝑉2 − 𝑉𝑉1 + 6 = 0.28

𝑉𝑉1 + 3𝑖𝑖1

Current source set to zero.

𝑉𝑉1 𝑉𝑉2 𝑉𝑉2 + 6

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

6 volts

5 Ω2 Ω 2 Ω

2 Ω

𝑖𝑖1


Superposition: add the voltages with the sources acting alone.

Current source acting alone: 𝑣𝑣𝑎𝑎 = −2.87 Voltage source acting alone: 𝑣𝑣𝑎𝑎 = 0.28

Both sources acting together:

𝑣𝑣𝑎𝑎 = −2.87 + 0.28 = −2.59 volts

Homework: Solve the circuit with both sources “acting together” and show that we get the same answer.

Voltage source set to zero.

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

2 amps

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1


Current source set to zero.

𝑉𝑉2 +

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

6 volts

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1


Homework: Alexander and Sadiku Figure 4.92

Find 𝑉𝑉𝑥𝑥 by superposition.

Find 𝑉𝑉𝑥𝑥 with:1) The 40 volt source acting alone and the 3 A

source set to zero, an open circuit.2) The 3 A source acting alone and the 40 V

source set to zero, a short circuit.

Add up 𝑉𝑉𝑥𝑥 with the 40 volt source acting alone plus 𝑉𝑉𝑥𝑥 with the 3 A source acting alone to find 𝑉𝑉𝑥𝑥 with both sources acting together.

Note that for both parts 1) and 2), the 2𝑉𝑉𝑥𝑥dependent source is active.

Superposition: DC source and AC source

This circuit has a 10 volt DC source and a 0.1-volt-amplitude AC source, 0.1 cos𝜔𝜔𝜔𝜔.

Use superposition to find the output voltage 𝑣𝑣𝑜𝑜.

Method:1.Solve the circuit with the DC source alone.2.Solve the circuit with the AC source alone.3.Add up the answers.

Solve for 𝑣𝑣𝑜𝑜 with the DC source acting alone.

+𝑣𝑣𝑜𝑜−

10 volts

0.1cos 𝜔𝜔𝜔𝜔

at 𝑓𝑓 = 1000 Hz

𝑖𝑖

𝑖𝑖

10𝑖𝑖5 kΩ

1 kΩ

10 kΩ10 volts

5 kΩ𝑖𝑖𝑖𝑖

10𝑖𝑖

1 kΩ

10 kΩ+𝑣𝑣𝑜𝑜−

Solve for 𝑣𝑣𝑜𝑜 with the AC source acting alone.



10𝑖𝑖

1 kΩ


Superposition: DC source acting alone.

Solve for 𝑣𝑣𝑜𝑜 with the DC source acting alone.Remark: work with resistors in kilohms, and then currents are in milliamps.

Work with node voltage 𝑉𝑉. Write a KCL equation at V:𝑖𝑖 − 𝑉𝑉

1+ 10𝑖𝑖 = 0

where

𝑖𝑖 =−𝑉𝑉5

Solve for 𝑉𝑉 with 𝑖𝑖 = −𝑉𝑉5

:−𝑉𝑉5− 𝑉𝑉

1+ 10 −𝑉𝑉

5= 0

Multiply by 5:−𝑉𝑉 − 5𝑉𝑉 − 10𝑉𝑉 = 0So𝑉𝑉 = 0And so𝑖𝑖 = 0Since i = 0, we have 10𝑖𝑖 = 0 and the voltage across the10 kΩ resistor is zero.

KVL for the output loop:+𝑣𝑣𝑥 + 10 10𝑖𝑖 − 10 = 0And since 𝑖𝑖 = 0 we have𝑣𝑣𝑜𝑜 = 10 volts.

You have to be confident in your circuit analysis to “believe” this solution!

10 volts5 kΩ

𝑖𝑖𝑖𝑖

10𝑖𝑖

1 kΩ

10 kΩ+𝑣𝑣𝑜𝑜−𝑉𝑉

+10(10𝑖𝑖)

−

Superposition: AC source acting alone.

Solve for 𝑣𝑣𝑜𝑜 with the AC source acting alone.

Write a KCL equation at 𝑉𝑉:

𝑖𝑖 −𝑉𝑉1 + 10𝑖𝑖 = 0

where

𝑖𝑖 =0.1 cos𝜔𝜔𝜔𝜔 − 𝑉𝑉

5

Substitute for 𝑖𝑖 and solve for 𝑉𝑉:0.1 cos𝜔𝜔𝜔𝜔 − 𝑉𝑉

5−𝑉𝑉1

+ 100.1 cos𝜔𝜔𝜔𝜔 − 𝑉𝑉

5= 0

110.1 cos𝜔𝜔𝜔𝜔 − 𝑉𝑉

5− 𝑉𝑉 = 0

Multiply by 5:11(0.1 cos𝜔𝜔𝜔𝜔 − 𝑉𝑉) − 5𝑉𝑉 = 0−11𝑉𝑉 − 5𝑉𝑉 = −1.1 cos𝜔𝜔𝜔𝜔16𝑉𝑉 = 1.1 cos𝜔𝜔𝜔𝜔𝑉𝑉 = 0.06875 cos𝜔𝜔𝜔𝜔

𝑖𝑖 =0.1 cos𝜔𝜔𝜔𝜔 − 𝑉𝑉

5 =0.1 cos𝜔𝜔𝜔𝜔 − 0.06875 cos𝜔𝜔𝜔𝜔

5𝑖𝑖 = −6.25𝑥𝑥10−3 cos𝜔𝜔𝜔𝜔 ampsKVL for the output loop:𝑣𝑣𝑥 + 10(10𝑖𝑖) = 0𝑣𝑣𝑥 = −100𝑖𝑖where 𝑖𝑖 = −6.25𝑥𝑥10−3 cos𝜔𝜔𝜔𝜔𝑣𝑣𝑜𝑜 = −100𝑖𝑖 = 0.625 cos𝜔𝜔𝜔𝜔 volts

The “gain” of this circuit is 𝑣𝑣𝑜𝑜𝑣𝑣𝑖𝑖

= 𝑥.625 cos 𝜔𝜔𝜔𝜔𝑥.1 cos 𝜔𝜔𝜔𝜔

=6.25



10𝑖𝑖

1 kΩ


𝑉𝑉

+10(10𝑖𝑖)

−

Superposition: add the two voltages to find the response with the sources acting together.

𝑣𝑣𝑜𝑜 = 0.625 cos𝜔𝜔𝜔𝜔 volts𝑣𝑣𝑜𝑜 = 10 volts.

Superposition: add the two voltages𝑣𝑣𝑜𝑜 = 10 + 0.625 cos𝜔𝜔𝜔𝜔 volts

Solve for 𝑣𝑣𝑜𝑜 with the DC source acting alone. Solve for 𝑣𝑣𝑜𝑜 with the AC source acting alone.



