chapter 4curt.nelson/engr351/lecture/chap4.pdf · engr228 - chapter 4, nilsson 9e 12 voltage...

Engr228 - Chapter 4, Nilsson 9E 1

Chapter 4

Engr228

Circuit Analysis

Dr Curtis Nelson

Chapter 4 Objectives

• Understand and be able to use the node-voltage method to solve a circuit.

• Understand and be able to use the mesh-current method to solve a circuit.

• Be able to determine which technique is best for a particular circuit.

• Understand source transformations and be able to use them to solve a circuit.

• Understand the concept of Thevenin and Norton equivalent circuits and be able to construct one.

• Know the condition for maximum power transfer to a resistive load and be able to calculate the value of the load resistor that satisfies this condition.


Circuit Analysis

• As circuits get more complicated, we need an organized

method of applying KVL, KCL, and Ohm’s law.

• Nodal analysis assigns voltages to each node, and then we

apply Kirchhoff's Current Law to solve for the node voltages.

• Mesh analysis assigns currents to each mesh, and then we

apply Kirchhoff’s Voltage Law to solve for the mesh currents.


Review - Nodes, Paths, Loops, Branches

• These two networks are equivalent.

• There are three nodes and five branches:

– Node: a point at which two or more elements have a common connection point.

– Branch: a single path in a network composed of one simple element and the node at each end of that element.

• A path is a sequence of nodes.

• A loop is a closed (circular) path.

Review - Kirchhoff’s Current Law

• Kirchhoff’s Current Law (KCL) states that the algebraic sum

of all currents entering a node is zero.

iA + iB + (−iC) + (−iD) = 0


Review - Kirchhoff’s Voltage Law

• Kirchhoff’s Voltage Law (KVL) states that the algebraic sum

of the voltages around any closed path is zero.

-v1 + v2 + -v3 = 0

Node Example

• Node = every point along the same wire

10V

4K

6K V

3 nodes


Nodes

How many nodes are there in the circuit(s) below?

Notes on Writing Nodal Equations

• All terms in the equations are in units of current.

• Everyone has their own style of writing nodal equations

– The important thing is that you remain consistent.

• Probably the easiest method if you are just getting started is to

remember that:

Current entering a node = current leaving the node

• Current directions can be assigned arbitrarily, unless they are

previously specified.


The Nodal Analysis Method

• Assign voltages to every node relative to a reference node.

Choosing the Reference Node

• By convention, the bottom node is often the reference node.

• If a ground connection is shown, then that becomes the

reference node.

• Otherwise, choose a node with many connections.

• The reference node is most often assigned a value of 0.0 volts.


Apply KCL to Find Voltages

• Assume reference voltage = 0.0 volts

• Assign current names and directions

• Apply KCL to node v1 ( Σ out = Σ in)

• Apply Ohm’s law to each resistor:

v1

2+

v1 − v2

5= 3.1

Apply KCL to Find Voltages

)4.1(1

0

5

221 −+−

=− vvv

We now have two equations for the two unknowns v1

and v2 and we can solve them simultaneously:

v1 = 5V and v2 = 2V

• Apply KCL to node v2 ( Σ out = Σ in)

• Apply Ohm’s law to each resistor:


Example: Nodal Analysis

Find the current i in the circuit below.

Answer: i = 0 (since v1=v2=20 V)

Nodal Analysis: Dependent Source Example

Determine the power supplied

by the dependent source.

Key step: eliminate i1 from the

equations using v1=2i1

Answer: 4.5 kW being generated

2

0

115 121 −

+−

=vvv

3

03

1

21

21 −=+

− vi

vv

2

011

−=

vi


Example #1

6

132

1

022

5

21

31

=

−=+

−

=

V

VVV

V

0V

V1 V2

Example #2

• How many nodes are in this circuit?

• How many nodal equations must you write to solve for the

unknown voltages?

