Chapter 03, Resistive Network Analysis

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3

3.1

Chapter 3 Instructor Notes Chapter 3 presents the principal topics in the analysis of resistive (DC) circuits. The presentation of node voltage and mesh current analysis is supported by several solved examples and drill exercises, with emphasis placed on developing consistent solution methods, and on reinforcing the use of a systematic approach. The aim of this style of presentation, which is perhaps more detailed than usual in a textbook written for a non-majors audience, is to develop good habits early on, with the hope that the orderly approach presented in Chapter 3 will facilitate the discussion of AC and transient analysis in Chapters 4 and 5. Make The Connection sidebars (pp. 75-77) introduce analogies between electrical and thermal circuit elements. These analogies are encountered again in Chapter 5. A brief discussion of the principle of superposition precedes the discussion of Thèvenin and Norton equivalent circuits. Again, the presentation is rich in examples and drill exercises, because the concept of equivalent circuits will be heavily exploited in the analysis of AC and transient circuits in later chapters. The Focus on Methodology boxes (p.76 – Node Analysis; p. 86 – Mesh Analysis; pp. 103, 107, 111 – Equivalent Circuits) provide the student with a systematic approach to the solution of all basic network analysis problems. After a brief discussion of maximum power transfer, the chapter closes with a section on nonlinear circuit elements and load-line analysis. This section can be easily skipped in a survey course, and may be picked up later, in conjunction with Chapter 9, if the instructor wishes to devote some attention to load-line analysis of diode circuits. Finally, those instructors who are used to introducing the op-amp as a circuit element, will find that sections 8.1 and 8.2 can be covered together with Chapter 3, and that a good complement of homework problems and exercises devoted to the analysis of the op-amp as a circuit element is provided in Chapter 8. The homework problems present a graded variety of circuit problems. Since the aim of this chapter is to teach solution techniques, there are relatively few problems devoted to applications. We should call the instructor's attention to the following end-of-chapter problems: 3.8 and 3.19 on the Wheatstone bridge; 3.21, 3.22, 3.23, on three-wire residential distribution service; 3.24, 3.25, 3.26 on AC three-phase electrical distribution systems; 3.28-3.31 on fuses; 3.62-66 on various nonlinear resistance devices.

Learning Objectives 1. Compute the solution of circuits containing linear resistors and independent and

dependent sources using node analysis. 2. Compute the solution of circuits containing linear resistors and independent and

dependent sources using mesh analysis. 3. Apply the principle of superposition to linear circuits containing independent sources. 4. Compute Thévenin and Norton equivalent circuits for networks containing linear

resistors and independent and dependent sources. 5. Use equivalent circuits ideas to compute the maximum power transfer between a source

and a load. 6. Use the concept of equivalent circuit to determine voltage, current and power for

nonlinear loads using load-line analysis and analytical methods.


Sectio

P

SolutioKnown quCircuit shoFind: The branc

a) Rb) Rc) R

Analysis: a) Assumnode A:

KCL: −

I

This can adirection o

b) Assumnode B:

KCL: I

ns 3.1, 3.2, 3.3, 3.4: Nodal and Mesh Analysis I

cus on Methodology: Node Voltage Analysis Method 1. Select a reference node(usually ground). This node usually has most elements tied to it.

All other nodes will be referenced to this node. 2. Define the remaining n–1 node voltages as the independent or dependent variables.

Each of the m voltage sources in the circuit will be associated with a dependent variable. If a node is not connected to a voltage source, then its voltage is treated as an independent variable.

3. Apply KCL at each node labeled as an independent variable, expressing each current in terms of the adjacent node voltages.

3.2

roblem 3.1

n: antities: wn in Figure P2.1 with mesh currents: I1 = 5 A, I2 = 3 A, I3 = 7 A.

h currents through: 1, 2, 3.

e a direction for the current through 1R (e.g., from node A to node B). Then summing currents at

0311 =++ III R

A2311 −=−= IIR

lso be done by inspection noting that the assumed direction of the current through R1 and the f I1 are the same.

e a direction for the current through R2 (e.g., from node B to node A). Then summing currents at

0322 =−+ II R

Focus on Methodology: Mesh Current Analysis Method 1. Define each mesh current consistently. Unknown mesh currents will be always defined

in the clockwise direction; known mesh currents (i.e., when a current source is present) will always be defined in the direction of the current source.

2. In a circuit with n meshes and m current sources, n–m independent equations will result. The unknown mesh currents are the n–m independent variables.

3. Apply KVL to each mesh containing an unknown mesh current, expressing each voltage in terms of one or more mesh currents..

4. Solve the linear system of n–m unknowns.


3.3

A4232 =−= III R

This can also be done by inspection noting that the assumed direction of the current through R2 and the direction of I3 are the same.

c) Only one mesh current flows through R3. If the current through R3 is assumed to flow in the same direction, then:

A731 == II R .

______________________________________________________________________________________

Problem 3.2

Solution: Known quantities: Circuit shown in Figure P3.1 with source and node voltages:

VVVVVVV BASS 107,103,11021 −==== .

Find: The voltage across each of the five resistors. Analysis: Assume a polarity for the voltages across R1 and R2 (e.g., from ground to node A, and from node B to ground). R1 is connected between node A and ground; therefore, the voltage across R1 is equal to this node voltage. R2 is connected between node B and ground; therefore, the voltage across R2 is equal to the negative of this voltage.

V107,V103 21 +=−=== BRAR VVVV The two node voltages are with respect to the ground which is given. Assume a polarity for the voltage across R3 (e.g., from node B to node A). Then: KVL: 03 =++ BRA VVV

V2103 =−= BAR VVV

Assume polarities for the voltages across R4 and R5 (e.g., from node A to ground , and from ground to node B):

KVL: 041 =++− ARS VVV

V714 =−= ASR VVV

KVL: 052 =−−− RBS VVV

V325 −=−−= BSR VVV

______________________________________________________________________________________

Problem 3.3

Solution: Known quantities: Circuit shown in Figure P3.3 with known source currents and resistances, Ω=Ω=Ω= 6,1,3 321 RRR .


3.4

Find: The currents I1, I2 using node voltage analysis. Analysis:

At node 1:

( ) 11131

21 =−⋅+

+⋅ vv

At node 2:

( ) 26111 21 −=

+⋅+−⋅ vv

Solving, we find that:

V3V5.1

2

1

−=−=

vv

Then,

A5.06

A5.03

22

11

−==

−==

vi

vi

______________________________________________________________________________________

Problem 3.4

Solution: Known quantities: Circuit shown in Figure P3.3 with known source currents and resistances, Ω=Ω=Ω= 6,1,3 321 RRR .

Find: The currents I1, I2 using mesh analysis. Analysis:

At mesh (a):

A 1=ai

At mesh (b):

( ) ( ) 0 6 3 =−++− cbbab iiiii

At mesh (c):

A 2=ci

Solving, we find that:

A 5.1=bi

Then, ( )( ) A5.0

A5.0

2

1

−=−=−=−=

cb

ba

iiiiii


3.5

______________________________________________________________________________________

Problem 3.5

Solution: Known quantities: Circuit shown in Figure P3.5 with resistance values, current and voltage source values. Find: The current, i, through the voltage source using node voltage analysis. Analysis:

At node 1:

01005200

31211 =−+−+ vvvvv

At node 2:

02.05

12 =++− ivv

At node 3:

050100

313 =+−+− vvvi

For the voltage source we have: V 5023 =−vv

Solving the system, we obtain: V 53.451 −=v , V 69.482 −=v , V 31.13 =v and, finally, mA 491=i .

______________________________________________________________________________________

Problem 3.6

Solution: Known quantities: The current source value, the voltage source value and the resistance values for the circuit shown in Figure P3.6. Find: The three node voltages indicated in Figure P3.6 using node voltage analysis. Analysis:

0.2 A 200 Ω 25 Ω

75 Ω

100 Ω

50 Ω

+

10V

i

V1V2

V3


3.6

At node 1:

A 2.075200

211 =−+ vvv

At node 2:

0502575

32212 =+−++− ivvvvv

At node 3:

010050

323 =+−+− vvvi

For the voltage source we have:

23 10 vv =+

Solving the system, we obtain: V 24.141 =v , V58.42 =v , V 42.53 −=v and, finally, mA 254−=i .

______________________________________________________________________________________

Problem 3.7

Solution: Known quantities: The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.7. Find: The current, i, drawn from the independent voltage source using node voltage analysis. Analysis:

At node 1:

025.05.05.0

3 2111 =−++− vvvv

At node 2:

05.075.025.0212 =++− vvv

Solving the system, we obtain: V 125.11 =v , V 75.02 =v

Therefore, A 75.35.0

3 1 =−= vi .

