Circuit Theorems

Circuit Theorems Overview

• Linearity

• Superposition

• Source Transformation

• Thevenin and Norton Equivalents

• Maximum Power Transfer

Portland State University ECE 221 Circuit Theorems Ver. 1.36 1

Linearity Defined

Given a function f(x)

y = f(x)y1 = f(x1)y2 = f(x2)

the function f(x) is linear if and only if

f(a1x1 + a2x2) = a1y1 + a2y2

for any two inputs x1 and x2 and any constants a1 and a2


Example 1: Linearity & Ohm’s Law

Is Ohm’s law linear?

v = f(i)= iR

v1 = i1R

v2 = i2R

f(a1i1 + a2i2) = (a1i1 + a2i2)R

= a1(i1R) + a2(i2R)= a1v1 + a2v2


Example 2: Linearity & Power of Resistors

Is the power dissipated by a resistor a linear function of the current?

p = f(i)= i2R

p1 = i21R

p2 = i22R

f(a1i1 + a2i2) = (a1i1 + a2i2)2 R

= a21i

21R + 2a1a2i1i2 + a2

2i22R

= a1p1 + a2p2


Linear Circuits

• A linear circuit is one whose output is linearly related (or directlyproportional) to its input

• In this class we will only consider circuits in which the voltage andcurrents are linearly related to the independent sources

• For circuits, the inputs are represented by independent sources

• The current through and voltage across each circuit element islinearly proportional to the independent source amplitude

• Will focus on how to apply this principle


Example 3: Linearity & Circuit Analysis

Vs

-

+vo

2 kΩ

4 kΩ4 kΩ

Solve for vo as a function of Vs.


Example 3: Continued

vo =12Vs

• Is vo a linear function of Vs?

• If we had solved the circuit for Vs = 10 V, could we find vo forVs = 20 V without having to reanalyze the circuit?


Example 4: Linearity & Circuit Analysis

Is

-

+voVs 2 kΩ

2 kΩ

Solve for vo as a function of Vs and Is.


Example 4: Continued

vo =12Vs + 1k Is

• If Is = 0, then vo is a linear function of Vs

• If Vs = 0, then vo is a linear function of Is

• This holds true in general

• When used for circuit analysis, this is called superposition


Superposition

• The superposition principle states that the voltage across (orcurrent through) an element in a linear circuit is the algebraic sumof the voltages across (or currents through) that element due toeach independent source acting alone

• To apply this principle for analysis, we follow these steps:

1. Turn off all independent sources except one. Find the output(voltage or current) due to that source.

2. Repeat Step 1 for each independent source.

3. Add the contribution of each source to find the total output.


Example 5: Superposition

10 V-

+vo 2 mA2 kΩ

2 kΩ

Solve for vo using superposition.

• First, find the contribution due to the 10 V source.

• This means we must turn off the current source

• How do you turn off a current source?

• What is the equivalent of turning off a current source?


Example 5: Continued (1)

-

+vo10 V 2 kΩ

2 kΩ

Solve for vo due to the 10 V source.

• Second, find the contribution due to the 2 mA source.

• This means we must turn off the voltage source

• How do you turn off a voltage source?

• What is the equivalent of turning off a voltage source?



-

+vo 2 mA2 kΩ

2 kΩ

Solve for vo due to the 2 mA source.



10 V-

+vo 2 mA2 kΩ

2 kΩ

• Finally, solve for vo by adding the contributions due to bothsources

• What if the 10 V source had been a 20 V source. Is there an easyway to find vo in this case?


Example 6: Superposition

35 V 7 mA

-

+

vo

5 kΩ

20 kΩ

5 iφ

iφ

Use the principle of superposition to find vo.

• We’ll find the contribution due to the 35 V source first

• So we must first turn off the current source



35 V

-

+

vo

5 kΩ

20 kΩ

5 iφ

iφ

Solve for vo with the 7 mA source turned off.



7 mA

-

+

vo

5 kΩ

20 kΩ

5 iφ

iφ

Solve for vo with the 35 V source turned off.


Superposition Final Remarks

• Superposition is based on circuit linearity

• Must analyze as many circuits as there are independent sources

• Dependent sources are never turned off

• As with the examples, is usually more work than combiningresistors, the node voltage analysis, or mesh current analysis

• Is an important idea

• If you want to consider a range of values for an independentsource, is sometimes easier than these methods

• Although multiple circuits must be analyzed, each is simpler thanthe original because all but one of the independent sources isturned off

• Will be necessary when we discuss sinusoidal circuit analysis


Source Transformation Introduction

• Recall that we discussed how to combine networks of resistors tosimplify circuit analysis

– Series combinations

– Parallel combinations

– Delta ↔ Wye Transformations

• We can also apply this idea to certain combinations of sources andresistors


Source Transformation Concept

Is

Vs

R

R

• Source Transformation: the replacement of a voltage source inseries with a resistor by a current source in parallel with a resistoror vice versa

• The two circuits are equivalent if they have the samecurrent-voltage relationship at their terminals


Source Transformation Proof

CircuitElement

No. 1V

I

I

-

+

VCircuit

ElementNo. 2

• A two-terminal circuit element is defined by its voltage-currentrelationship

• Relationship can be found by applying a voltage source to theelement and finding the relationship to current

• Equivalently, can apply a current source and find relationship tovoltage

• If two elements have the same relationship, they are interchangable


Source Transformation Proof Continued

IsVs

R1

R2V

I

I

-

+

V

V − Vs = I R1V

R2= Is + I

V = R1I + Vs V = R2I + R2Is

y = mx + b y = mx + b

Equivalent if and only if R1 = R2 = R and Vs = RIs.


