This document provides an errata listing corrections to errors found in the 6th edition of the textbook "Introduction to Electric Circuits" by R.C. Dorf and J.A. Svoboda. The errata is organized by chapter and page number and provides corrections to issues such as incorrect equations, figures, answers to problems, and textual errors. A link is also provided to access the errata online which contains additional corrections not listed in the printed document.
The document contains 34 problems involving circuit analysis techniques like Kirchhoff's laws, voltage and current division. The problems involve finding unknown currents, voltages, powers, and components in circuits. Suggested solutions are provided that apply circuit analysis methods to solve for the unknown quantities in 3 steps or less.
This document discusses a project about understanding teenagers and the major issues they face. It begins by providing definitions of teenagers and adolescence. It then describes 10 common social groups teenagers belong to, including jocks, geeks, skaters, outsiders, hipsters, scenesters, preps, nerds, mean girls, and emo kids. The major issues facing teenagers are discussed, such as internet/gaming addiction, violence in media, cyberbullying, violence in video games, and violence at home. Parents are provided advice on how to address these issues and help their teenagers.
This document provides an errata listing corrections to errors found in the 6th edition of the textbook "Introduction to Electric Circuits" by R.C. Dorf and J.A. Svoboda. The errata is organized by chapter and page number and provides corrections to issues such as incorrect equations, figures, answers to problems, and typos or grammatical errors in the text. A link is also provided to access the errata online which contains additional corrections not listed in the printed document.
Here are the steps to verify this circuit:
1) Identify branches that do not adhere to the passive sign convention: branches with voltages and currents of the same sign. These are the 3 V, 3 A branch and the -2 V, -4 A branch.
2) For branches that adhere to the passive sign convention, calculate power as voltage × current. For non-passive branches, calculate power as -voltage × current.
3) Sum the powers for each branch:
- (3 V)(3 A) = -9 W (non-passive)
- (-2 V)(-4 A) = 8 W (non-passive)
- (5 V)(2 A) = 10 W
Este documento presenta una guía para resolver ecuaciones diferenciales de orden uno y dos, sujetas a condiciones iniciales. Incluye problemas para determinar soluciones de ecuaciones homogéneas e inhomogéneas usando métodos como reducción de orden. También cubre ecuaciones diferenciales de orden superior y condiciones de frontera.
The document contains 34 problems involving circuit analysis techniques like Kirchhoff's laws, voltage and current division. The problems involve finding unknown currents, voltages, powers, and components in circuits. Suggested solutions are provided that apply circuit analysis methods to solve for the unknown quantities in 3 steps or less.
This document discusses a project about understanding teenagers and the major issues they face. It begins by providing definitions of teenagers and adolescence. It then describes 10 common social groups teenagers belong to, including jocks, geeks, skaters, outsiders, hipsters, scenesters, preps, nerds, mean girls, and emo kids. The major issues facing teenagers are discussed, such as internet/gaming addiction, violence in media, cyberbullying, violence in video games, and violence at home. Parents are provided advice on how to address these issues and help their teenagers.
This document provides an errata listing corrections to errors found in the 6th edition of the textbook "Introduction to Electric Circuits" by R.C. Dorf and J.A. Svoboda. The errata is organized by chapter and page number and provides corrections to issues such as incorrect equations, figures, answers to problems, and typos or grammatical errors in the text. A link is also provided to access the errata online which contains additional corrections not listed in the printed document.
Here are the steps to verify this circuit:
1) Identify branches that do not adhere to the passive sign convention: branches with voltages and currents of the same sign. These are the 3 V, 3 A branch and the -2 V, -4 A branch.
2) For branches that adhere to the passive sign convention, calculate power as voltage × current. For non-passive branches, calculate power as -voltage × current.
3) Sum the powers for each branch:
- (3 V)(3 A) = -9 W (non-passive)
- (-2 V)(-4 A) = 8 W (non-passive)
- (5 V)(2 A) = 10 W
Este documento presenta una guía para resolver ecuaciones diferenciales de orden uno y dos, sujetas a condiciones iniciales. Incluye problemas para determinar soluciones de ecuaciones homogéneas e inhomogéneas usando métodos como reducción de orden. También cubre ecuaciones diferenciales de orden superior y condiciones de frontera.
The document describes an experiment to determine the thermal conductivity (k) of copper using a modified version of Searle's method. Thermal conductivity is calculated in two ways: (1) by measuring the rate of heat extracted from the copper bar by circulating water, and (2) by measuring the rate of heat supplied to the bar electrically. Measurements are taken at different water flow rates and power inputs. The values of k calculated by the two methods are compared to each other and literature values to evaluate accuracy and sources of error.