V1

V2

V3

0V


132332417

0241436331396

03

213

4

318

−=−−

=−++−+

=−

++−

+

VVV

VVVV

VVVV

V1 V2 V3

0V

At node V1

Example #2 – node V1

183321112

0263323181222

01

02

2

323

3

12

=−+−

=+−+−−

=−

+−

+−−

VVV

VVVVV

VVVVVAt node V2


V1 V2 V3

0V


50031921015

0345001535210310

05

0325

4

13

2

23

=+−−

=+−−+−

=−

+−−

+−

VVV

VVVVV

VVVVVAt node V3


V1 V2 V3

0V

All 3 Equations

50031921015

183321112

132332417

=+−−

=−+−

−=−−

VVV

VVV

VVV

Answer: V1 = 0.956V

V2 = 10.576V

V3 = 32.132V


Voltage Sources and the Supernode

If there is a DC voltage source between two non-reference

nodes, you can get into trouble when trying to use KCL

between the two nodes because the current through the voltage

source may not be known, and an equation cannot be written

for it. Therefore, we use a supernode instead.

The Supernode Analysis Technique

22

2551

343

8334

23

323121

2131

=−

−++−=−

+−

−−=−

+−

vv

vvvvvv

vvvv

• Apply KCL at Node v1.

• Apply KCL at the supernode.

• Add the equation for the

voltage source inside the

supernode.

v1 = 1.0714V

v2 = 10.5V

v3 = 32.5V


Modified Example

V1 V2 V3

0V

132332417

0241436331396

03

213

4

318

−=−−

=−++−+

=−

++−

+

VVV

VVVV

VVVVAt node V1

Modified Example – node V1

V1 V2 V3

0V


132

1680327280135

02603121500115315180120220

01

02

5

0325

4

133

3

12

=−

=++−

=++−−+−−

=−

+−

+−−

+−−

VV

VVV

VVVVVV

VVVVVVAt node V2

Supernode

Modified Example – node V2, supernode

supernode

V1 V2 V3

0V

132

1680327280135

132332417

=−

=++−

−=−−

VV

VVV

VVV

V1 = -4.952 V

V2 = 14.333 V

V3 = 13.333 V

Solving Simultaneous Equations


Mesh Analysis: Nodal Alternative

• A mesh is a loop that does not contain any other loops within it.

• In mesh analysis, we assign mesh currents and solve using KVL.

• All terms in the equations are in units of voltage.

• Remember – voltage drops in the direction of current flow except

for sources that are generating power.

• The circuit below has four meshes:

Mesh Example


The Mesh Analysis Method

Mesh currents

Branch currents

Mesh: Apply KVL

Apply KVL to mesh 1

( Σ drops = 0 ):

-42 + 6i1 +3(i1-i2) = 0

Apply KVL to mesh 2

( Σ drops = 0 ):

3(i2-i1) + 4i2 -10 = 0

i1 = 6A

i2 = 4A


Example: Mesh Analysis

Determine the power supplied by the 2 V source.

Applying KVL to the meshes:

−5 + 4i1 + 2(i1 − i2) − 2 = 0

+2 + 2(i2 − i1) + 5i2 + 1 = 0

i1 = 1.132 A i2 = −0.1053 A

Answer: -2.474 W (the source is generating power)

A Three Mesh Example

Follow each

mesh clockwise

Simplify

Solve the equations:

i1 = 3 A, i2 = 2 A, and i3 = 3 A


Example

• Use mesh analysis to determine Vx

I1

I2

I3

6362312

03)23(36)13(2

033261

0)32(322)12(1

132213

0)31(26)21(17

=+−−

=+−+−−

=−+−

=−++−

=−−

=−++−+−

III

IIIII

III

IIIII

III

IIII

I1 = 3A, I2 = 2A, I3 = 3A

Vx = 3(I3-I2) = 3V

Equation I

Equation II

Equation III

Example - continued

I1

I2

I3


Current Sources and the Supermesh

• If a current source is present in the network and shared

between two meshes, then you must use a supermesh formed

from the two meshes that have the shared current source.

We can eliminate the need for

introducing a voltage variable

by applying KVL to the

supermesh formed by joining

meshes 1 and 3.

Supermesh Example

• Use mesh analysis to evaluate Vx

I1

I2

I3


033261

0)32(322)12(1

=−+−

=−++−

III

IIIIILoop 2:

I1

I2

I3

Equation I

Supermesh Example - continued

I1

I2

I3

Supermesh

731

734241

03)23(3)21(17

=−

=+−

=+−+−+−

II

III

IIIII

Equation II

Equation III

Supermesh Example - continued

I1 = 9A

I2 = 2.5A

I3 = 2A

Vx = 3(I3-I2) = -1.5V


Node or Mesh: How to Choose?