10𝑖𝑖

1 kΩ


𝑉𝑉

+𝑣𝑣1𝑥−

10 volts5 kΩ

𝑖𝑖𝑖𝑖

10𝑖𝑖

1 kΩ

10 kΩ+𝑣𝑣𝑜𝑜−𝑉𝑉

𝑉𝑉𝑠𝑠

𝑉𝑉2𝑉𝑉1+𝑉𝑉2−

𝑅𝑅1𝑅𝑅2

𝑅𝑅3𝑅𝑅4

1.Find the response 𝑉𝑉21 when the source voltage is 𝑉𝑉𝑠𝑠 = 5 volts, and the response 𝑉𝑉22 when the source voltage is 𝑉𝑉𝑠𝑠 = 100 volts. Use these values to demonstrate that the circuit is linear.

To follow up on the lecture material, solve the problems that were solved in class.

1 v 2 v2 Ω

3 Ω

1 Ω

𝑉𝑉𝑥

+𝑉𝑉𝑥−

2.Find the response 𝑉𝑉𝑥1 when the 1 volt source acts alone, and the response 𝑉𝑉𝑥2 when the 2 volt source acts alone. Show that 𝑉𝑉𝑥1 + 𝑉𝑉𝑥2 is equal to the response of the circuit when both sources act together.

3𝑖𝑖1

2𝑣𝑣𝑎𝑎

2 amps 6 volts

5 Ω 2 Ω 2 Ω

2 Ω

𝑖𝑖1


3.1 Find 𝑣𝑣𝑎𝑎 using the Superposition Theorem.3.2 Solve the circuit with both sources “acting together and show that 𝑣𝑣𝑎𝑎is the same as in part 1.

4. This circuit has a 10 volt DC source and a 0.1-volt-amplitude AC source, 0.1 cos𝜔𝜔𝜔𝜔.4.1 Solve the circuit with the DC source alone.4.2Solve the circuit with the AC source alone.4.3Use superposition to find the output voltage 𝑣𝑣𝑜𝑜 with both sources acting together.

+𝑣𝑣𝑜𝑜−

10 volts


at 𝑓𝑓 = 1000 Hz

𝑖𝑖

𝑖𝑖

10𝑖𝑖5 kΩ

1 kΩ

10 kΩ

Thevenin’ s TheoremAlexander and Sadiku page 137, 6th edition

Charles Levin Thevenin (1857-1925) was a French telegrapher. He published Thevenin’s Theorem in 1883.

Thevenin’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, and resistors is equivalent to a voltage source 𝑉𝑉𝑇𝑇 in series with a resistor 𝑅𝑅𝑇𝑇.We can evaluate 𝑉𝑉𝑇𝑇 and 𝑅𝑅𝑇𝑇 as follows:1) 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜, where 𝑉𝑉𝑜𝑜𝑜𝑜 is the open-circuit voltage, And2) 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

𝐼𝐼𝑠𝑠𝑜𝑜where 𝐼𝐼𝑠𝑠𝑜𝑜 is the short-circuit current.

Two TerminalCircuit

A

B

A

B

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇equivalence

What does “equivalent” mean?

Equivalent means that the voltages and currents in a load are the same using the original circuit and using the Thevenin Equivalent Circuit.

Circuit #1 is connected to circuit #2. Solve circuit #1 and circuit #2 together to find the terminal voltage V and terminal current I.

Replace circuit #1 by its Thevenin Equivalent with 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜, and 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜

for circuit #1.

Then connect the Thevenin Equivalent to circuit #2 and solve for the terminal voltage V and current I.

You will get the same V and I for both configurations.

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇Circuit

#1Circuit

#2Circuit

#2+𝑉𝑉−

+𝑉𝑉−

𝐼𝐼 𝐼𝐼

Find 𝑉𝑉 and 𝐼𝐼. 𝑉𝑉 and 𝐼𝐼 are the same as with the original circuit #1.

Procedure for finding the Thevenin Equivalent CircuitA

B

Problem: Find the Thevenin Equivalent Circuit at terminals AB.

Procedure

1.Open-circuit testFind the voltage 𝑉𝑉𝑜𝑜𝑜𝑜 with an open circuit at terminals AB.

2.Short-circuit testFind the current 𝐼𝐼𝑠𝑠𝑜𝑜 with a short circuit at terminals AB.

3.Find the Thevenin Equivalent voltage source 𝑉𝑉𝑇𝑇𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

4.Find the Thevenin Equivalent resistance 𝑅𝑅𝑇𝑇𝑅𝑅𝑇𝑇 =

𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜

A

B

A

B

+𝑉𝑉𝑜𝑜𝑜𝑜−

𝐼𝐼𝑠𝑠𝑜𝑜

Circuit

Circuit

Circuit

A

B

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇

Example 11. Find the Thevenin Equivalent Circuit at terminals AB.2. If a 5 ohm resistor is connected across AB, then show that the voltage is the same with the original circuit and with the Thevenin Equivalent Circuit.

Open-circuit test: find the voltage with terminals AB open-circuited.1 − 𝑉𝑉𝑜𝑜𝑜𝑜

1 −𝑉𝑉𝑜𝑜𝑜𝑜2 = 0

2 − 2𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉𝑜𝑜𝑜𝑜 = 03𝑉𝑉𝑜𝑜𝑜𝑜 = 2𝑉𝑉𝑜𝑜𝑜𝑜 = 2

3volts so the Thevenin-equivalent voltage source is 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 2

3volts

Short-circuit test: put a short circuit across terminals AB and find the current in the short circuit:𝐼𝐼𝑠𝑠𝑜𝑜 = 1

1= 1 amp

So

𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜

= ⁄2 31

= 23

ohm.

A

B

1 v

1 Ω

2 Ω

A

B

1 v

1 Ω

2 Ω

A

B

1 v

1 Ω

2 Ω



A

B

𝑅𝑅𝑇𝑇 =23𝑉𝑉𝑇𝑇 =

23

Example 1, continuedCheck: The “dead circuit” method.What is the resistance with the source set to zero?The resistance should be the same as 𝑅𝑅𝑇𝑇 found above.

The resistance is 𝑅𝑅𝑇𝑇 = 1𝑥𝑥21+2

= 23

ohm.

So the value of 𝑅𝑅𝑇𝑇 we found above agrees with the ‘dead circuit’ method.

1 Ω

2 Ω 𝑅𝑅𝑇𝑇

A

B


231

1 Ω

2 Ω

equivalence

Example 1, continued

𝑅𝑅𝑇𝑇 = 23

ohm. A

B

2. If a 5 ohm resistor is connected across AB, then show that the voltage is the same with the original circuit and with the Thevenin Equivalent Circuit.