______________________________________________________________________________________

Problem 3.8

Solution: Known quantities: The voltage source value, 15 V, and the four resistance values, indicated in Figure P3.8. Find: The voltage at nodes a and b, aV and bV , and their difference, ba VV − using node voltage analysis.


3.7

Analysis:

Using nodal analysis at the two nodes a and b, we write the equations

02036

15

02018

15

=+−

=+−

aa

bb

VV

VV

Rearranging the equations,

07514030038

=−=−

a

b

VV

Solving for the two unknowns, V89.7andV36.5 == ba VV

Therefore, V54.2−=− ba VV

______________________________________________________________________________________

Problem 3.9

Solution: Known quantities: The voltage source value, 15 V, and the four resistance values, indicated in Figure P3.8. Find: The voltage at nodes a and b, aV and bV , and their difference, ba VV − using mesh analysis.

Analysis:

Using mesh analysis at the two meshes a and b, we write the equations ( ) ( )

( ) ( ) 0 36 20201815 20 36

=−+−++=−+−

ababbb

baba

iiiiiiiiii

Rearranging the equations,

056945615

=−

+=

ab

ba

ii

ii

Solving for the two unknowns, mA 662=ai and mA 395=bi

Therefore, ( ) V 36.5 20 =−= aba iiV , V 89.720 == bb iV and V54.2−=− ba VV .

______________________________________________________________________________________

Problem 3.10

Solution: Known quantities: Circuit of Figure P3.10 with voltage source, SV , current source, IS, and all resistances.


3.8

Find: a. The node equations required to determine the node voltages. b. The matrix solution for each node voltage in terms of the known parameters. Analysis: a) Specify the nodes (e.g., A on the upper left corner of the circuit in Figure P3.10, and B on the right corner). Choose one node as the reference or ground node. If possible, ground one of the sources in the circuit. Note that this is possible here. When using KCL, assume all unknown current flow out of the node. The direction of the current supplied by the current source is specified and must flow into node A. KCL:

2112

12

111

0

RV

IR

VRR

V

RVV

RVVI

SSba

baSaS

+=

−+

+

=−

+−

+−

KCL:

34311

431

1111

00

RV

RRRV

RV

RV

RVV

RVV

Sba

bSbab

=

+++

−

=−

+−

+−

b) Matrix solution:

−

−−

++

+

−

−

++

+

=

++−

−+

++

−+

=

1143121

134312

4311

121

4313

12

1111111

1111

1111

111

111

1

RRRRRRR

RRV

RRRRVI

RRRR

RRR

RRRRV

RRVI

V

SSS

S

SS

a

−

−−

++

+

+

−−

+

=

++−

−+

−

++

=

1143121

21321

4311

121

31

221

1111111

111

1111

111

1

11

RRRRRRR

RVI

RRV

RR

RRRR

RRR

RV

R

RVI

RR

V

SS

SS

SS

b

Notes: 1. The denominators are the same for both solutions. 2. The main diagonal of a matrix is the one that goes to the right and down. 3. The denominator matrix is the "conductance" matrix and has certain properties:

a) The elements on the main diagonal [ i(row) = j(column) ] include all the conductance connected to node i = j.


3.9

b) The off-diagonal elements are all negative. c) The off-diagonal elements are all symmetric, i.e., the i j-th element = j i-th element. This is

true only because there are no controlled (dependent) sources in this circuit. d) The off-diagonal elements include all the conductance connected between node i [row] and

node j [column]. ______________________________________________________________________________________

Problem 3.11

Solution: Known quantities: Circuit shown in Figure P3.11

mΩ333mΩ200mΩ700

mΩ167mΩ500V110

54

3

21

21

===

====

RRR

RRVV SS

Find: a. The most efficient way to solve for the voltage across R3. Prove your case. b. The voltage across R3. Analysis: a) There are 3 meshes and 3 mesh currents requiring the solution of 3 simultaneous equations. Only one of these mesh currents is required to determine, using Ohm's Law, the voltage across R3. In the terminal (or node) between the two voltage sources is made the ground (or reference) node, then three node voltages are known (the ground or reference voltage and the two source voltages). This leaves only two unknown node voltages (the voltages across R1, VR1, and across R2, VR2). Both these voltages are required to determine, using KVL, the voltage across R3, VR3. A difficult choice. Choose node analysis due to the smaller number of unknowns. Specify the nodes. Choose one node as the ground node. In KCL, assume unknown currents flow out. b)

KCL: 00

3

21

1

1

4

11 =−+−+−

RVV

RV

RVV RRRSR

KCL: ( ) 00

3

12

2

2

5

22 =−+−+−−R

VVR

VR

VV RRRSR

5

2

3252

31

4

1

32

4311

1111

1111

RV

RRRV

RV

RV

RV

RRRV

SRR

SRR

−=

+++

−

=

−+

++


3.10

A33010333

110A55010200

110

Ω43.110700

11

Ω42.1010700

1101671

103331111

Ω43.810200

110700

110500

1111

35

23

4

1

1-3

3

1-333

325

1-333

431

=⋅

==⋅

=

=⋅

=

=⋅

+⋅

+⋅

=++

=⋅

+⋅

+⋅

=++

−−

−

−−−

−−−

RV

RVR

RRR

RRR

SS

( ) ( )( ) ( )

( ) ( ) V26.2380.85

7862782790.85

330429.1550429.8

V30.6104.284.87

4725731

42.1043.143.143.842.1033043.1550

1

1

−=−−−=−−

=

=−−=

−−

−−

=

R

R

V

V

KVL: V59.840 213231 =−==++− RRRRRR VVVVVV ______________________________________________________________________________________

Problem 3.12


V524.2kΩ24kΩ10kΩ3

kΩ12V42CV/10kkT

34

32

11

2

−====

===°==

R

SS

S

VRRR

RRVV

.

The voltage across R3, which is given, indicates the temperature. Find: The temperature, T. Analysis: Specify nodes (A between R1 and R3, C between R3 and R2) and polarities of voltages (VA from ground to A, Vc from ground to C, and VR3 from C to A). When using KCL, assume unknown currents flow out.

KVL: 3

3 0

RAC

CRA

VVVVVV

−==++−

Now write KCL at node C, substitute for VC, solve for VA:

KCL: 0432

1 =+−+−RV

RVV

RVV CACSC


3.11

( )

( )V14.18

10241

1031

10241

10101

1031524.2

10324

11

111

0111

33

3333

42

4323

2

1

4323

2

1

3

=

⋅+

⋅

⋅+

⋅+

⋅−+

⋅=+

+++

=

=

++−+−−

RR

RRRV

RV

V

RRRVV

RV

RV

RS

A

RASA

( ) V66.20524.214.183 =−−=−= RAC VVV Now write KCL at node A and solve for VS2 and T:

KCL: 03

2

1

1 =−+−+−R

VVR

VVR

VV CA

S

SASA

( ) ( )

( ) ( )

C926.01026.9

kT

V26.966.2014.18101010122414.18

1012101214.18

2

3

3

3

33

11

2

°===

=−⋅⋅+−

⋅⋅+=

=−+−+=

S

CAS

SAS

AS

V

VVRRVV

RRVV

______________________________________________________________________________________

Problem 3.13


Ω220kΩ8.6kΩ8.1kΩ2.270V5

432

1

======

RRRRAV VS

Find: The voltage across R4 using KCL and node voltage analysis. Analysis: A node analysis is not a method of choice because the dependent source is [1] a voltage source and [2] a floating source. Both factors cause difficulties in a node analysis. A ground is specified. There are three unknown node voltages, one of which is the voltage across R4. The dependent source will introduce two additional unknowns, the current through the source and the controlling voltage (across R1) that is not a node voltage. Therefore 5 equations are required:


3.12

[ ]

[ ]

[ ]

[ ][ ] ( )13132213

1111

4

3

3

13

2

12

2

21

3

31

1

1

0504

03

02

01

VVAVVAVVVVAVKVLVVVVVVKVL

RVI

RVVKCL

IR

VVKCL

RVV

RVV

RVVKCL

SVRVRV

SRRS

CS

CS

S

−+=+==+−−−==++−

=++−

=−−

=−+−+−

Substitute using Equation [5] into Equations [1], [2] and [3] and eliminate V2 (because it only appears twice in these equations). Collect terms:

( )

( )

( ) 01111

111

011111

433

31

223

221

21233

22311

=++

++

−

−=−+

+

−−

+=+

−−+

+++

CS

VSCS

V

VSSCS

V

IRR

VR

V

RAVI

RV

RA

RV

RAV

RVI

RRV

RA

RRRV

1-3-333

2231

1-3-3

22

1-3-33

43

1-6-33

23

1-6-3

3

1-6-3

2

Ω1040.05101.8701

106.81

102.21111

Ω1039.44101.87011Ω104.69

100.221

106.8111

Ω10702.6101.8

1106.8

111

Ω10147.1106.8

11Ω10555.6101.8

11

⋅=⋅

++⋅

+⋅

=+++

⋅=⋅

+=+⋅=⋅

+⋅

=+

⋅=⋅

+⋅

=+

⋅=⋅

=⋅=⋅

=

RA

RRR

RA

RRR

RR

RR

V

V

( ) ( ) mA196.7

101.870(5)

102.25mA194.4

101.870(5)

3321

32

=⋅

+⋅

=+=⋅

=RAV

RV

RAV VSSVS

Solving, we have: mV5.134 == VVR Note: 1. This solution was not difficult in terms of theory, but was terribly long and arithmetically cumbersome.