• Use the one with fewer equations, or

• Use the method you like best, or

• Use both (as a check).

Example

Does nodal analysis or loop analysis have fewer equations?

0V

7V

V1 V2

V3


Find i1

Dependent Source Example

Answer: i1 = - 250 mA.

Dependent Source Example

Find Vx

I1

I2

I3


I1

I2

I3

I1=15A, I2=11A, I3=17A

Vx = 3(17-11) = 18V

)23(3

9

113

033261

0)32(322)12(1

151

IIVx

VxII

III

IIIII

AI

−=

=−

=−+−

=−++−

= Equation I

Equation II

Equation III

Equation IV

Dependent Source Example - continued

Textbook Problem 4.52 Hayt 7E

• Obtain a value for the current labeled i10 in the circuit below

I10 = -4mA


Linear Elements and Circuits

• A linear circuit element has a linear voltage-current

relationship:

– If i(t) produces v(t), then Ki(t) produces Kv(t)

– If i1(t) produces v1(t) and i2(t) produces v2(t), then i1(t) + i2(t) produces

v1(t) + v2(t),

• Resistors and sources are linear elements

– Dependent sources need linear control equations to be linear elements

• A linear circuit is one with only linear elements

The Superposition Concept

For the circuit shown below, the question is:

• How much of v1 is due to source ia, and how much is due to

source ib?

We will use the superposition principle to answer this question.


The Superposition Theorem

In a linear network, the voltage across or the current through

any element may be calculated by adding algebraically all the

individual voltages or currents caused by the separate

independent sources acting “alone”, i.e. with

– All other independent voltage sources replaced by short circuits (i.e. set

to a zero value) and

– All other independent current sources replaced by open circuits (also

set to a zero value).

Applying Superposition

• Leave one source ON and turn all other sources OFF:

– Voltage sources: set v=0

These become short circuits.

– Current sources: set i=0

These become open circuits.

• Then, find the response due to

that one source

• Add the responses from the other

sources to find the total response


Superposition Example (Part 1 of 4)

Use superposition to solve for the current ix


First, turn the current source off:

′ i x =3

6+9= 0.2



Then, turn the voltage source off:

ix′′ =

6

6 + 9(2) = 0.8


Finally, combine the results:

ix = ix′ + ix

′′ = 0.2 + 0.8 =1.0


Example: Power Ratings

Determine the maximum positive current to which the source Ix

can be set before any resistor exceeds its power rating.

Answer: Ix < 42.49 mA

Superposition with a Dependent Source

• When applying superposition to circuits with dependent

sources, these dependent sources are never “turned off.”


Superposition with a Dependent Source

Current source off

Voltage source off

ix = ix’+ix

’’=2 + (−0.6) = 1.4 A

Superposition Example

Find voltage Vx. First, solve by nodal analysis.

(42-Vx)/6 = (Vx-(-10))/4 + Vx/3

Vx = 54/9 = 6 volts


V

Vx V

333.9

42)7/12(6

)7/12(42

)4||3(6

)4||3()42(

=

×+

=×+

=

Example - Continued

V

Vx V

333.3

1042

210

4)3||6(

)3||6()10(

−=

×+

−=×+

−=

Example - Continued


V

VxVxVx VV

6333.3333.9

)10()42(

=−=

+=

Example - Continued

Source Transformation and Equivalent Sources

The sources are equivalent if

Rs=Rp and vs=isRs


Source Transformation

• The circuits (a) and (b) are

equivalent at the terminals.

• If given circuit (a), but circuit (b) is

more convenient, switch them.

• This process is called source

transformation.

Example: Source Transformation

We can find the current I in the circuit below using a source

transformation, as shown.

I = (45-3)/(5+4.7+3) = 3.307 mA

I = 3.307 mA


Example

Use source transformations to find Ix

Example - continued


Example - continued

AI X3

1

6

2

321

2==

++=

Example - continued



• Using source transformations, determine the power

dissipated by the 5.8 kΩ resistor.