𝑉𝑉𝑇𝑇 = 23

volts

A

B


23

𝑅𝑅𝐿𝐿 = 5 Ω1 11 Ω

2 Ω

1 Ω

2 Ω 𝑅𝑅𝐿𝐿 = 5 Ω+𝑉𝑉−

+𝑉𝑉−

Connect a5 ohm load

1 − 𝑉𝑉1 −

𝑉𝑉2 −

𝑉𝑉5 = 0

10 1 − 𝑉𝑉 − 5𝑉𝑉 − 2𝑉𝑉 = 010 − 10𝑉𝑉 − 5𝑉𝑉 − 2𝑉𝑉 = 017𝑉𝑉 = 10𝑉𝑉 = 1𝑥

17volts

23 − 𝑉𝑉

23

−𝑉𝑉5 = 0

𝑥𝑥103 : 5

23 − 𝑉𝑉 −

23𝑉𝑉 = 0

𝑥𝑥3: 10 − 15𝑉𝑉 − 2𝑉𝑉 = 017𝑉𝑉 = 10𝑉𝑉 = 1𝑥

17volts

Example 2 Find the Thevenin Equivalent Circuit at terminals AB. 1.Open-circuit test: Find 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: Find 𝐼𝐼𝑠𝑠𝑜𝑜.3.𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜4. 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜


Open circuit test:10 − 𝑉𝑉𝑜𝑜𝑜𝑜

5−𝑉𝑉𝑜𝑜𝑜𝑜3−𝑉𝑉𝑜𝑜𝑜𝑜 − 5

2= 0

𝑥𝑥30: 60 − 6𝑉𝑉𝑜𝑜𝑜𝑜 − 10𝑉𝑉𝑜𝑜𝑜𝑜 − 15𝑉𝑉𝑜𝑜𝑜𝑜 + 75 = 031𝑉𝑉𝑜𝑜𝑜𝑜 = 135

𝑉𝑉𝑜𝑜𝑜𝑜 =13531

Short circuit test:

𝑖𝑖1 =105 = 2

𝑖𝑖2 =52

𝐼𝐼𝑠𝑠𝑜𝑜 = 𝑖𝑖1 + 𝑖𝑖2 = 2 + 52

= 92

amps


=1353192

= 13531𝑥𝑥 29

= 27𝑥279

= 3𝑥31

ohms

So 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 13531

volts

A

B

10

5 Ω

3 Ω5

2 Ω

A

B

10

5 Ω

3 Ω5

2 Ω +𝑉𝑉𝑜𝑜𝑜𝑜−

10

5 Ω

3 Ω5

2 Ω𝐼𝐼𝑠𝑠𝑜𝑜

𝑖𝑖1𝑖𝑖2

Example 2, continuedFind the Thevenin Equivalent Circuit at terminals AB.

𝑅𝑅𝑇𝑇= 2 parallel 5 parallel 3

2 parallel 5 = 2𝑥𝑥52+5

= 1𝑥7

= 1𝑥7

1𝑥7

parallel 3 =107 𝑥𝑥3107 +3

=307317

= 3𝑥31

This agrees with 𝑅𝑅𝑇𝑇 = 3𝑥31

ohms found above.

𝑉𝑉𝑇𝑇 =13531

𝑅𝑅𝑇𝑇 =3031

Check: use the dead circuit method to find the Thevenin Equivalent Resistance.

A

B

10

5 Ω

3 Ω5

2 Ω A

B

𝑅𝑅𝑇𝑇

5 Ω

3 Ω2 Ω

Example 3: A circuit with a dependent source.

Find the Thevenin Equivalent Circuit at terminals AB.

10 volts

10 Ω 100 Ω

1 Ω10 Ω

10𝑖𝑖𝑖𝑖

A

B

Method:1. Open-circuit test: leave the terminal open-circuited and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2. Short circuit the terminals and find the short circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜.3. The Thevenin equivalent voltage source is 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜.4. The Thevenin equivalent resistance is 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜


Open-circuit test:

Open Circuit Test

Solve the equations.Multiply by 100:First equation:100 − 10𝑉𝑉1 − 100𝑉𝑉1 − 𝑉𝑉1 + 𝑉𝑉𝑜𝑜𝑜𝑜 = 0111𝑉𝑉1 − 𝑉𝑉𝑜𝑜𝑜𝑜 = 100Second equation: 𝑖𝑖 = 𝑉𝑉1 (how can amps=volts? Answer: 𝑖𝑖 = 𝑉𝑉1

1 Ω)

𝑉𝑉1 − 𝑉𝑉𝑜𝑜𝑜𝑜 − 1000𝑉𝑉1 − 10𝑉𝑉𝑜𝑜𝑜𝑜 = 0−999𝑉𝑉1 − 11𝑉𝑉𝑜𝑜𝑜𝑜 = 0

𝑉𝑉1 = −11

999𝑉𝑉𝑜𝑜𝑜𝑜111𝑉𝑉1 − 𝑉𝑉𝑜𝑜𝑜𝑜 = 100

111−11999 𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉𝑜𝑜𝑜𝑜 = 100

−119 𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉𝑜𝑜𝑜𝑜 = 100

−11𝑉𝑉𝑜𝑜𝑜𝑜 − 9𝑉𝑉𝑜𝑜𝑜𝑜 = 900−20𝑉𝑉𝑜𝑜𝑜𝑜 = 900𝑉𝑉𝑜𝑜𝑜𝑜 = −9𝑥𝑥

2𝑥= −45 volts

𝑖𝑖 =𝑉𝑉11

10 − 𝑉𝑉110 −

𝑉𝑉11 −

𝑉𝑉1 − 𝑉𝑉𝑜𝑜𝑜𝑜100 = 0

𝑉𝑉1 − 𝑉𝑉𝑜𝑜𝑜𝑜100 − 10𝑖𝑖 −

𝑉𝑉𝑜𝑜𝑜𝑜10 = 0

10 volts

10 Ω 100 Ω

1 Ω10 Ω

10𝑖𝑖𝑖𝑖

A

B


𝑉𝑉1

Short-circuit test:

Short Circuit Test

Solve the equation:100 − 100𝑉𝑉1 − 10𝑉𝑉1 − 𝑉𝑉1 = 0

111𝑉𝑉1 = 100

𝑉𝑉1 =100111

Find 𝐼𝐼𝑆𝑆𝑆𝑆 using a KCL equation:𝑉𝑉1

100 − 10𝑖𝑖 − 𝐼𝐼𝑠𝑠𝑜𝑜 = 0𝑉𝑉1

100 − 10𝑉𝑉1 − 𝐼𝐼𝑠𝑠𝑜𝑜 = 0

𝐼𝐼𝑠𝑠𝑜𝑜 =𝑉𝑉1

100 − 10𝑉𝑉1

𝐼𝐼𝑠𝑠𝑜𝑜 =𝑉𝑉1 − 1000𝑉𝑉1

100 =−999𝑉𝑉1

100𝐼𝐼𝑠𝑠𝑜𝑜 = −999𝑉𝑉1

1𝑥𝑥= −999

1𝑥𝑥𝑥𝑥 1𝑥𝑥111

= −9 amps

𝑖𝑖 =𝑉𝑉11

10 − 𝑉𝑉110 −

𝑉𝑉11 −

𝑉𝑉1100 = 0

10 volts

10 Ω 100 Ω

1 Ω10 Ω10𝑖𝑖𝑖𝑖

A

B


𝑉𝑉1

Thevenin Equivalent Circuit

Short circuit test: 𝐼𝐼𝑠𝑠𝑜𝑜 = −9 ampsOpen circuit test: 𝑉𝑉𝑜𝑜𝑜𝑜 = −9𝑥𝑥2𝑥

= −45 volts

𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = −45 volts


= −45−9

= 5 ohms

A

B

𝑅𝑅𝑇𝑇 = 5𝑉𝑉𝑇𝑇 = −45

10 volts

10 Ω 100 Ω

1 Ω10 Ω10𝑖𝑖𝑖𝑖

A

B


𝑉𝑉1

10 volts

10 Ω 100 Ω

1 Ω10 Ω

10𝑖𝑖𝑖𝑖

A

B


𝑉𝑉1

Remarks:1.The dead-circuit method does not work with dependent sources.2.It is possible that 𝑅𝑅𝑇𝑇 < 0 when the circuit contains a dependent sources, see the homework problems.