This was because the wrong method was used. There are only 2 mesh currents in the circuit; the sources were voltage sources; therefore, a mesh analysis is the method of choice.

2. In general, a node analysis will have fewer unknowns (because one node is the ground or reference node) and will, in such cases, be preferable.

______________________________________________________________________________________


3.13

Problem 3.14

Solution: Known quantities: The values of the resistors and of the voltage sources (see Figure P3.14). Find: The voltage across the Ω10 resistor in the circuit of Figure P3.14 using mesh current analysis. Analysis:

For mesh (a): ( ) ( ) ( ) 122020202050 =−−++ cba iii

For mesh (b): ( ) ( ) ( ) 0510102020 =+−++− cba iii

For mesh (c): ( ) ( ) ( ) 01510201020 =+++−− cba iii

Solving,

mA 6.41mA 8.67mA 5.127

=−=

=

c

b

a

iii

and ( ) ( ) V09.1109.0 10 104

−=−=−= cbR iiv . ______________________________________________________________________________________

Problem 3.15

Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.15. Find: The voltage across the current source using mesh current analysis. Analysis: For mesh (a):

( ) ( ) 3303020 =−++ ba ii

For meshes (b) and (c):

( ) ( ) ( ) 02030301030 =++++− cba iii

For the current source:

5.0=− bc ii

Solving,


3.14

mA 133−=ai , mA 322−=bi and mA 178=ci .

Therefore, ( ) V89.82030 =+= civ .

______________________________________________________________________________________

Problem 3.16

Solution: Known quantities: The values of the resistors and of the voltage source in the circuit of Figure P3.16. Find: The current i through the resistance 4R mesh current analysis. Analysis: For mesh (a):

( ) ( ) 6.51200120050 =−++ ba ii


( ) ( ) ( ) 044033012001200 =+++− cba iii


( )( ) ( )babaxbc iiiivii −=−==− 240 1200 2.02.0

Solving,

mA 136=ai , mA 137=bi and mA 106−=ci .

Therefore, mA 106−== cii .

______________________________________________________________________________________

Problem 3.17

Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.5. Find: The current through the voltage source using mesh current analysis. Analysis: For mesh (a):

( ) ( ) 05055100 =+−++ ba ii

For the current source: 2.0=− cb ii


3.15

For meshes (b) and (c): ( ) ( ) ( ) 505052005 =+++− cba iii

Solving, mA 465−=ai , mA 226=bi and mA 26=ci .

Therefore, mA 491=−= ac iii .

______________________________________________________________________________________

Problem 3.18

Solution: Known quantities: The values of the resistors and of the current source in the circuit of Figure P3.6. Find: The current through the voltage source in the circuit of Figure P3.6 using mesh current analysis. Analysis:

For mesh (a):

( ) 010100 =+ai

For mesh (b): ( ) ( ) ( ) 0200 2.0252575200 =−+−+++ cb ii

For mesh (c): ( ) ( ) 10255025 =++− cb ii

Solving, mA 100−=ai , mA 148=bi and mA 183=ci .

Therefore, mA 283=−= ac iii

______________________________________________________________________________________

Problem 3.19

Solution: Known quantities: The values of the resistors in the circuit of Figure P3.19.

0.2 A 200 Ω 25 Ω

75 Ω

50 Ω

+

10V

i

ic

ib

ia

100 Ω


3.16

Find: The current in the circuit of Figure P3.19 using mesh current analysis. Analysis:

Since I is unknown, the problem will be solved in terms of this current.

For mesh #1, it is obvious that: Ii =1

For mesh #2:

( ) 051

51

2111 321 =

−+

+++− iii

For mesh #3:

051

31

41

51

41

321 =

+++

−+

− iii

Solving,

IiIi

483.0645.0

3

2

==

Then, 23 iii −=

and IIIi 163.0645.0483.0 −=−= ______________________________________________________________________________________

Problem 3.20

Solution: Known quantities: The values of the resistors of the circuit in Figure P3.20. Find:

The voltage gain, 1

2

vvAV = , in the circuit of Figure P3.20

using mesh current analysis. Analysis:

Note that 2

21 iiv −=

For mesh #1:

I1/5 Ω

1/2 Ωi2

i3

i1

1/4 Ω 1/3 Ω

1 Ω

i


3.17

( ) 1321 021

211 viii =+

−+

+

For mesh #2:

( ) ( ) ( ) 025.025.1

241

41

41

21

21

321

321

=−++−

=

−+

+++

−

iiior

viii

For mesh #3:

( )

( ) ( ) ( ) 05.025.11

241

41

410

321

321

=+−+

−=

++

−+

iiior

viii

Solving, 13 16.0 vi −=

from which 132 04.041 viv −==

and 04.01

2 −==vvAV

______________________________________________________________________________________

Problem 3.21

Solution: Known quantities: Circuit in Figure P3.21 and the values of the voltage sources, V45021 == SS VV , and the values of the 5 resistors:

Ω23Ω.250

Ω5Ω8

354

21

=====

RRRRR

Find: The voltages across R1, R2 and R3 using KCL and node analysis. Analysis: Choose a ground/reference node. The node common to the two voltage sources is the best choice. Specify polarity of voltages and direction of the currents.

KCL: 00

3

21

1

1

4

11 =−+−+−R

VVR

VR

VV RRRSR

KCL: ( ) 00

3

12

2

2

5

22 =−+−+−−R

VVR

VR

VV RRRSR

Collect terms in terms of the unknown node voltages:

5

2

3252

31

4

1

32

3141

1111

1111

RV

RRRV

RV

RV

RV

RRRV

SRR

SRR

−=

+++

−

=

−+

++


3.18

Evaluate the coefficients of the unknown node voltages:

1

325

1

341

1

35

2

4

1

Ω4.23321

51

0.251111

Ω4.14321

81

0.251111

Ω03125.03211kA1.8

0.25450

−

−

−

=++=++

=++=++

=====

RRR

RRR

RRV

RV SS

V422.217.59

18000131.2518004.156

V429.5

4.230131.250131.254.16

4.2318000131.251800

3

2

3

3

3

1

−=−⋅−

=

=

⋅−⋅−

−⋅−

=

−

−

−

−

R

R

V

V

KVL: V.0852

0

213

231

=−==++−

RRR

RRR

VVVVVV

______________________________________________________________________________________

Problem 3.22

Solution: Known quantities: Circuit in Figure P3.22 with the values of the voltage sources, V51121 == SS VV , and the values of the 5 resistors:

mΩ002

Ω10Ω5

54

321

=====

RRRRR

Find: The new voltages across R1, R2 and R3, in case F1 "blows" or opens using KCL and node analysis. Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero.

KCL: 0003

21

1

1 =−+−+R

VVR

V RRR

KCL: ( ) 00

3

12

2

2

5

22 =−+−+−−R

VVR

VR

VV RRRSR

Collect terms in unknown node voltages:


3.19

5

2

3252

31

32

311

1111

0111

RV

RRRV

RV

RV

RRV

SRR

RR

−=

+++

−

=

−+

+

1

3253-

5

2

1

31

1

3

Ω5.3111A57510200

115

Ω3.011Ω1.01011

−

−−

=++=⋅

=

=+==

RRRRV

RRR

S

( ) ( )( ) ( )

( ) ( ) V109.21.58

0172.51.58

5751.003.0

V36.390.011.59

57.50

5.31.01.03.0

5.35751.00

2

1

−=−−=−−−

=

−=−

−=

−−

−−

=

R

R

V

V

KVL: V81.72

0

213

231

=−==++−

RRR

RRR

VVVVVV

KVL: ( ) V151.436.39015100 414141

=−−−====+++−

F

RRFRS

VRIVVVVV

Note the voltages are strongly dependent on the loads (R1, R2 and R3) connected at the time the fuse blows. With other loads, the result will be quite different. ______________________________________________________________________________________

Problem 3.23

Solution: Known quantities: Circuit in Figure P3.22 and the values of the voltage sources, V12021 == SS VV , and the values of the 5 resistors:

mΩ250

Ω8Ω2

54

321

=====

RRRRR

Find: The voltages across R1, R2, R3, and F1 in case F1 "blows" or opens using KCL and node analysis. Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero.