42.97 mW


(a) Determine the individual contributions of each of the two

current sources to the nodal voltage v1

(b) Determine the power dissipated by the 2Ω resistor

v17A = 6.462V, v14A = -2.154V, v1tot = 4.31V, P2Ω= 3.41W



Determine the current labeled i after first transforming the

circuit such that it contains only resistors and voltage

sources.

i = -577µA


Using source transformations, reduce the circuit to a voltage

source in series with a resistor, both of which are in series

with the 6 MΩ resistor.

Vs = -6.284V Rs = 825.4 kΩ



P7v = 17.27W (generating)

Find the power generated by the 7V source.

Thévenin Equivalent Circuits

Thévenin’s theorem: a linear network can be replaced by its

Thévenin equivalent circuit, as shown below:


Thévenin Equivalent using Source Transformations

• We can repeatedly

apply source

transformations on

network A to find its

Thévenin equivalent

circuit.

• This method has

limitations - not all

circuits can be source

transformed.

Finding the Thévenin Equivalent

• Disconnect the load.

• Find the open circuit voltage voc.

• Find the equivalent resistance Req of the network with all

independent sources turned off.

Then:

VTH = voc and

RTH = Req


Thévenin Example

Example

Find Thevenin’s equivalent circuit and the current

passing thru RL given that RL = 1Ω


0V

10V

0V 0V

6V 6V

VVTH 61032

3=×

+=

Find VTH

Example - continued

Short voltage source

RTH

Ω=

++

×+=

++=

2.13

232

3210

23||210THR

Example - continued

Find RTH


Thevenin’s equivalent circuit

The current thru RL = 1Ω is A423.012.13

6=

+

Example - continued

Example

Find Thevenin’s equivalent circuit


Find VTH

0V

5V

0V 0V

3V 3V

VVTH 331 =×=

Example - continued

Open circuit

current source

RTH

Ω=

++=

15

2310THR

Find RTH

Example - continued



Example - continued

Example: Bridge Circuit

Find Thevenin’s equivalent circuit as seen by RL


Find VTH

0V

10V

8V 2V

VTH = 8-2 = 6V

Example - continued

Find RTH

RTH

Example - continued


KKK

KKKKRTH

4.28.06.1

1||48||2

=+=

+=

Example - continued


Example - continued


Thevenin’s Equivalent Circuit - Recap

Norton Equivalent Circuits

Norton’s theorem: a linear network can be replaced by its Norton

equivalent circuit, as shown below:


Finding the Norton Equivalent

• Replace the load with a short circuit.

• Find the short circuit current isc.

• Find the equivalent resistance Req of the network with all

independent sources turned off.

Then:

IN = isc and RN = Req

Source Transformation: Norton and Thévenin

• The Thévenin and Norton equivalents are source

transformations of each other.

RTH=RN =Req and vTH=iNReq


Example: Norton and Thévenin

Find the Thévenin and Norton equivalents for the network faced

by the 1-kΩ resistor.

The load

resistor

This is the circuit we will simplify

Example: Norton and Thévenin

NortonThévenin

Source

Transformation


One method to find the Thévenin equivalent of a circuit with a

dependent source: find VTH and IN and solve for RTH=VTH / IN

Thévenin Example: Handling Dependent Sources


Finding the ratio VTH / IN fails when both quantities are zero!

Solution: apply a test source.



Solve: vtest =0.6 V, and so RTH = 0.6 Ω

v test

2+

v test − (1.5i)

3=1

i = −1

Example

• Find Norton’s equivalent circuit and the current

through RL if RL = 1Ω


Find IN

Find R total:

Find I total:

Current divider:

Ω=+

×+=++ 4.4

123

1232)210(||32

AR

VI 27.2

4.4

10===

AI SC 45.027.2123

3=×

+=

Example - continued

Ω=

++

×+=

++=

2.13

232

3210

23||210THR

Find Rn

Example - continued


Norton’s Equivalent Circuit

The current thru RL = 1Ω is A418.045.012.13

2.13=×

+

Relationship Between Thevenin and Norton

I

VVOC

ISC

Slope = - 1/RTH

THNTH

NTH

RIV

RR

×=

=


Norton’s equivalent circuitThevenin’s equivalent circuit

Same R value

2.1345.06 ×=

THNTH

NTH

RIV

RR

×=

=

Recap: Thevenin and Norton


Find the Thevenin equivalent of the circuit below.