Homework problem:Alexander and Sadiku problem 4.49.

Find the Thevenin equivalent at terminals ab. 1) Find the open circuit voltage at

terminals ab.

2) Find the short circuit current at terminals ab.

3) Then 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 and 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜

.

4) Verify your value for 𝑅𝑅𝑇𝑇 by finding the dead-circuit resistance at terminals ab.

Example 4: Thevenin with a Dependent Source

Find the Thevenin Equivalent Circuit at terminals AB.

Method:1. Open-circuit test: leave the terminal open-circuited and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2. Short circuit the terminals and find the short circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜.3. The Thevenin equivalent voltage source is 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜.4. The Thevenin equivalent resistance is 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜


A

B

10 Ω1 Ω

200 Ω 𝑖𝑖 100𝑖𝑖10volts

Open-Circuit Test Controlling current:

𝑖𝑖 =10 − 𝑉𝑉1

200

KCL equation at 𝑉𝑉1:10 − 𝑉𝑉1

200−𝑉𝑉11

+ 100𝑖𝑖 = 0KCL equation at 𝑉𝑉𝑜𝑜𝑜𝑜:

−𝑉𝑉𝑜𝑜𝑜𝑜10

− 100𝑖𝑖 = 0

Solve the equations: 10 − 𝑉𝑉1

200 −𝑉𝑉11 + 100

10 − 𝑉𝑉1200 = 0

Multiply by 200: 10 − 𝑉𝑉1 − 200𝑉𝑉1 + 1000 − 100𝑉𝑉1 = 0−𝑉𝑉1 − 200𝑉𝑉1 − 100𝑉𝑉1 = −10 − 1000301𝑉𝑉1 = 1010

𝑉𝑉1 =1010301 = 3.355

KCL equation at 𝑉𝑉𝑜𝑜𝑜𝑜:

−𝑉𝑉𝑜𝑜𝑜𝑜10 − 100𝑖𝑖 = 0

𝑉𝑉𝑜𝑜𝑜𝑜 = −1000𝑖𝑖 = −100010 − 𝑉𝑉1

200 = −50 + 5𝑉𝑉1 = −33.22

Find the voltage 𝑉𝑉𝑜𝑜𝑜𝑜 across the terminals AB of the circuit.

Use node analysis with two node voltages, 𝑉𝑉1 and 𝑉𝑉𝑜𝑜𝑜𝑜.

A

B

10 Ω1 Ω


𝑉𝑉1

𝑉𝑉𝑜𝑜𝑜𝑜


10

Short-circuit test

Hence 𝑖𝑖 = 1𝑥−𝑉𝑉1

2𝑥𝑥= 1𝑥−3.355

2𝑥𝑥= 33.23 mA

The short circuit current is 𝐼𝐼𝑠𝑠𝑜𝑜 = −100𝑖𝑖 = −100𝑥𝑥33.23 = −3.323 amps

Put a short circuit across the terminals and find the short-circuit current 𝐼𝐼𝑠𝑠𝑜𝑜.

The KCL equation at 𝑉𝑉1 is the same as before:10 − 𝑉𝑉1

200 −𝑉𝑉11 + 100𝑖𝑖 = 0

where

𝑖𝑖 =10 − 𝑉𝑉1

200So the solution is the same:𝑉𝑉1 = 3.355

A

B

10 Ω1 Ω


𝑉𝑉1

0 volts


10

Example 4: Find the Thevenin Equivalent Circuit:

Short circuit test;𝐼𝐼𝑠𝑠𝑜𝑜 = −3.323 Amps

Open circuit test: 𝑉𝑉𝑜𝑜𝑜𝑜 = −33.22 volts

𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = −33.22 volts


= −33.22−3.323

= 9.996 ≈ 10 ohms

A

B

10 Ω1 Ω


𝑉𝑉1

0 volts


10A

B

10 Ω1 Ω


𝑉𝑉1



10

A

B

𝑅𝑅𝑇𝑇 = 10𝑉𝑉𝑇𝑇= −33.22

Homework problem: dependent sourceAlexander and Sadiku problem 4.57

Find the Thevenin equivalent circuit at terminals ab.

1) Find the open circuit voltage at terminals ab.

2) Find the short circuit current at terminals ab.

3) Then 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 and 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜

.

4) Verify your value for 𝑅𝑅𝑇𝑇 by finding the dead-circuit resistance at terminals ab. To do set the 50 v source to zero (hence “dead circuit”) and connect a 1 volt source to terminals ab. Find the current 𝑖𝑖 that flows. Then 𝑅𝑅𝑇𝑇 = 1 𝑣𝑣𝑜𝑜𝑣𝑣𝜔𝜔

𝑖𝑖

ELEC273 Lecture Notes Set 5 Circuit TheoremsThe course web site is:http://users.encs.concordia.ca/~trueman/web_page_273.htm

Homework on network theorems.

Mid-Term ExamSaturday October 20, 2018 from 2:00 to 3:30.The mid-term will cover chapters 1 to 4 of Sadiku and Alexander.The mid-term includes homework assignments #1 to 6.

The tentative final exam schedule is available.ELEC 273 is December 14, from 2:00 to 5:00.

http://users.encs.concordia.ca/%7Etrueman/web_page_273.htm

Thevenin’ s TheoremAlexander and Sadiku page 137, 6th edition

Charles Levin Thevenin (1857-1925) was a French telegrapher. He published Thevenin’s Theorem in 1883.

Thevenin’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, and resistors is equivalent to a voltage source 𝑉𝑉𝑇𝑇 in series with a resistor 𝑅𝑅𝑇𝑇.We can evaluate 𝑉𝑉𝑇𝑇 and 𝑅𝑅𝑇𝑇 as follows:1) 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜, where 𝑉𝑉𝑜𝑜𝑜𝑜 is the open-circuit voltage, And2) 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

𝐼𝐼𝑠𝑠𝑜𝑜where 𝐼𝐼𝑠𝑠𝑜𝑜 is the short-circuit current.

Two TerminalCircuit

A

B

A

B

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇equivalence

Review

Procedure for finding the Thevenin Equivalent CircuitA

B

Problem: Find the Thevenin Equivalent Circuit at terminals AB.

Procedure

1.Open-circuit testFind the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜 with an open circuit at terminals AB.

2.Short-circuit testFind the short-circuit current 𝐼𝐼𝑠𝑠𝑜𝑜 with a short circuit at terminals AB.