KCL: 0003

21

1

1 =−+−+R

VVR

V RRR


3.20

KCL: ( ) 00

3

12

2

2

5

22 =−+−+−−R

VVR

VR

VV RRRSR

5

2

3252

31

32

311

1111

0111

RV

RRRV

RV

RV

RRV

SRR

RR

−=

+++

−

=

−+

+

1

3253-

5

2

1

31

1

3

Ω256.4111A48010250

120

Ω256.011Ω125.0811

−

−−

=++=⋅

=

=+==

RRRRV

RRR

S

( ) ( )( ) ( )

( ) ( ) V104.352.87

03002.87

480125.000.625

V20.870.0162.89600

4.625125.0125.00.625

4.625480125.00

2

1

−=−−=−−

=

−=−−=

−−

−−

=

R

R

V

V

KVL: V84.83

0

213

231

=−==++−

RRR

RRR

VVVVVV

KVL: ( ) V140.920.87012000 414141

=−−−====+++−

F

RRFRS

VRIVVVVV

______________________________________________________________________________________

Problem 3.24

Solution: Known quantities: The values of the voltage sources, V170321 === SSS VVV , and the values of the 6 resistors in the circuit of Figure P3.24:

Ω11Ω3.2Ω9.1

Ω7.0

321

321

======

RRRRRR WWW

Find: a. The number of unknown node voltages and mesh currents. b. Unknown node voltages. Analysis: If the node common to the three sources is chosen as the ground/reference node, and the series resistances are combined into single equivalent resistances, there is only one unknown node voltage. On the other


3.21

hand, there are two unknown mesh currents. A node analysis is the method of choice! Specify polarity of voltages and direction of currents.

Ω7.11Ω0.3Ω6.2

333

222111

=+=

=+==+=

RRRRRRRRR

Weq

WeqWeq

KCL: ( )

V28.94

11.71

3.01

2.61

11.7170

3.0170

2.6170

111

0

321

3

3

2

2

1

1

3

3

2

2

1

1

=++

+−=

++

+−=

=−+−−+−

eqeqeq

eq

S

eq

S

eq

S

N

eq

SN

eq

SN

eq

SN

RRR

RV

RV

RV

V

RVV

RVV

RVV

KVL:

A54.26

2.628.94170

0

11

11

11111

=−=+−

=

=+++−

RRVVI

VRIRIV

W

NS

NWS

______________________________________________________________________________________

Problem 3.25

Solution: Known quantities: The values of the voltage sources, V701321 === SSS VVV , the common node voltage,

V94.28=NV ,and the values of the 6 resistors in the circuit of Figure P3.24:

Ω11Ω2.3Ω1.9

Ω.70

321

321

======

RRRRRR WWW

Find: The current through and voltage across R1. Analysis: KVL:

A54.262.6

28.941700

11

11

11111

=−=+−

=

=+++−

RRVVI

VRIRIV

W

NS

NWS

OL: ( )( ) V103.11.954.26111 === RIVR ______________________________________________________________________________________

Problem 3.26

Solution: Known quantities: The values of the voltage sources, V170321 === SSS VVV , and the values of the 6 resistors in the circuit of Figure P3.24:


3.22

Ω11Ω3.2Ω9.1

Ω7.0

321

321

======

RRRRRR WWW

Find: The mesh (or loop) equations and any additional equation required to determine the current through R1 in the circuit shown in Figure P3.24. Analysis: KVL: ( ) ( ) 0222122111111 =−−+−+++− SWWS VRIIRIIRIRIV

KVL: ( ) ( ) 0332322122122 =+++−+−+ SWWS VRIRIRIIRIIV

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )( ) ( )332222

222211

332232

2221

11

3233222221

2122222111

WWW

WWW

WWSS

WSS

R

SSWWW

SSWWW

RRRRRRRRRRRRRRRRVV

RRVV

II

VVRRRRIRRIVVRRIRRRRI

++++−+−+++++++−

+−+

==

−−=++++−−+=−−++++

______________________________________________________________________________________

Problem 3.27

Solution: Known quantities: The values of the voltage sources, V011,V90 321 === SSS VVV , and the values of the 6 resistors in the circuit of Figure P3.24:

Ω.73Ω.97

Ω3.1

321

321

======

RRRRRR WWW

Find: The branch currents, using KVL and loop analysis. Analysis: Three equations are required. Voltages will be summed around the 2 loops that are meshes, and KCL at the common node between the resistances. Assume directions of the branch currents and the associated polarities of the voltages. After like terms are collected:

KVL: ( ) ( ) 0222122111111 =−−+−+++− SWWS VRIIRIIRIRIV

KVL: ( ) ( ) 0332322122122 =−++−+−+ SWWS VRIRIRIIRIIV

KCL: 213 III −= Plugging in the given parameters results in the following system of equations: 2000.52.14 21 =− II

00.100.5 21 =− II

213 III −= Solving the system of equations gives: 09.171 =I A

55.82 =I A


3.23

55.83 =I A

Hence, the assumed polarity of the second and third branch currents is actually reversed. ______________________________________________________________________________________

Problem 3.28

Solution: Known quantities: The values of the voltage sources, V11521 == SS VV , and the values of the 5 resistors in the circuit of Figure P3.22:

R1 R2 5 R3 10 R4 R5 200 m

Find: The voltages across R1, R2 and R3, under normal conditions, i.e., no blown fuses using KVL and a mesh analysis. Analysis: KVL:

I1 R1 R4 R1I3 Vs1

I2 R2 R5 R2I3 Vs2

-I1R1 I2R2 R1 R2 R3 I3 0

Rearranging the above equations:

( )( )

V213V5.106

V5.106

A3.21A6.428.280

11960

205552.50502.5205052.511550115

2

0205511552.511552.5

333

2322

3111

321

3121

321

32

31

==−=−=

=−=

==−−=

−−−−−−

==

==

=−+=−=−

IRVIIRV

IIRV

III

IIII

IIIIIII

R

R

R

______________________________________________________________________________________


3.24

Problem 3.29


Ω13

Ω70Ω22Ω100

54

321

=====

RRRRR

Find: The voltage across R1 using KVL and mesh analysis. Analysis: KVL:

( )( )

( ) 033212211

232522

131411

=+++−=−+

=−+

IRRRRIR-IVIRRRI

VIRRRI

S

S


( ) V89.75

A88.1A64.2354668935660

1922210022350

10001131922202235110

1000110

0192221001102235

110100113

3111

31

321

32

31

=−=

==−−=

−−−−−−

=

=−+=−

=−

IIRV

II

IIIII

II

R

______________________________________________________________________________________

Problem 3.30


Ω2.0

Ω10Ω5Ω5

54

321

=====

RRRRR

Find: The voltage across R1 using KVL and mesh analysis.


3.25

Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero. 01 =I

KVL: ( ) 0522322 =+−+− RIRIIVS

KVL: ( ) 02233313 =−++ RIIRIRI

( ) ( )( ) ( ) 0321322

223522

=+++−=−++RRRIRIVRIRRI S

Ω20

Ω2.5

321

52

=++=+

RRRRR

( ) ( )( ) ( )

( ) ( ) A7.2879

5750

20555.2

051155.2

A29.1125104

02300

20555.2

2005115

3

2

=−−=

−−

−=

=−−=

−−

−

=

I

I

( )V72.78

V109.16V36.39

33333

223222

13111

===−=−==

−=−==

RIRIVRIIRIV

RIRIV

RR

RR

RR

KVL: V39.151

01411

==+++−

F

RFS

VVVRIV

______________________________________________________________________________________

Problem 3.31


Ω1

Ω5.12Ω5.7Ω4

54

321

=====

RRRRR

Find: The voltages across R1, R2, R3, and across the open fuse using KVL and mesh analysis. Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero. 01 =I


3.26

KVL:

( )

( ) 0332122

232522

=+++−=−+

IRRRRIVIRRRI S


( )

V35.138V73

V3.96V35.23

A84.5A68.1875.147

2760

245.75.75.8

2405.7115

0245.71155.75.8

11

333

2322

131

32

32

32

=−===

−=−=−=−=

==−−=

−−−−

=

=−=−

RSF

R

R

R

VVVIRV

IIRVRIV

II

IIII

______________________________________________________________________________________


3.27

Section 3.5: Superposition

Problem 3.32

Solution: Known quantities: The values of the voltage sources, V1101 =SV , V902 =SV and the values of the 3 resistors in the circuit of Figure P3.32: Ω810Ω5.3Ω560 321 === RkRR

Find: The current through R1 due only to the source VS2. Analysis:

Suppress VS1. Redraw the circuit. Specify polarity of VR1. Choose ground.