RTH = 8.523 kΩ

VTH = 83.5 V


Equivalent Circuits with Dependent Sources

We cannot find RTH in circuits with dependent sources using

the total resistance method.

But we can use

SC

OCTH

I

VR =

Example

Find Thevenin and Norton’s equivalent circuits


Find Voc

I1 I2

12400014250

0)21(400012501

=−

=−++−

II

IIIKVL

loop1

024060801404000

)21(4000

010028022000)12(4000

=+−

−=

=−++−

II

IIVx

VxIIIIKVL

loop2

Example - continued

I1 I2

Solve equations

I1 = 3.697mA I2 = 3.678mA

V

mA

IIIVxIVOC

3.7

)678.3697.3(400000)678.3(80

)21(400000280100280

−=

−−=

−−=−=

Example - continued


Find Isc

I1 I2 I3

12400014250

0)21(400012501

=−

=−++−

II

IIIKVL

loop1

038024060801404000

)21(4000

0100)32(8022000)12(4000

=−+−

−=

=−−++−

III

IIVx

VxIIIIIKVL

loop2

KVL

loop3 038024000801400000

0100)23(80

=+−

=+−

III

VxII

Example - continued

Find Isc

I1 I2 I3

I1 = 0.632mA

I2 = 0.421mA

I3 = -1.052 A

Isc = I3 = -1.052 A

Example - continued


Ω=−

−== 94.6

052.1

28.7

SC

OCTH

I

VR

Thevenin’s equivalent circuit Norton’s equivalent circuit

Example - continued

Maximum Power Transfer

Thevenin’s or Norton’s equivalent circuit, which has an RTH

connected to it, delivers a maximum power to the load RL for

which RTH = RL


2

22

2

)( LTH

LTHL

LTH

TH

L

RR

RVR

RR

VP

RIP

+=⋅

+=

=LTH

TH

RR

VI

+=and

Plug it in

0)(

)(2)(4

222

=+

+⋅−+=

LTH

LTHLTHTHLTH

L RR

RRRVVRR

dR

dP

Maximum Power Theorem Proof

LTH

LLTH

LTHLTHTHLTH

RR

RRR

RRRVVRR

=

=+

+⋅=+

2)(

)(2)(222

0)(

)(2)(4

222

=+

+⋅−+=

LTH

LTHLTHTHLTH

L RR

RRRVVRR

dR

dP

For maximum power transfer

Maximum Power Theorem Proof - continued


Example

Evaluate RL for maximum power transfer and find the power.


RL should be set to 13.2Ω to get max. power transfer

Max. power is WR

V68.0

2.13

)2/6( 22

==

Example - continued


Practical Voltage Sources

• Ideal voltage sources: a first approximation model for a

battery.

• Why do real batteries have a current limit and experience

voltage drop as current increases?

• Two car battery models:

Practical Source: Effect of Connecting a Load

For the car battery example:

VL = 12 – 0.01 IL

This line represents

all possible RL


Practical Voltage Source

s

sLsc

soc

R

vi

vv

=

=

LssL iRvv −=

OR

s

L

s

sL

R

v

R

vi −=

Practical Current Source

sLsc

spLoc

ii

iRv

=

=

p

LsL

R

vii −=


Equivalent Practical Sources

p

LsL

R

vii −=

s

L

s

sL

R

v

R

vi −=

• These two practical sources are equivalent if

RS = RP S

SS

R

vi =

Chapter 4 Summary

• Illustrated the node-voltage method to solve a circuit.

• Illustrated the mesh-current method to solve a circuit.

• Practiced choosing which technique is better for a particular

circuit.

• Explained source transformations and how to use them to

solve a circuit.

• Illustrated the techniques of constructing Thevenin and Norton

equivalent circuits.

• Explained the principle of maximum power transfer to a

resistive load and showed how to calculate the value of the

load resistor that satisfies this condition.

chapter 4curt.nelson/engr351/lecture/chap4.pdf · engr228 - chapter 4, nilsson 9e 12 voltage...

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