3.Find the Thevenin Equivalent voltage source 𝑉𝑉𝑇𝑇𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

4.Find the Thevenin Equivalent resistance 𝑅𝑅𝑇𝑇𝑅𝑅𝑇𝑇 =


A

B

A

B



Circuit

Circuit

Circuit

A

B

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇

Review

Example:1. Find the Thevenin Equivalent Circuit at terminals AB.2. If a 5 ohm resistor is connected across AB, then show that the voltage is the same with the original circuit and with the Thevenin Equivalent Circuit.

Open-circuit test: find the voltage with terminals AB open-circuited.1 − 𝑉𝑉𝑜𝑜𝑜𝑜

1 −𝑉𝑉𝑜𝑜𝑜𝑜2 = 0

2 − 2𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉𝑜𝑜𝑜𝑜 = 03𝑉𝑉𝑜𝑜𝑜𝑜 = 2𝑉𝑉𝑜𝑜𝑜𝑜 = 2

3volts so the Thevenin-equivalent voltage source is 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 2

3volts

Short-circuit test: put a short circuit across terminals AB and find the current in the short circuit:𝐼𝐼𝑠𝑠𝑜𝑜 = 1

1= 1 amp

So


= ⁄2 31

= 23

ohm.

A

B

1 v

1 Ω

2 Ω

A

B

1 v

1 Ω

2 Ω

A

B

1 v

1 Ω

2 Ω



A

B


23

Review

Norton’s TheoremAlexander and Sadiku page 145, 5th edition

Norton’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, and resistors is equivalent to a current source 𝐼𝐼𝑁𝑁 in parallel with a conductance 𝐺𝐺𝑁𝑁.We can evaluate 𝐼𝐼𝑁𝑁 and 𝐺𝐺𝑁𝑁 as follows:1) 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜, where 𝐼𝐼𝑠𝑠𝑜𝑜 is the short-circuit current, and2) 𝐺𝐺𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜where 𝑉𝑉𝑜𝑜𝑜𝑜 is the open-circuit voltage.

Also, 𝐺𝐺𝑁𝑁 is the conductance of the circuit with the independent sources set equal to zero.

E.L. Norton was an engineer at Bell Telephone Laboratories and published his theorem 43 years after Thevenin’ Theorem.

Two Terminal Circuit

A

B

A

B

𝐼𝐼𝑁𝑁 𝐺𝐺𝑁𝑁

equivalence

Example #1: Find the Norton equivalent circuit. This is the Thevenin Equivalent of some circuit.

Convert it into a Norton equivalent Circuit.

Short circuit test: The short circuit current is𝐼𝐼𝑠𝑠𝑜𝑜 = 1𝑥

5= 2 amps

So 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝑆𝑆 = 2 amps

Open circuit test: The open circuit voltage is𝑉𝑉𝑜𝑜𝑜𝑜 = 10 voltsSo 𝐺𝐺𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜= 2

1𝑥= 1

5Siemen

The Norton Equivalent Circuit is a current generator 𝐼𝐼𝑁𝑁= 2 amps in parallel with a conductance of 𝐺𝐺𝑁𝑁 =15

Siemen.

A

B

𝑅𝑅𝑇𝑇 = 5 Ω𝑉𝑉𝑇𝑇 = 10volts

105 Ω



5 Ω10

The Norton equivalent circuit:

Check: dead circuit methodWith the voltage source set to 0 volts, the conductance at the terminals is 𝐺𝐺𝑁𝑁 = 1

5Siemen

A

B

𝑅𝑅𝑇𝑇 = 5 Ω𝑉𝑉𝑇𝑇 = 10volts

A

B

𝐼𝐼𝑁𝑁=2 A 𝐺𝐺𝑁𝑁 = 15

S

𝐼𝐼𝑁𝑁 = 2 amps𝐺𝐺𝑁𝑁 = 1

5Siemen

𝑅𝑅𝑇𝑇 = 5 Ω

𝐺𝐺𝑁𝑁 = 15

S

Example #2: Find the Norton Equivalent Circuit.

Short circuit test: Write a mesh equation for 𝐼𝐼𝑠𝑠𝑜𝑜. 13

3 − 𝐼𝐼𝑠𝑠𝑜𝑜 −14𝐼𝐼𝑠𝑠𝑜𝑜 = 0

Multiply by 12:12 − 4𝐼𝐼𝑠𝑠𝑜𝑜 − 3𝐼𝐼𝑠𝑠𝑜𝑜 = 07𝐼𝐼𝑠𝑠𝑜𝑜 = 12The short circuit current is𝐼𝐼𝑠𝑠𝑜𝑜 = 12

7amps

So 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝑆𝑆 = 127

amps

Open circuit test: The open circuit voltage is the voltage across the 3 S conductance with 3 amps flowing.𝑉𝑉𝑜𝑜𝑜𝑜 = 1

3𝑆𝑆𝑥𝑥3𝑆𝑆 = 1 volt

So 𝐺𝐺𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜𝑉𝑉𝑜𝑜𝑜𝑜

=1271

= 127

Siemen

Find the Norton Equivalent Circuit at terminals AB.Note that the circuit diagram shows conductances, and 𝑅𝑅 = 𝐼𝐼

𝐺𝐺.

3 A

3 A

3 A

2 S

3 S

4 S A

B

2 S

3 S

4 S A

B

2 S

3 S

4 S A

B



3 𝐼𝐼𝑠𝑠𝑜𝑜

3

Check: find the dead circuit conductance

With the voltage source set to 0 volts, the conductance at the terminals is 4 S in series with 3s.Conductances in series combine like resistances in parallel, using product over sum:𝐺𝐺𝑁𝑁 = 3𝑥𝑥4

3+4= 12

7Siemens. YES!

The Norton Equivalent Circuit is a current generator 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝑆𝑆 = 12

7amps

in parallel with a conductance of 𝐺𝐺𝑁𝑁 = 12

7Siemen.

A

B

𝐼𝐼𝑁𝑁 = 127

A 𝐺𝐺𝑁𝑁 = 127

S

𝐺𝐺𝑁𝑁

2 S

3 S

4 S

Example #3: Find the Norton Equivalent Circuit at terminals AB:

Short-circuit test: find the short-circuit current.Use one node voltage

−𝑉𝑉210 −

𝑉𝑉210 + 1 = 0

Multiply by 10:-2𝑉𝑉2 = −10𝑉𝑉2 = 5 voltsUse KCL to find the short-circuit current: 𝐼𝐼𝑠𝑠𝑜𝑜 = 1𝑥

4+ 𝑉𝑉2

1𝑥= 5

2+ 5

1𝑥= 3 amps

Find the Norton Equivalent Circuit at terminals AB.

10

4 Ω 10 Ω

10 Ω 1A

B

10

4 Ω 10 Ω

10 Ω1

A

B𝐼𝐼𝑠𝑠𝑜𝑜

0 volts 𝑉𝑉2

Open-circuit test: find the open-circuit voltage.