KCL: ( )

000

3

221

2

21

1

21 =−−−

+−−

+−− −−−

RVV

RV

RV SRRR

V61.33

8101

35001

5601

81090

111

321

3

2

21 =++

=++

=−

RRR

RV

V

S

R

OL: mA02.60560

61.33

1

2121 === −

− RVI R

R

______________________________________________________________________________________

Problem 3.33

Solution: Known quantities: The values of the current source, of the voltage source and of the resistors in the circuit of Figure P3.33:

Ω23.0Ω3.0V12

Ω1A12

=====

RRVRI

GG

BB

Find: The voltage across 1R using superposition.

Analysis:

Specify a ground node and the polarity of the voltage across R. Suppress the

voltage source by replacing it with a short circuit. Redraw the circuit.

KCL: 0=+++− −−−

RV

RV

RVI IR

G

IR

B

IRB


3.28

V38.1

23.01

3.01

11

12111

=++

=++

=−

RRR

IV

GB

BIR

Suppress the current source by replacing it with an open circuit.

KCL: 0=+−+ −−−

RV

RVV

RV VR

G

GVR

B

VR

V99.5

V61.4

23.01

3.01

11

3.012

111

=+=

=++

=++

=

−−

−

VRIRR

GB

G

G

VR

VVVRRR

RV

V

Note: Superposition essentially doubles the work required to solve this problem. The voltage across R can

easily be determined using a single KCL. ______________________________________________________________________________________

Problem 3.34

Solution: Known quantities: The values of the voltage sources and of the resistors in the circuit of Figure P3.34:

kΩ1

V12

321

21

=====RRR

VV SS

Find: The voltage across 2R using superposition.

Analysis: Specify the polarity of the voltage across 2R . Suppress the voltage source 1SV by replacing it with a short

circuit. Redraw the circuit.

kΩ5.0kΩ121

31 === RRReq

( )( ) V85001000

100012

2

2222 =

+=

+=−

eqSR RR

RVV

Suppress the voltage source 2SV by replacing it with a short circuit. Redraw the

circuit.

Ωk5.0Ωk121

32 === RRReq

( )( ) V4kΩ5.0kΩ1kΩ5.0V12

1112 −=

+=

+−=−

eq

eqSR RR

RVV

V4V8V422122 =+−=+= −− RRR VVV


3.29

Note: Although superposition is necessary to solve some circuits, it is a very inefficient and very

cumbersome way to solve a circuit. This method should, if at all possible, be avoided. It must be used when

the sources in a circuit are AC sources with different frequencies, or where some sources are DC and others

are AC. ______________________________________________________________________________________

Problem 3.35


Ω1Ω10Ω5Ω7

V450

54321

21

=======

RRRRRVV SS

Find: The component of the current through 3R that is due to VS2, using superposition.

Analysis:

Suppress VS1 by replacing it with a short circuit. Redraw the circuit. A solution using equivalent resistances looks reasonable. 1R and 4R are in parallel:

( )( ) Ω875.0

1717

41

4114 =

+=

+=

RRRRR

14R is in series with 3R :

( ) ( )( ) Ω4.42510.875510.87551

Ω10.87510875.0

1432

1432514325

314143

=+

+=+

+=+=

=+=+=

RRRRRRRRR

RRR

eq

OL: A695.101425.4

4502 ===eq

SS R

VI

CD: ( )( ) A03.32

10.87555695.101

1432

223 =

+=

+=− RR

RII SR

______________________________________________________________________________________

Problem 3.36


Ω11Ω3.2Ω9.1Ω7.0

V170

321

321

321

======

===

RRRRRR

VVV

WWW

SSS

Find: The current through 1R , using superposition.


3.30

Analysis: Ω7.11Ω3Ω6.2 333222111 =+==+==+= RRRRRRRRR WeqWeqWeq

Specify the direction of I1. Suppress VS2 and VS3. Redraw circuit.

A08.34

Ω99.4

11

32

321

==

=+

+=

−eq

SI

eqeq

eqeqeqeq

RV

I

RRRR

RR

Suppress VS1 and VS3. Redraw circuit.

KCL: ( ) 0

312

2 =++−−

eq

A

eq

A

eq

SA

RV

RV

RVV

A13.27

V54.701

12

3

2

1

2

2

=−=

−=++

−=

−eq

AI

eq

eq

eq

eq

SA

RVI

RR

RR

VV

Suppress VS1 and VS2. Redraw circuit.

KCL: ( )

000

213

3 =−+−+−−

eq

A

eq

A

eq

SA

RV

RV

RVV

A96.6

V09.181

13

2

3

1

3

3

−=−=

=++

−=

−eq

AI

eq

eq

eq

eq

SA

RVI

RR

RR

VV

A25.54321 =++= −−− III IIII

Note: Superposition should be used only for special conditions, as stated in the solution to Problem 3.34. In

the problem above a better method is:

a. mesh analysis using KVL (2 unknowns)

b. node analysis using KCL (1 unknown but current must be obtained using OL). ________________________________________________________________________

Problem 3.37

Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.5. Find: The current through the voltage source using superpoistion. Analysis: (1) Suppress voltage source V. Redraw the circuit.


3.31

For mesh (a):

( ) ( ) 055100 =−++ ba ii

For the current source: 2.0=− cb ii

For meshes (b) and (c): ( ) ( ) ( ) 05052005 =+++− cba iii

Solving, mA 2=ai , mA 39=bi and mA 161−=ci .

Therefore, mA 1631 −=−= ac iii .

(2) Suppress current source I. Redraw the circuit. For mesh (a):

( ) ( ) 05055100 =+−++ ba ii

For mesh (b): ( ) ( ) 505052005 =+++− ba ii

Solving, mA467−=ai and mA 187=bi .

Therefore, mA 6542 =−= ab iii .

Using the principle of superposition, mA 49121 =+= iii

________________________________________________________________________

Problem 3.38

Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.6. Find: The current through the voltage source using superposition. Analysis: (1) Suppress voltage source V. Redraw the circuit.

0.2 A 200 Ω 25 Ω

75 Ω

100 Ω

50 Ω

ib

ia

ic

i1


3.32

For mesh (a):

0=ai

For mesh (b): ( ) ( ) 040252575200 =−−+++ cb ii

For mesh (c): ( ) ( ) 01002525 =++− cb ii

Solving, mA 136=bi and mA 27=ci .

Therefore, mA 271 == cii .

(2) Suppress current source I. Redraw the circuit.

For mesh (a):

( ) 01050 =−ai

For mesh (b): ( ) ( ) 0252575200 =−+++ cb ii

For mesh (c): ( ) ( ) 101002525 −=++− cb ii

Solving, mA 200=ai , mA 8.6−=bi and mA 81−=ci .

Therefore, mA 2812 −=−= ac iii .

Using the principle of superposition, mA 25421 −=+= iii

________________________________________________________________________

200 Ω 25 Ω

75 Ω

100 Ω

50 Ω

+

10V

ib

ic

ia

i2


3.33

Problem 3.39

Solution: Known quantities: The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.7. Find: The current, i, drawn from the independent voltage source using superposition. Analysis: (1) Suppress voltage source V. Redraw the circuit.

At node 1:

025.05.05.0

2111 =−++ vvvv

At node 2:

05.075.025.0212 =++− vvv

Solving the system, we obtain: V 075.01 −=v , V 15.02 −=v

Therefore, mA 1505.01

1 =−= vi .

(2) Suppress current source I. Redraw the circuit.

At node 1:

( ) 025.05.025.05.05.0

3 111 =++

++− vvv

Solving, V 2.11 =v

Therefore, A 6.35.0

3 11 =−= vi .

Using the principle of superposition,

A 75.321 =+= iii ________________________________________________________________________

Problem 3.40

Solution: Known quantities: Circuit in Figure P3.12

V524.2kΩ24kΩ10kΩ3

kΩ12V42CV/10kkT

34

32

11

2

−====

===°==

R

SS

S

VRRR

RRVV

.


3.34

The voltage across R3, which is given, indicates the temperature. Find: The temperature, T using superposition. Analysis:

(1) Suppress voltage source 2SV . Redraw the circuit. For mesh (a):

( ) ( ) ( ) 24121224 =−+−+ kikiki cba

For mesh (b): ( ) ( ) ( ) 0104612 =−++− kikiki cba

For mesh (c): ( ) ( ) ( ) 0251012 =+−+− kikiki cba

Solving, mA 08.2=ai , mA 83.0=bi and mA 33.1=ci .