Source Transformation Dependent Sources

Is

Vs

R

R

• Also works with dependent sources

• Arrow of the current source must point towards the positiveterminal of the voltage source

• Does not work if R = 0


Voltage Sources & Resistor Series Equivalent

12 V

5 V 8 V

2 V

13 V

1 Ω

4 Ω

8 Ω

13 Ω

• Recall: Voltage sources in series add

• Recall: Resistors in series add

• Mixture of both in series also has an equivalent

– Equivalent voltage source = sum of the voltages

– Equivalent resistance = sum of the resistors

• Proof possible by KVL (left as exercise)


Current Sources & Resistor Parallel Equivalent

9 A 2 A 5 A 12 A8 Ω 12 Ω 4.8 Ω

• Recall: Current sources in parallel add

• Recall: The conductance of resistors in parallel adds

• Mixture of both in parallel also has an equivalent

– Equivalent current source = sum of the currents

– Equivalent resistance = parallel combination

• Proof possible by KCL (left as exercise)


Example 7: Source Transformation

10 A

300 V

-

+

vo

4 Ω

6 Ω

8 Ω

10 Ω

24 Ω

40 Ω

Use source transformations to find vo.


Example 7: Workspace


Example 8: Source Transformation

4 A

10 V

10 A

io1 Ω

2 Ω

4 Ω

5 Ω

40 Ω

Use source transformations to find io.


Thevenin’s Theorem

VTh

Req

LinearCircuit

• Thevenin’s theorem: a linear two-terminal circuit is electricallyequivalent to a voltage source in series with a resistor

• This applies to any two terminals in a circuit

• This is a surprising result

• Proof is in text; we will focus on how to apply

• Better model of physical power supplies like batteries


Norton’s Theorem

IN Req

LinearCircuit

• Norton’s theorem: a linear two-terminal circuit is electricallyequivalent to a current source in parallel with a resistor

• The Norton equivalent can be obtained by a source transformationof the Thevenin equivalent and vice versa

• This implies RTh = RN and VTh = RThIN

• In lectures, I will denote RN and RTh as simply Req


Finding Thevenin & Norton Equivalents

VTh

Req

IN Req

LinearCircuit

• Recall: Two terminal circuits are only equivalent if they have thesame voltage-current relationship

• This means regardless of what is connected to the terminals, allthree devices must behave the same

• Consider

– Open-Circuit Voltage

– Short-Circuit Current

• This is sufficient, but there are two other methods


Finding Thevenin & Norton Equivalent Resistance

LinearCircuit

AllIndependentSources Set

to Zero

Req

Req

• If we set all of the independent sources equal to zero in all threecircuits, they should all have the same resistance

• With the independent sources removed, it should be relatively easyto find the internal resistance of the circuit

• If the circuit has dependent sources, this can be tricky


Thevenin & Norton Equivalent Resistance Continued

LinearCircuit


to Zero

V

ILinearCircuit


to Zero

I

-

+

V

• If the circuit has dependent sources, we need to find thevoltage-current relationship for the circuit

• Easiest to hook up a voltage source (or current source) andcalculate the current (or voltage)

• The source can have any value

• Then Req = VI

• If the circuit has dependent sources, Req may be negative


Thevenin & Norton Equivalents: Summary

To find the Thevenin or Norton equivalent of a two-terminal circuit,must do two of three tasks

1. Find the open-circuit voltage: Voc

2. Find the short-circuit current: Isc

3. Find the internal resistance: Ri

Then you can find the equivalent values by the following equations

VTh = Voc VTh = IscRi

IN = Isc IN =Voc

Ri

Req = Ri Req =Voc

Isc


Example 9: Thevenin & Norton Equivalents

50 V

a

b

2 kΩ 5 kΩ

20 kΩ

Find the Thevenin equivalent with respect to the terminals a,b.



30 V

1.5 A

a

b

5 Ω

20 Ω

25 Ω

40 Ω

60 Ω

Find the Norton & Thevenin equivalents with respect to the terminalsa,b.



30 V

1.5 A

a

b

5 Ω

20 Ω

25 Ω

40 Ω

60 Ω



50 V

a

b

2 kΩ 5 kΩ 10 kΩ

20 kΩ 40 kΩ50 kΩ

20iβ

iβ

Find the Norton & Thevenin equivalents with respect to the terminalsa,b.



50 V

a

b

2 kΩ 5 kΩ 10 kΩ

20 kΩ 40 kΩ50 kΩ

20iβ

iβ


Maximum Power Transfer

VTh

Req

RL

What load resistance RL will maximize the power absorbed by theresistor?


Maximum Power Transfer Derivation

Goal: Find the value of RL that maximizes the power absorbed by RL.

p = i2RL

=(

VTh

Req + RL

)2

RL

dp

dRL= V 2

Th

(Req + RL) − 2RL

(Req + RL)3

= V 2Th

Req − RL

(Req + RL)3

= 0

This can only be true if RL = Req


Maximum Power Transfer Summary

VTh

Req

RL

• Finding the load resistance that maximizes power transfer isusually a two-step process

1. Find the Thevenin or Norton equivalent

2. Find the load resistance RL


Example 12: Maximum Power Transfer

RL

+ -

100 V

2 Ω 4 Ω

5 Ω

iφ

13iφ

v

v

The variable resistor (RL) is adjusted until it absorbs maximum powerfrom the circuit. Find RL, the maximum power absorbed by RL, andthe percentage of total power developed that is delivered to RL.


Example 12: Workspace (1)

+ -

100 V

2 Ω 4 Ω

5 Ω

iφ

13iφ

v

v



+ -

100 V

2 Ω 4 Ω

5 Ω

iφ

13iφ

v

v


Circuit Theorems

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