10 − 𝑉𝑉𝑜𝑜𝑜𝑜4

−𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉2

10= 0

𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉210

−𝑉𝑉210

+ 1 = 0Multiply 1st equation by 20:50 − 5𝑉𝑉𝑜𝑜𝑜𝑜 − 2𝑉𝑉𝑜𝑜𝑜𝑜 + 2𝑉𝑉2 = 07𝑉𝑉𝑜𝑜𝑜𝑜 − 2𝑉𝑉2 = 50And multiply 2nd equation by 10:𝑉𝑉𝑜𝑜𝑜𝑜 − 𝑉𝑉2 − 𝑉𝑉2 = −10𝑉𝑉𝑜𝑜𝑜𝑜 − 2𝑉𝑉2 = −10Subtract7𝑉𝑉𝑜𝑜𝑜𝑜 − 2𝑉𝑉2 = 50𝑉𝑉𝑜𝑜𝑜𝑜 − 2𝑉𝑉2 = −106𝑉𝑉𝑜𝑜𝑜𝑜 = 60𝑉𝑉𝑜𝑜𝑜𝑜 = 10 volt

Open-circuit test.

Or, you can consider 4 and 10 to be in series and write one node equation for 𝑉𝑉2, then find 𝑉𝑉2 directly, and then find 𝑉𝑉𝑜𝑜𝑜𝑜 from 𝑉𝑉2. Try it!

Or, write one mesh equation. Try it!

10

4 Ω 10 Ω

10 Ω1

A

B


𝑉𝑉2𝑉𝑉𝑜𝑜𝑜𝑜

Conclusion

Check: use the dead circuit method.

𝐼𝐼𝑠𝑠𝑜𝑜 = 3 amps𝑉𝑉𝑜𝑜𝑜𝑜 = 10 volts

𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝑆𝑆 = 3 ampsin parallel with a conductance of𝐺𝐺𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜

𝑉𝑉𝑜𝑜𝑜𝑜= 3

1𝑥= 3

1𝑥Siemen

𝑅𝑅𝑇𝑇 = 4𝑥𝑥2𝑥4+2𝑥

= 8𝑥24

= 1𝑥3

ohms

𝐺𝐺𝑁𝑁 = 1𝑅𝑅𝑇𝑇

= 31𝑥

SiemensA

B

4 Ω 10 Ω

10 Ω𝐺𝐺𝑁𝑁

A

B

𝐺𝐺𝑁𝑁

10 Ω

10 Ω 4 Ω

10

4 Ω 10 Ω

10 Ω1

A

B𝐼𝐼𝑠𝑠𝑜𝑜

0 volts 𝑉𝑉2

10

4 Ω 10 Ω

10 Ω1

A

B


𝑉𝑉2𝑉𝑉𝑜𝑜𝑜𝑜

A

B

𝐼𝐼𝑁𝑁 = 3 A 𝐺𝐺𝑁𝑁 = 31𝑥

S

A

B

1 v

1 Ω

2 Ω

Follow up on the lecture by re-solving the problems that were done in class.

6. Find the Thevenin Equivalent Circuit at terminals AB. Find the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜Find the short-circuit current 𝐼𝐼𝑆𝑆𝑆𝑆Find the Thevenin equivalent voltage source 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 and the Thevenin equivalent resistance 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

𝐼𝐼𝑠𝑠𝑜𝑜Use the dead-circuit method to verify that your value for 𝑅𝑅𝑇𝑇 is correct.


𝐼𝐼𝑠𝑠𝑜𝑜Use the dead-circuit method to verify that your value for 𝑅𝑅𝑇𝑇 is correct.

A

B

10

5 Ω

3 Ω5

2 Ω



10 volts

10 Ω 100 Ω

1 Ω10 Ω

10𝑖𝑖𝑖𝑖

A

B



A

B

10 Ω1 Ω


10. Find the Norton Equivalent Circuit at terminals AB. Find the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜Find the short-circuit current 𝐼𝐼𝑆𝑆𝑆𝑆Find the Norton equivalent current source 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜 and the Norton equivalent conductance 𝐺𝐺𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜


3 A

2 S

3 S

4 S A

B

10

4 Ω 10 Ω

10 Ω 1A

B

11. Find the Norton Equivalent Circuit at terminals AB. Find the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜Find the short-circuit current 𝐼𝐼𝑆𝑆𝑆𝑆Find the Norton equivalent current source 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜 and the Norton equivalent conductance 𝐺𝐺𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜


Power and the Maximum Power Transfer TheoremAlexander and Sadiku page 150, 5th edition.

Power delivered to a component:𝑝𝑝 𝜔𝜔 = 𝑣𝑣 𝜔𝜔 𝑖𝑖(𝜔𝜔) wattsEnergy delivered to a component starting at 𝜔𝜔 = 0:𝑤𝑤 𝜔𝜔 = ∫𝑥

𝜔𝜔 𝑝𝑝 𝜔𝜔′ 𝑑𝑑𝜔𝜔𝑑 = ∫𝑥𝜔𝜔 𝑣𝑣 𝜔𝜔′ 𝑖𝑖(𝜔𝜔′)𝑑𝑑𝜔𝜔𝑑 Joules

In D.C. circuits, the voltage and the current are constant so power is constant and is simply𝑝𝑝 = 𝑣𝑣𝑖𝑖 watts

For a resistor in a D.C. circuit, 𝑣𝑣 = 𝑖𝑖𝑅𝑅 so 𝑝𝑝 = 𝑣𝑣𝑖𝑖 = 𝑖𝑖𝑅𝑅 𝑖𝑖 = 𝑖𝑖2𝑅𝑅Or also

𝑝𝑝 = 𝑣𝑣𝑖𝑖 = 𝑣𝑣 𝑣𝑣𝑅𝑅

= 𝑣𝑣2

𝑅𝑅watts

Power delivered by a voltage source of value 𝑉𝑉𝑆𝑆:𝑝𝑝 = 𝑣𝑣𝑖𝑖 = 𝑉𝑉𝑠𝑠𝑖𝑖 watts, where 𝑖𝑖 flows out of the “+” terminal of 𝑉𝑉𝑆𝑆.

Power delivered by a current sources of value 𝐼𝐼𝑆𝑆:𝑝𝑝 = 𝑣𝑣𝑖𝑖 = 𝑣𝑣𝐼𝐼𝑠𝑠 wattswhere 𝑣𝑣 is the voltage across the current source.

𝑣𝑣𝑖𝑖

𝑖𝑖 𝑣𝑣

+𝑉𝑉𝑠𝑠−

𝐼𝐼𝑠𝑠

+𝑣𝑣−

+𝑣𝑣−

𝑖𝑖

𝑖𝑖

The Maximum Power Transfer Theorem

A load resistor 𝑅𝑅 is connected to the terminals of a circuit.What value of 𝑅𝑅 should be used so that 𝑅𝑅 absorbs the largest possible amount of power?

We can replace the circuit by its Thevenin Equivalent Circuit.

The Maximum Power Transfer Theorem: the largest possible power is dissipated in 𝑹𝑹 when 𝑹𝑹 = 𝑹𝑹𝑻𝑻.