Therefore, ( ) V 5100002,3 −=−= cbSR iiV .

(2) Suppress voltage source 1SV . Redraw the circuit. For mesh (a):

( ) ( ) ( ) 010121224 =+−+−+ Tkikiki cba

For mesh (b): ( ) ( ) ( ) Tkikiki cba 10104612 =−++−

For mesh (c): ( ) ( ) ( ) 0251012 =+−+− kikiki cba

Solving, mA 52.0 Tia −= , mA 029.0 Tib = and mA 2381.0 Tic −= .

Therefore, ( ) V 671.2100001,3 TiiV cbSR =−= .

Using the principle of superposition, V 524.2671.251,32,33 −=+−=+= TVVV SRSRR

Therefore, 926.0=T °C.

________________________________________________________________________

Problem 3.41

Solution: Known quantities: The values of the resistors and of the voltage sources (see Figure P3.14). Find: The voltage across the Ω10 resistor in the circuit of Figure P3.14 using superposition.


3.35

Analysis:

(1) Suppress voltage source 1SV . Redraw the circuit.

For mesh (a): ( ) ( ) ( ) 02020202050 =−−++ cba iii

For mesh (b): ( ) ( ) ( ) 0510102020 =+−++− cba iii

For mesh (c): ( ) ( ) ( ) 01510201020 =+++−− cba iii

Solving, mA 8.73−=ai , mA 245−=bi and mA 2.87−=ci .

Therefore, ( ) V 578.1101,10 −=−=Ω cbS iiV .

(2) Suppress voltage source 2SV . Redraw the circuit.

For mesh (a): ( ) ( ) ( ) 122020202050 =−−++ cba iii

For mesh (b): ( ) ( ) ( ) 010102020 =−++− cba iii

For mesh (c): ( ) ( ) ( ) 01510201020 =+++−− cba iii

Solving, mA 201=ai , mA 177=bi and

mA 129=ci .

Therefore, ( ) V 48.0101,10 =−=Ω cbS iiV .

Using the principle of superposition, V 09.11,102,1010 −=+= ΩΩΩ SS VVV .

________________________________________________________________________

Problem 3.42

Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.15. Find: The voltage across the current source using superposition. Analysis: (1) Suppress voltage source. Redraw the circuit.


3.36

For mesh (a):

( ) ( ) 0303020 =−++ ba ii


( ) ( ) ( ) 02030301030 =++++− cba iii


5.0=− bc ii

Solving,

mA 208−=ai , mA 347−=bi and mA 153=ci .

Therefore, ( ) V65.72030 =+= cV iv .

(2) Suppress current source. Redraw the circuit. For mesh (a):

( ) ( ) 3303020 =−++ ba ii

For mesh (b):

( ) ( ) 09030 =+− ba ii

Solving,

mA 75=ai and mA 25=bi .

Therefore, ( ) V25.12030 =+= bI iv .

Using the principle of superposition, V9.8=+= IV vvv .


3.37

Section 3.6: Equivalent circuits

Problem 3.43

Solution: Known quantities: The schematic of the circuit (see Figure P3.1). Find: The Thévenin equivalent resistance seen by resistor 3R , the Thévenin (open-circuit) voltage and the

Norton (short-circuit) current when 3R is the load.

Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.

52

52

41

415241 ||||

RRRR

RRRRRRRRRT +

++

=+=

(2) Remove the load, leaving the load terminals open circuited. Redraw the circuit.

For node #1:

04

11

1

1 =−+R

VvRv S

For node #2:

Focus on Methodology: Computing the Norton current 1. Replace the load with a short circuit. 2. Define the short-circuit current iSC to be the Norton equivalent current. 3. Apply any preferred method (e.g.: nodal analysis) to solve for iSC. 4. The Norton current is iN = iSC.

Focus on Methodology: Computation of equivalent resistance of a one-port network that does not contain dependent sources

1. Remove the load, leaving the load terminals open circuited. 2. Zero all independent voltage and current sources 3. Compute the total resistance between load terminals, with the load removed. This

resistance is equivalent to that which would be encountered by a current source connected to the circuit in place of the load.

Focus on Methodology: Computing the Thevenin voltage 1. Remove the load, leaving the load terminals open circuited. 2. Define the open-circuit voltage vOC across the open load terminals. 3. Apply any preferred method (e.g.: nodal analysis) to solve for vOC. 4. The Thevenin voltage is vT = vOC.


3.38

05

22

2

2 =++R

VvRv S

Solving the system,

252

22

141

11

S

S

VRR

Rv

VRR

Rv

+−=

+=

Therefore,

252

21

41

121 SSOC V

RRRV

RRRvvv

++

+=−=

(3) Replace the load with a short circuit. Redraw the circuit.

For mesh (a):

( ) 1141 Sca ViRRRi =−+

For mesh (b):

( ) 2252 Scb ViRRRi =−+

For mesh (c):

( ) 02121 =++−− RRiiRiR cba

Solving the system,

( )( ) ( )

( )( ) ( )

( ) ( )( ) ( )41525241

24121521

41525241

2424121121

41525241

2211525121

RRRRRRRRVRRRVRRRi

RRRRRRRRVRRRRRRVRRi

RRRRRRRRVRRVRRRRRRi

SSc

SSb

SSa

++++++=

++++++=

++++++=

Therefore, ( ) ( )

( ) ( )41525241

24121521

RRRRRRRRVRRRVRRRii SS

cSC ++++++==

________________________________________________________________________

Problem 3.44

Solution: Known quantities: The schematic of the circuit (see Figure P3.6). Find: The Thévenin equivalent resistance seen by resistor 5R , the Thévenin (open-circuit) voltage and the




3.39

( ) Ω=Ω+ΩΩ= 92.22 200 75|| 25TR


For node #1:

2.075200

211 =−+ vvv

For node #2:

0502575 10

32212 =+−++−Vivvvvv

For node #3:

Vivv10

23

50=−

For the voltage source:

23 10 vv =+

Solving the system, V 33.131 =v , V 33.32 =v and V 67.63 −=v .

Therefore, V 67.63 −== vvOC .


200 Ω 25 Ω

75 Ω

50 Ω

RT

0.2 A 200 Ω 25 Ω

75 Ω

50 Ω

+

10VV1V2 V3

+voc

_


3.40

For mesh (a):

( ) 1050 =ai

For mesh (b):

( ) ( ) 4025300 =− cb ii

For mesh (c):

( ) ( ) 102525 =− cb ii

Solving the system,

mA 200=ai , mA 109=bi and mA 291−=ci . Therefore,

mA 291−== cSC ii .

________________________________________________________________________

Problem 3.45

Solution: Known quantities: The schematic of the circuit (see Figure P3.7). Find: The Thévenin equivalent resistance seen by resistor

5R , the Thévenin (open-circuit) voltage and the



( ) Ω=ΩΩ+Ω+Ω= 1 5.0|| 5.0 25.0 5.0TR


For node #1:

025.05.05.0

3 2111 =−++− vvvv

For node #2:

05.025.0

12 =+−vv

Solving the system,

0.2 A 200 Ω 25 Ω

75 Ω

50 Ω

+

10V

iscicib

ia


3.41

V 375.11 =v and V 25.12 =v . Therefore,

V 25.12 == vvOC .


For mesh (a):

( ) ( ) 35.05.05.0 =−+ ba ii


( ) ( ) ( ) 05.025.05.05.0 =+++− cba iii


5.0=− cb ii

Solving the system,

A 875.3=ai , A 75.1=bi and A 25.1=ci . Therefore,

A 25.1== cSC ii .

________________________________________________________________________

Problem 3.46

Solution: Known quantities: The schematic of the circuit (see Figure P3.12). Find: The Thévenin equivalent resistance seen by resistor 3R , the Thévenin (open-circuit) voltage and the Norton (short-circuit) current when 3R is the load.

Assumption: As in P3.12, we assume 926.0=T °C, so that V 26.92 =SV .


Ω=ΩΩ+ΩΩ= k 67.8k 24||k 3k 12||k 12TR


For node #1:

012000

26.912000

24 11 =−+− vv

For node #2:

0240003000

24 22 =+− vv

Solving the system, V 63.161 =v and V 33.212 =v .

Therefore, V 7.421 −=−= vvvOC .


3.42


For mesh (a):

( ) ( ) ( ) 26.924121224 −=−− kikiki cba

For mesh (b):

( ) ( ) 26.93612 =+− kiki ba

For mesh (c):

( ) ( ) 01512 =+− kiki ca

Solving the system,

mA 71.1=ai , mA 83.0=bi and

mA 37.1=ci . Therefore,

mA 54.0−=−= cbSC iii .

