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Chapter 5Chapter 5

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Chapter 5 Source Transformation

By KVL:Vs=iRs + v

By KCL: is=i + v/Rp

Vs/Rs=i + v/Rs

is=Vs/RsRs=Rp

is=i + v/Rp

Two circuits have the same terminal voltage and current

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Chapter 5 Source Transformation

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Chapter 5 Source Transformation

Example 1: Find the values of is and R in two circuits if they are equivalent

R=10 Ωis=12/R=1.2 A

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Chapter 5 Source Transformation

Example 2: Find current i in circuit (a)

i=(5V-1.2V)/(5Ω+12Ω)=0.224A

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Chapter 5 Superposition

Example1:(a) A circuit containing two independent sources. (b) The circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter im.

De-activate the voltage sourceDe-activate the current source

i1=6/(3+6)=0.67 Α i2=[3/(3+6)]% 2=0.67 Α im=i1+ i2=1.33 Α

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Chapter 5 Superposition

Example2: a) A circuit containing two independent sources. (b) The circuit after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter vm.

With only the voltage source

With only the current source

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Chapter 5 Superposition

Example3: (a) A circuit containing two independent sources. (b) The circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter im.

With only the voltage source

With only the current source

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Chapter 5 Superposition

Example4: (a) A circuit containing two independent sources. (b) The circuit after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter vm.

With only the voltage source

With only the current source

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Chapter 5 Superposition

Example5:Find current i.

De-activate the voltage sourceDe-activate the current source

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Chapter 5 Superposition

Example5: Find current i.

For circuit (a): by KVL, we have

24-(3+2)i1-3i1=0

Then i1=3 (A)

For circuit (b): we use node voltage analysis at node a

-i2 –7+ (va-3i2)/2=0For 3Ω resistor, we have

-i2=va/3

Then i2=-7/4i=i1 +i2 =1.25 (A)

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Chapter 5 Thévenin’s Theorem

(a) A circuit partitioned into two parts: circuit A and circuit B. (b) Replacing circuit A by its Thévenin equivalent circuit.

A: Driving circuitB: Load

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Chapter 5 Thévenin’s Theorem

The Thévenin equivalent circuit involves three parameters:(a) the open-circuit voltage, voc, (b) the short-circuit current isc, and (c) the Thévenin resistance, Rt.

voc=Rt isc

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Chapter 5 Thévenin’s Theorem

(a) The Thévenin resistance, Rt, (b) A method for measuring or calculating the Thévenin

resistance, Rt.

Rt=vt/it

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Chapter 5 Thévenin’s Theorem

Example1: Find current Example1: Find current ii usingusingThévenin’sThévenin’s TheoremTheorem

Steps for determining the Thévenin equivalent circuit for the circuit left of the terminals

Rt=4+5//20=8Ωvoc

+

-

Voc=20

20+5% 50

=40 V

40i= R+8

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Chapter 5 Thévenin’s Theorem

The Thévenin equivalent circuit involves three parameters:

(a) the open-circuit voltage, voc, (b) the short-circuit current isc, and (c) the Thévenin resistance, Rt.

Example2: Find the Example2: Find the ThThééveninvenin equivalent circuit for:equivalent circuit for:

Keep in mind:

voc=Rt isc

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Chapter 5 Thévenin’s Theorem

First, find Rt:

Rt =10//40 + 4=12Ω

Circuit reduction by de-activate all ideal sourcesThen find the equivalent resistant

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Chapter 5 Thévenin’s Theorem

Then, find voc:

voc

+

-Using Node voltage mothod to find vc, since 1-b is open circuit, no voltage drop for 4Ω resistor, voc= vc

vc -1010

+ 40vc + 2=0

Solve for vc

vc =-8V

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Chapter 5 Thévenin’s Theorem

For circuit with dependent sources, we can not directly obtain the Rt from simple circuit reduction.

The procedure to get Rt :• Find open circuit voltage voc, • Find the short-circuit current isc, Rt =

voc

iscExample 3: Find the Example 3: Find the Thévenin’s Thévenin’s equivalent circuit for the following circuit: equivalent circuit for the following circuit:

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Chapter 5 Thévenin’s Theorem

First, find open circuit voltage Voc

Voc

For the left loop, apply KVL:

20-6i+2i-6i=0

i=2 (A)

Voc =6i=12 (V)

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Chapter 5 Thévenin’s Theorem

Make a-b a short circuit, and find isc,

Rt =voc

isc

By mesh current method, we have

20-6i1+2i1-6(i1-i2)=0

And -6(i2-i1)-10i2 =0 i2 = isc=120/136 (A)

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Chapter 5 Thévenin’s Theorem

The Thévenin’s resistance is

Rt =voc

isc

=120/136

10 =13.6 Ω

The Thévenin’s equivalent circuit is

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Chapter 5 Norton’s equivalent Circuit

(a) A circuit partitioned into two parts: circuit A and circuit B. (b) Replacing circuit A by its Norton equivalent circuit.

Norton equivalent is simply the source transformation of the Norton equivalent is simply the source transformation of the ThéveninThévenin equivalentequivalent

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Chapter 5 Norton’s equivalent Circuit

Example1: Find the Norton Equivalent Circuit for

Rn =6x126+12

= 4 kΩ

Find Rn by replacing the voltage source with a short circuit

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Chapter 5 Norton’s equivalent Circuit

Find the short circuit current isc,

Apply KVL for the large loop:

15-12000isc=0

(Note: No current for the 6kΩ resistor—it is shorted)

isc=1.25mA

isc=1.25mARn =4000Ω

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Chapter 5

First, find the open circuit voltage:

Norton’s equivalent Circuit

Example2: Find the Norton Equivalent Circuit for

vocApply KVL for the close loop:

12+6ia-2ia=0

ia=-3(A) voc=2ia=-6(V)

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Chapter 5

Then, find the short circuit current:

Norton’s equivalent Circuit

iscApply KVL for the left loop:

12+6ia-2ia=0

ia=-3(A)

isc=2ia3 =-2 (A)

Rt =voc

isc= -6 V

-2 A =3Ω

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Chapter 5

For a circuit A and load resistor RL

Maximum Power Transfer

Circuit A contains resistors and independent and dependent sources.

The Thévenin equivalent is substituted for circuit A. Here we use vs for theThévenin source voltage.

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Chapter 5

It can be proved that when RRtt=R=RLL

maximum power transferred from circuit A to the load resistor, and the power is

Maximum Power Transfer

We can also use Norton’s equivalent circuit to substitute circuit A. Here we use is as the Norton source current.

Pmax=vs

2

4Rt

Again, the maximum power occurs at RRtt=R=RLLand the maximum power is

Pmax=Rt is

2

4

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Chapter 5

Example: Find the Load RL that result in maximum power delivered to the load. Also determine Pmax

First, we use the circuit (b) to obtain theThévenin equivalent circuit.

Maximum Power Transfer

Find the open circuit voltage voc. Apply KVL to the close loop:

voc6-6i+2vab-4i=0

And vab=4i

i=3Avoc= vab= 12 (V)

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Chapter 5 Maximum Power Transfer

Find the short circuit current for circuit (c), isc.

6-6isc+2vab=0

isc=1 (A)

Since ab is short, vab=0.Apply KVL to the close loop:

Rt =voc

isc

Find the equivalent resistance:

=12 Ω

RRLL==RRtt=12 =12 ΩΩΩΩΩΩΩΩ Pmax=voc

2

4Rt=

122

4(12)=3 W