The largest power is

𝑝𝑝 = 𝑉𝑉𝑇𝑇2

4𝑅𝑅𝑇𝑇watts

Linear Circuit

A

B

𝑅𝑅 𝑅𝑅𝑅𝑅𝑇𝑇𝑉𝑉𝑇𝑇

A

B

How does the power to R vary with the value of R?To illustrate the variation in power, use𝑉𝑉𝑇𝑇 = 1 volt𝑅𝑅𝑇𝑇 = 1 ohmAnd plot the power delivered to R as a function of R.

𝑝𝑝 = 𝑣𝑣𝑖𝑖 = 𝑅𝑅𝑉𝑉𝑇𝑇𝑅𝑅𝑇𝑇+𝑅𝑅

𝑉𝑉𝑇𝑇𝑅𝑅𝑇𝑇+𝑅𝑅

watts.

• The power increases rapidly from R=0.• The power has a maximum at R=1 ohm, equal to

the value of 𝑅𝑅𝑇𝑇.• The power then gradually declines to zero

asymptotically as R goes to infinity.

For this specific example we have demonstrated that for the maximum power to be dissipated, choose 𝑅𝑅 = 𝑅𝑅𝑇𝑇.

𝑅𝑅𝑅𝑅𝑇𝑇𝑉𝑉𝑇𝑇

Prove the Maximum Power Transfer Theorem

The current is 𝑖𝑖 = 𝑉𝑉𝑇𝑇𝑅𝑅𝑇𝑇+𝑅𝑅

The voltage is 𝑣𝑣 = 𝑅𝑅𝑖𝑖 = 𝑅𝑅𝑉𝑉𝑇𝑇𝑅𝑅𝑇𝑇+𝑅𝑅

The power is 𝑝𝑝 = 𝑣𝑣𝑖𝑖 = 𝑅𝑅𝑉𝑉𝑇𝑇𝑅𝑅𝑇𝑇+𝑅𝑅

𝑉𝑉𝑇𝑇𝑅𝑅𝑇𝑇+𝑅𝑅

= 𝑅𝑅𝑉𝑉𝑇𝑇2

𝑅𝑅𝑇𝑇+𝑅𝑅 2 watts

To find the maximum value of 𝑝𝑝 as 𝑅𝑅 varies, set the derivative of 𝑝𝑝 with respect to 𝑅𝑅 equal to zero, 𝑑𝑑𝑑𝑑𝑑𝑑𝑅𝑅

= 0:

𝑝𝑝 =𝑅𝑅𝑉𝑉𝑇𝑇2

𝑅𝑅𝑇𝑇 + 𝑅𝑅 2 = 𝑅𝑅𝑉𝑉𝑇𝑇2 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2

So use the product rule for 𝑑𝑑𝑑𝑑𝑑𝑑𝑅𝑅

, where the product rule states 𝑑𝑑𝑑𝑑𝑅𝑅

𝑢𝑢 𝑅𝑅 𝑣𝑣(𝑅𝑅) = 𝑢𝑢 𝑑𝑑𝑣𝑣𝑑𝑑𝑅𝑅

+ 𝑣𝑣 𝑑𝑑𝑑𝑑𝑑𝑑𝑅𝑅

𝑑𝑑𝑝𝑝𝑑𝑑𝑅𝑅 =

𝑑𝑑𝑑𝑑𝑅𝑅 𝑅𝑅𝑉𝑉𝑇𝑇2 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2 = 𝑅𝑅

𝑑𝑑𝑑𝑑𝑅𝑅 𝑉𝑉𝑇𝑇2 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2 + 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2 𝑑𝑑

𝑑𝑑𝑅𝑅 𝑅𝑅𝑉𝑉𝑇𝑇2

𝑅𝑅

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇𝑖𝑖

+𝑣𝑣−

Proof of the MPT, continued:𝑑𝑑𝑝𝑝𝑑𝑑𝑅𝑅

= 𝑅𝑅𝑑𝑑𝑑𝑑𝑅𝑅

𝑉𝑉𝑇𝑇2 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2 + 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2 𝑑𝑑𝑑𝑑𝑅𝑅

𝑅𝑅𝑉𝑉𝑇𝑇2

𝑑𝑑𝑝𝑝𝑑𝑑𝑅𝑅

= 𝑅𝑅 −2𝑉𝑉𝑇𝑇2 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −3 𝑑𝑑𝑑𝑑𝑅𝑅

(𝑅𝑅𝑇𝑇 + 𝑅𝑅) + 𝑅𝑅𝑇𝑇 + 𝑅𝑅 −2 𝑉𝑉𝑇𝑇2

𝑑𝑑𝑝𝑝𝑑𝑑𝑅𝑅

= 𝑅𝑅−2𝑉𝑉𝑇𝑇2

𝑅𝑅𝑇𝑇 + 𝑅𝑅 3 +𝑉𝑉𝑇𝑇2

𝑅𝑅𝑇𝑇 + 𝑅𝑅 2

Set the derivative equal to zero:

𝑅𝑅−2𝑉𝑉𝑇𝑇2

𝑅𝑅𝑇𝑇 + 𝑅𝑅 3 +𝑉𝑉𝑇𝑇2

𝑅𝑅𝑇𝑇 + 𝑅𝑅 2 = 0

Divide by 𝑉𝑉𝑇𝑇2

𝑅𝑅−2

𝑅𝑅𝑇𝑇 + 𝑅𝑅 3 +1

𝑅𝑅𝑇𝑇 + 𝑅𝑅 2 = 0

Multiply by 𝑅𝑅𝑇𝑇 + 𝑅𝑅 3

𝑅𝑅 −2 + (𝑅𝑅𝑇𝑇 + 𝑅𝑅) = 0−2𝑅𝑅 + 𝑅𝑅𝑇𝑇 + 𝑅𝑅 = 0−𝑅𝑅 + 𝑅𝑅𝑇𝑇 = 0𝑅𝑅 = 𝑅𝑅𝑇𝑇 We have proven that 𝑑𝑑𝑑𝑑

𝑑𝑑𝑅𝑅= 0 when 𝑅𝑅 = 𝑅𝑅𝑇𝑇.

How much power is dissipated in 𝑅𝑅when 𝑅𝑅 = 𝑅𝑅𝑇𝑇?

𝑖𝑖 =𝑉𝑉𝑇𝑇

𝑅𝑅𝑇𝑇 + 𝑅𝑅hence

𝑣𝑣 = 𝑅𝑅𝑖𝑖 =𝑅𝑅𝑉𝑉𝑇𝑇

𝑅𝑅𝑇𝑇 + 𝑅𝑅With 𝑅𝑅 = 𝑅𝑅𝑇𝑇𝑖𝑖 =

𝑉𝑉𝑇𝑇2𝑅𝑅𝑇𝑇

𝑣𝑣 =𝑅𝑅𝑇𝑇𝑉𝑉𝑇𝑇2𝑅𝑅𝑇𝑇

=𝑉𝑉𝑇𝑇2

𝑝𝑝 = 𝑣𝑣i = 𝑉𝑉𝑇𝑇2

𝑉𝑉𝑇𝑇2𝑅𝑅𝑇𝑇

= 𝑉𝑉𝑇𝑇2

4𝑅𝑅𝑇𝑇watts.