________________________________________________________________________

Problem 3.47



( )( ) ( )( ) Ω=ΩΩ+ΩΩ=+= 24.12 15|| 05 02|| 20|||| 5132 RRRRRT


For node #1:

02050

125

211 =+−+−Vivvv

For node #2:

02020

212 =+− vvv

For node #3:

015 5

3 =− Viv

For the 5-V voltage source:

531 =−vv Solving the system,

V 14.51 =v , V 57.22 =v , V 13.01 =v and mA 95.85 =Vi .


3.43

Therefore, V 44.232 =−= vvvOC .


For mesh (a):

( ) ( ) ( ) 12202090 =−− cba iii

For mesh (b):

( ) ( ) 052020 =++− ba ii

For mesh (c):

( ) ( ) 03520 =+− ca ii

Solving the system,

mA 5.119=ai , mA 5.130−=bi and

mA 3.68=ci . Therefore,

mA 8.198=−= bcSC iii .

________________________________________________________________________

Problem 3.48



( ) Ω=ΩΩ+Ω+Ω= 52 30|| 02 10 30TR


For node #1:

0103020

3 2111 =−++− vvvv

For node #2:

5.010

12 =−vv

Solving the system, V 8.71 =v and V 8.122 =v .

Therefore, V 8.122 == vvOC .



3.44

For mesh (a):

( ) ( ) 3303020 =−+ ba ii


( ) ( ) ( ) 030103030 =+++− cba iii


5.0=− bc ii

Solving the system,

mA 92−=ai , mA 254−=bi and mA 246=ci . Therefore,

mA 246== cSC ii .

________________________________________________________________________

Problem 3.49

Solution: Known quantities: The schematic of the circuit (see Figure P3.33). Find: The Thévenin equivalent resistance seen by resistor R , the Thévenin (open-circuit) voltage and the Norton (short-circuit) current when R is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.

Ω=ΩΩ= 23.0 3.0|| 1TR


For node #1:

123.012

111 =−+ vv

Solving, V 121 =v .

Therefore, V 121 == vvOC .


For mesh (a):

( ) ( ) ( ) 0121123.03.01 =+−−+ ba ii

For mesh (b):

( ) ( ) 123.03.0 =+− ba ii

Solving the system,

A 12=ai and A 52=bi . Therefore,


3.45

A 52== bSC ii .

________________________________________________________________________

Problem 3.50

Solution: Known quantities: The schematic of the circuit (see Figure P3.35). Find: The Thévenin equivalent resistance seen by resistor 3R , the Thévenin

(open-circuit) voltage and the Norton (short-circuit) current when 3R is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.

Ω=ΩΩ+ΩΩ= 71.1 5|| 1 7|| 1TR


For node #1:

071

450 11 =+− vv

For node #2:

051

450 22 =++ vv

Solving the system, V 75.3931 =v and V 3752 −=v .

Therefore, V 75.76821 =−= vvvOC .


For mesh (a):

( ) ( ) 450771 =−+ ca ii

For mesh (b):

( ) ( ) 450515 =−+ cb ii

For mesh (c):

( ) ( ) ( ) 05757 =++−− cba iii

Solving the system,

A 450=ai , A 450=bi and A 450=ci . Therefore,

A 450== cSC ii .

________________________________________________________________________


3.46

Problem 3.51

Solution: Known quantities: The values of the voltage source, V110=SV , and the values of the 4 resistors in the circuit of Figure P3.51:

mΩ19

mΩ100mΩ930 321

====

SRRRR

Find: The change in the voltage across the total load, when the customer connects the third load R3 in parallel with the other two loads. Analysis: Choose a ground. If the node at the bottom is chosen as ground (which grounds one terminal of the ideal source), the only unknown node voltage is the required voltage. Specify directions of the currents and polarities of voltages.

Without R3: KCL: 021 =++ IIIS

OL:

000

0

21

2

2

1

1

=−+−+−

=++

RV

RV

RVV

RV

RV

RV

OiOi

S

SOi

RR

S

RS

V7.10504086.1110

111

21

==++

=S

S

S

S

S

Oi RR

RRR

RV

V

With R3: KCL: 0321 =+++ IIII S

OL:

0000

0

321

3

3

2

2

1

1

=−

+−

+−

+−

=+++

RV

RV

RV

RVV

RV

RV

RV

RV

OfOfOf

S

SOf

RRR

S

RS

V37.8919.004086.1

1101111

321

=+

=+++

=S

S

S

S

S

Of RR

RRRR

RV

V

Therefore, the voltage decreased by: V33.16∆ −=−= OiOfo VVV

Notes:


3.47

1. "Load" to an EE usually means current rather than resistance.

2. Additional load reduces the voltage supplied to the customer because of the additional voltage dropped

across the losses in the distribution system. ________________________________________________________________________

Problem 3.52

Solution: Known quantities: The values of the voltage source, V450=SV , and the values of the 4 resistors in the circuit of Figure P3.52

mΩ19

mΩ500Ω3.1 321

====

SRRRR

Find: The change in the voltage across the total load, when the customer connects the third load R3 in parallel with the other two loads. Analysis: See Solution to Problem 3.40 for a detailed mathematical analysis.

V6.15∆ −=−= OiOfo VVV

________________________________________________________________________

Problem 3.53

Solution: Known quantities: The circuit shown in Figure P3.53, the values of the terminal voltage, TV , before and after the application

of the load, respectively V20=TV and V18=TV , and the value of the load resistor kΩ7.2=LR .

Find: The internal resistance and the voltage of the ideal source. Analysis: KVL: 0=++− TSTS VRIV

If IT = 0: V20== TS VV

If VT = 18 V:

OL: Ω300andmA67.6 =−===T

TSS

L

TT I

VVRRVI

Note that RS is an equivalent resistance, representing the various internal losses of the source and is not physically a separate component. VS is the voltage generated by some internal process. The source voltage can be measured directly by reducing the current supplied by the source to zero, i.e., no-load or open-circuit conditions. The source resistance cannot be directly measured; however, it can be determined, as was done above, using the interaction of the source with an external load. ________________________________________________________________________


3.48

Problem 3.54

Solution: Known quantities: The values of the voltage source, V20=CCV , and the values of the 2 resistors in the circuit of Figure P3.54: kΩ220MΩ3.1 21 == RR

Find: The Thévenin equivalent circuit with respect to the port shown in Figure P3.54. Analysis: The Thévenin equivalent voltage is the open circuit voltage [with I = 0] between the terminals of the port. Specify the polarity of the voltage.

KCL:

V895.211

0

21

1

21

=+

=

=++−

RR

RV

V

IR

VR

VV

CC

TH

THCCTH

Note that, since I = 0, R1 and R2 are effectively in series,

using a VD relation would be easier. Suppress the ideal, independent voltage source, by shorting it. Determine the equivalent resistance with respect to the terminals of the port.

( )( ) Ωk2.18810220103.110220103.1

36

36

21 =⋅+⋅⋅⋅== RRREQ

______________________________________________________________________________________

Problem 3.55

Solution: Known quantities: The values of the battery voltage, V11=BV , the value of the generator voltage, V12=GV , and the values of the 3 resistors in the circuit of Figure P3.55: Ω2.7Ω3.0Ω7.0 === LGB RRR

Find: a. The Thévenin equivalent of the circuit to the right of the terminal pair of port x-x'. b. The terminal voltage of the battery, i.e., the voltage between x and x'.


3.49

Analysis: a. Specify the polarity of the voltage:

VD: ( )( ) V52.11

2.73.02.712 =

+=

+=

LG

LGTH RR

RVV

Suppress generator source: ( )( ) Ωm288

2.73.03.02.7 =

+== GLEQ RRR

b. Specify the polarity of the terminal voltage. Choose a ground.

KCL: 0=−+−

EQ

THT

B

BT

RVV

RVV

V37.1111 =+

+=

EQB

EQ

TH

B

B

T

RR

RV

RV

V

______________________________________________________________________________________

Problem 3.56

Solution: The values of the battery voltage, V11=BV , the value of the generator voltage, V12=GV , and the values of the 3 resistors in the circuit of Figure P3.56: Ω2.7Ω3.0Ω7.0 === LGB RRR

Find: a. The Thévenin equivalent of the circuit to the left of the terminal pair of port y-y'. b. The terminal voltage of the generator, i.e., the voltage between y and y'. Analysis: a. Specify the polarity of the Thévenin equivalent voltage:

VD: ( )( ) V03.10

2.77.02.711 =

+=

+=

LB

LBTH RR

RVV

Suppress generator source: ( )( ) Ωm638

2.77.07.02.7 =

+== BLT RRR

b. Specify the polarity of the terminal voltage. Choose a ground.