𝑅𝑅

𝑅𝑅𝑇𝑇

𝑉𝑉𝑇𝑇𝑖𝑖

+𝑣𝑣−

Example 1: Maximum Power Transfer Theorem1.What value of load resistor 𝑅𝑅 dissipates the largest possible power?2.With this value of 𝑅𝑅, how much power does 𝑅𝑅 dissipate?

Find the Thevenin Equivalent Circuit.Open circuit test:𝑉𝑉𝑜𝑜𝑜𝑜 = 13𝑥𝑥1𝑥

6+13= 13𝑥

19= 6.842 volts

𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 6.842 volts

Short circuit test:𝐼𝐼𝑠𝑠𝑜𝑜 = 1𝑥

6amps


=13019106

= 13𝑥19

61𝑥

= 4.105 ohms.

1.What value of load resistor 𝑅𝑅 dissipates the largest possible power? Answer: R = 𝑅𝑅𝑇𝑇 = 4.105 ohms

2.With this value of 𝑅𝑅, how much power does 𝑅𝑅 dissipate? Answer: 𝑝𝑝 = 𝑉𝑉𝑇𝑇2

4𝑅𝑅𝑇𝑇= 6.842 2

4𝑥𝑥4.1𝑥5= 2.851 watts

106 Ω

13 Ω𝑅𝑅

106 Ω

13 Ω

106 Ω

13 Ω



A

B

𝑅𝑅𝑇𝑇= 4.105 Ω𝑉𝑉𝑇𝑇

= 6.842volts

Example 2: Maximum Power Transfer Theorem1.What value of load resistor 𝑅𝑅 dissipates the largest possible power?2.With this value of 𝑅𝑅, how much power does 𝑅𝑅 dissipate?

Method:1.Find the Thevenin Equivalent Circuit.• Open circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜• Short circuit current 𝐼𝐼𝑠𝑠𝑜𝑜• 𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

𝐼𝐼𝑠𝑠𝑜𝑜• 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜2.Use the Maximum Power Transfer TheoremChoose 𝑅𝑅 = 𝑅𝑅𝑇𝑇Then the power delivered to 𝑅𝑅 is 𝑝𝑝 = 𝑉𝑉𝑇𝑇

2

4𝑅𝑅𝑇𝑇watts.

10

2 Ω

𝑖𝑖2𝑖𝑖

4 Ω

4 Ω𝑅𝑅

A

B

Find the open-circuit voltage:Consider the two 4-resistors to be in series and write a node equation for 𝑉𝑉1:

Open-Circuit Test: 10 − 𝑉𝑉1

2+ 2𝑖𝑖 −

𝑉𝑉18

= 0

where 𝑖𝑖 = 1𝑥−𝑉𝑉12

10 − 𝑉𝑉12 + 2

10 − 𝑉𝑉12 −

𝑉𝑉18 = 0

310 − 𝑉𝑉1

2 −𝑉𝑉18 = 0

Multiply by 8:12(10 − 𝑉𝑉1) − 𝑉𝑉1 = 0120 − 13𝑉𝑉1 = 0𝑉𝑉1 = 12𝑥

13=9.231 volts

Using voltage divider:𝑉𝑉𝑜𝑜𝑜𝑜 = 4𝑉𝑉1

4+4= 𝑉𝑉1

2= 6𝑥

13=4.615 volts

Replace the two 4-ohm resistors in series with a single 8-ohm resistor.

10

2 Ω

𝑖𝑖2𝑖𝑖

4 Ω

4 Ω

A

B

𝑉𝑉1 𝑉𝑉𝑜𝑜𝑜𝑜


4+4=8 ohms in series

Find the short-circuit current: Short-Circuit Test: 1𝑥−𝑉𝑉12

+ 2𝑖𝑖 − 𝑉𝑉14

= 0 “8” is replaced by “4” because the 4-ohm resistor across AB is shorted out.

where 𝑖𝑖 = 1𝑥−𝑉𝑉12

10 − 𝑉𝑉12

+ 210 − 𝑉𝑉1

2−𝑉𝑉14

= 0

310 − 𝑉𝑉1

2 −𝑉𝑉14 = 0

Multiply by 46(10 − 𝑉𝑉1) − 𝑉𝑉1 = 060 − 7𝑉𝑉1 = 0𝑉𝑉1 = 6𝑥

7=8.571 volts

then𝐼𝐼𝑠𝑠𝑜𝑜 = 𝑉𝑉1

4= 15

7=2.143 amps

10

2 Ω

𝑖𝑖2𝑖𝑖

4 Ω

4 Ω

A

B


Example 2, concluded. 1.What value of load resistor 𝑅𝑅 dissipates the largest possible power?2.With this value of 𝑅𝑅, how much power does 𝑅𝑅 dissipate?

We calculated:

𝐼𝐼𝑠𝑠𝑜𝑜 =2.143 amps𝑉𝑉𝑜𝑜𝑜𝑜 =4.615 volts

So 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 4.615 volts𝑅𝑅𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜

𝐼𝐼𝑠𝑠𝑜𝑜= 4.615

2.143= 2.154 ohms

1.The load resistor that dissipates maximum power is 𝑅𝑅 = 𝑅𝑅𝑇𝑇 = 2.154 ohms.

2.The amount of power dissipated is 𝑝𝑝 = 𝑉𝑉𝑇𝑇2

4𝑅𝑅𝑇𝑇= 4.615 2

4𝑥𝑥2.154= 2.472 watts

10

2 Ω

𝑖𝑖2𝑖𝑖

4 Ω

4 Ω𝑅𝑅

A

B

𝑅𝑅𝑅𝑅𝑇𝑇 =

2.154 Ω𝑉𝑉𝑇𝑇 = 4.615

A

B

Homework problemAlexander and Sadiku problem 4.69

1)Find the value of 𝑅𝑅 which dissipates the maximum amount of power. 2)With the value of 𝑅𝑅 from part 1), how much power does 𝑅𝑅 dissipate?

Method: Disconnect 𝑅𝑅 from terminals ab.Find the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜.Find the short-circuit current 𝐼𝐼𝑠𝑠𝑜𝑜.Find the Thevenin equivalent circuit:

𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝑅𝑅𝑇𝑇 =


The resistor which dissipates the maximum amount of power is 𝑅𝑅 = 𝑅𝑅𝑇𝑇.

The power that 𝑅𝑅 dissipates is 𝑃𝑃 = 𝑉𝑉𝑇𝑇2

4𝑅𝑅𝑇𝑇(𝑤𝑤𝑤𝑤𝑤? )

a

b

Follow up on the lecture by re-solving the problems that were done in class.

12. Find the value of 𝑅𝑅 which dissipates the maximum amount of power. With your chosen value, how much power does 𝑅𝑅 dissipate?

106 Ω

13 Ω𝑅𝑅

10

2 Ω

𝑖𝑖2𝑖𝑖

4 Ω

4 Ω𝑅𝑅

A

B

13. Find the value of 𝑅𝑅 which dissipates the maximum amount of power. With your chosen value, how much power does 𝑅𝑅 dissipate?

elec273 lecture notes set 5 circuit theoremstrueman/elec273files/elec273_5_2… · chapter 4...

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