KCL: 0=−+−

T

TT

G

GT

RvV

RVV

V37.1111

=+

+=

TG

T

T

G

G

T

RR

Rv

RV

V


3.50

Section 3.7: Maximum power transfer Problem 3.57

Solution: Known quantities: The values of the voltage and of the resistor in the equivalent circuit of Figure P3.57:

Ω8

V12==

eq

TH

RV

Assumptions: Assume the conditions for maximum power transfer exist. Find: a. The value of LR .

b. The power developed in LR . c. The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the

source. Analysis: a. For maximum power transfer: Ω8== eqL RR

b. VD: ( )( ) V6

88812 =

+=

+=

Leq

LTHRL RR

RVV

( ) W4.58

6 22===

L

RLRL R

VP

c. %505.0η 22

2

Re

0 ==+

=+

=+

==Leq

L

LSeqS

LS

RLq

RL

S RRR

RIRIRI

PPP

PP

________________________________________________________________________ Problem 3.58

Solution: Known quantities: The values of the voltage and of the resistor in the equivalent circuit of Figure P3.57:

Ω600V300

==

eq

TH

RV

Assumptions: Assume the conditions for maximum power transfer exist. Find: a. The value of LR .

b. The power developed in LR . c. The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the

source. Analysis: a. For maximum power transfer:


3.51

Ω600== eqL RR

b. VD: ( )( ) V5.17

60060060035 =

+=

+=

Leq

LTHRL RR

RVV

( ) mW510.4600

17.5 22===

L

RLRL R

VP

c. %505.0η 22

2

Re

0 ==+

=+

=+

==Leq

L

LSeqS

LS

RLq

RL

S RRR

RIRIRI

PPP

PP

________________________________________________________________________ Problem 3.59

Solution: Known quantities: The values of the voltage source, V12=SV , and of the resistance representing the internal losses of the

source, Ω3.0=SR , in the circuit of Figure P3.59.

Find: a. Plot the power dissipated in the load as a function of the load resistance. What can you conclude from

your plot? b. Prove, analytically, that your conclusion is valid in

all cases. Analysis:

a. KVL: RIP

RRVI

IRIRV

RS

S

SS

2

0

=+

=

=+=−

R [Ω] I [A] PR [W] 0 40 0.0

0.1 30 90.0 0.3 20 120.0 0.9 10 90.0 2.1 5 52.5

b. ( )( ) 22

2

22 −+=

+== SS

S

SR RRRV

RRRVRIP

( )( ) ( )( )( ) ( )

( ) SS

SSSSR

RRRRR

RRRVRRVdRdP

==−+

=+−++= −−

02

0121

1

3222

0 0.5 1 1.5 2 2.50

20

40

60

80

100

120Power dissipated in the load

Load Resistance [Ohm]

Pow

er [

W]


3.52

Section 3.8: Nonlinear circuit elements Problem 3.60

Solution: Known quantities: The two nonlinear resistors, in the circuit of Figure P3.60, are characterized by: bbbaa vvivi 102 33 +==

Find: The node voltage equations in terms of v1 and v2. Analysis:

At node 1, 1211

31

1 =+=+ aa vviv

At node 2, 2621026 33 =−+=− abbab vvvii

But 21 vvva −= and 2vvb = . Therefore, the node equations are

( )v v v1 1 2

32 1+ − =

and ( ) 26210 3

21232 =−−+ vvvv

________________________________________________________________________

Problem 3.61

Solution: Known quantities: The characteristic curve I-V shown in Figure P3.61, and the values of the voltage, V15=TV , and of the

resistance, Ω200=TR , in the circuit of Figure P3.61.

Find: a. The operating point of the element that has the characteristic curve shown in Figure P3.61. b. The incremental resistance of the nonlinear element at the operating point of part a. c. If TV were increased to 20 V, find the new operating point and the new incremental resistance.

Analysis:

a. KVL: ( ) 00025.020015020015

2 =++−=++−

VVVI

Solving for V and I, V57.6orV57.4mA2.52 −== VI

The second voltage value is physically impossible.

b. ( ) Ω8.430522.010 5.0inc == −R

c. Ω37V40.5mA73 inc === RVI

________________________________________________________________________


3.53

Problem 3.62

Solution: Known quantities: The characteristic curve I-V shown in Figure P3.62, and the values of the voltage, V450=SV , and of

the resistance, Ω9=R , in the circuit of Figure P3.62.

Find: The current through and the voltage across the nonlinear device. Analysis: The I-V characteristic for the nonlinear device is given. Plot the circuit I-V characteristic, i.e., the DC load line. KVL: 0=++− DDS VRIV

0ifA50V450ifA0

9450

====

=−=−=

D

D

DDSD

VV

VR

VVI

The DC load line [circuit characteristic] is linear. Plotting the two intercepts above and connecting them with a straight line gives the DC load line. The solution for V and I is at the intersection of the device and circuit characteristics:

V210A26 ≈≈ DQDQ VI .

________________________________________________________________________ Problem 3.63

Solution: Known quantities: The I-V characteristic shown in Figure P3.63, and the values of the voltage, V5.1== TS VV , and of the

resistance, Ω60== eqRR , in the circuit of Figure P3.63.

Find: The current through and the voltage across the nonlinear device. Analysis: The solution is at the intersection of the device and circuit characteristics. The device I-V characteristic is given. Determine and plot the circuit I-V characteristic. KVL: 0=++− DDS VRIV

0ifmA25V5.1ifA0

Ω60VV1.5 D

====

=−=−=

D

D

DSD

VV

RVVI

The DC load line [circuit characteristic] is linear. Plotting the two intercepts above and connecting them with a straight line gives the DC load line. The solution is at the intersection of the device and circuit characteristics, or "Quiescent", or "Q", or "DC operating" point:

V77.0mA12 ≈≈ DQDQ VI .

________________________________________________________________________


3.54

Problem 3.64

Solution: Known quantities: The I-V characteristic shown in Figure P3.64 as a function of pressure. Ω125V5.2 ==== eqTS RRVV

Find: The DC load line, the voltage across the device as a function of pressure, and the current through the nonlinear device when p = 30 psig. Analysis: Circuit characteristic [DC load line]: KVL: 0=++− DDS VRIV

0ifmA20V5.2ifA0

Ω125V5.2

====

=−=−=

D

D

DDSD

VV

VR

VVI

The circuit characteristic is a linear relation. Plot the two intercepts and connect with a straight line to plot the DC load line. Solutions are at the intersections of the circuit with the device characteristics, i.e.:

p [psig] VD [V] 10 2.14 20 1.43 25 1.18 30 0.91 40 0.60

The plot is nonlinear. At p = 30 psig: mA5.12V08.1 == DD IV . ________________________________________________________________________

Problem 3.65

Solution: Known quantities: The I-V characteristic of the nonlinear device in the circuit shown in Figure P3.65:

Ω60V5.1

mV26A10 150

0

====

==

=−

ea

TS

T

Vv

D

RRVV

VIeII T

D

Find: An expression for the DC load line. The voltage across and current through the nonlinear device.


3.55

Analysis: Circuit characteristic [DC load line]: KVL: 0=++− DDS VRIV

[ ]60

1.51 DDSD

VR

VVI −=−=

⋅=

= −15

0 10ln026.0ln[2] DD

TDI

IIVV

Iterative procedure: Initially guess mV750=DV . Note this voltage must be between zero and the value of the source voltage. Then: a. Use Equation [1] to compute a new ID. b. Use Equation [2] to compute a new VD. c. Iterate, i.e., go step a. and repeat.

VD [mV] ID [mA] 750 12.5

784.1 11.93 782.9 11.95 782.9 11.95

… …

mV9.782mA95.11 ≈≈ DQDQ VI .

________________________________________________________________________ Problem 3.66

Solution: Known quantities: The I-V characteristic shown in Figure P3.66 as a function of pressure.

Ω125V5.2 ==== eqTS RRVV

Find: The DC load line, and the current through the nonlinear device when p = 40 psig. Analysis: Circuit characteristic [DC load line]: KVL: 0=++− DDS VRIV

0ifmA20V5.2ifA0

Ω125V5.2

====

=−=−=

D

D

DDSD

VV

VR

VVI

The circuit characteristic is a linear relation that can be plotted by plotting the two intercepts and connecting them with a straight line. Solutions are at the intersections of the circuit and device characteristics. At p = 40 psig: mA2.15V60.0 == DD IV ________________________________________________________________________

Chapter 03, Resistive Network Analysis

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Transcript of Chapter 03, Resistive Network Analysis