Este documento presenta un resumen de un capítulo sobre capacitancia y dieléctricos. Explica que los capacitores permiten almacenar energía eléctrica de forma mecánica sin reacciones químicas, y consisten en dos conductores cargados con cargas opuestas separados por un dieléctrico. También define la capacitancia eléctrica como la habilidad de un conductor para almacenar carga sin un cambio sustancial en su potencial, y explica cómo la capacitancia depende de la geometría del conductor.
Download luận văn đồ án tốt nghiệp với đề tài: Tìm hiểu nội dung và phương pháp dạy học phép nhân, phép chia các số tự nhiên theo sách giáo khoa Toán lớp 3
Linear circuit analysis - solution manuel (R. A. DeCarlo and P. Lin) (z-lib.o...Ansal Valappil
This document provides solutions to multiple physics problems related to electricity and circuits. Solution 1.1 calculates the net charge on a sphere with 10^13 electrons. Solution 1.2 calculates the number of atoms and total charge in a sample of copper. Solution 1.3 calculates current flowing based on the charge and time.
The document advertises free solution manuals and textbooks for many university-level books. It states that the solution manuals contain fully solved and clearly explained solutions to all the exercises in the textbooks. It encourages visiting the website to download the files for free.
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARSLUIS POWELL
This document contains 15 examples that demonstrate concepts related to current, resistance, and electromotive force. The examples calculate values like charge, current, drift velocity, electric field, resistivity, and temperature coefficient of resistance using formulas like I=Q/t, J=I/A, vd=J/nq, E=ρJ, and R=R0[1+α(T-T0)]. The examples use materials like copper, silver, tungsten, and aluminum to illustrate how properties vary between conductors.
ECE 342 Problem Set #9 Due 5 P.M. Wednesday, October 28.docxjacksnathalie
ECE 342
Problem Set #9
Due: 5 P.M. Wednesday, October 28, 2015
Fall Semester 2015 Prof. E. Rosenbaum
Prof. T. Trick
Reading Assignment: Sections 6.6.3, 6.6.4, 6.6.6, 5.6.6
1. Consider the circuit shown below,
(a) Find Gv (vout/vsig) for IBIAS values of 0.1, 0.2, 0.5, 1 and 1.25 mA.
(b) Why isn’t Gv a linear function of IBIAS?
(c) For IBIAS = 1mA, what is the maximum allowed value of Vsig, where vsig = Vsig*sin( t)? The circuit
must provide linear amplification.
vsig
∞
VCC
vout
∞
IBIA S
RB=480kΩ
-VEE
RL=100kΩ
∞
Rsig=10kΩ
RC=10kΩ
ß = 100
VA=25V
2. Consider the circuit shown below,
(a) Choose the value of Re such that Gv is maximized, subject to the constraint that vbe ≤ 10mV and the
transistor stays in the active mode. What is the value of Gv?
(b) If β drops by 10%, what is the new value of Gv? All other design variables are unchanged.
vsig=0.05sin(ωt) ∞
+5V
vout
∞
0.2mA
100kΩ
20kΩ
∞
20kΩ
20kΩ
ß = 100
VA=∞
Re
-5V
10V
-10V
3. Find Rin, Rout, Gv, and the overall current gain io/isig (“Gi”). You are given that β=100 and VA=∞.
VBE,ON = 0.7V.
100kΩ
vsig
∞
5V
-5V
3.3kΩ 2kΩ
vo
io
ii
4. Choose the value of IBIAS such that vout is a sinusoidal signal with amplitude 200 mV or larger.
5V
10kΩ
vout
75Ω
∞
vsig=0.5sin(ωt)
VIN=3V
IBIAS
ß = 100
VA=∞
5. Consider the circuit shown below,
(a) Find the dc bias point and small-signal model parameters. Assume λ = 0.
(b) Find Rin, Rout, Avo and Gv.
(c) Repeat part (b) for the case that λ = 0.03. You will have to recalculate ro.
(d) Finally, you will consider the body effect; i.e., you will no longer assume that V B = VS but, instead,
that VB = -VSS, which is -5V in this circuit. You are given γ = 0.4V
1/2
and 2ϕF = 0.6V. You may
assume λ = 0 for simplicity. First, you need to recalculate the dc bias point. You may iteratively
solve for VS and Vt, using
( )
and (√ √ ).
Do not iterate more than 2 or 3 times, as this should be sufficient to obtain the value of V t with less
than 10mV error. Then, redraw the small-signal model, now including the gmb current source shown
in Fig. 5.62 of the textbook. The value of gmb can be found using equations (5.110) and (5.111). You
are to calculate the value of Gv.
1MΩ
vsig
∞
5V
5kΩ
vout
4.7MΩ
-5V
∞
0.5mA
Vto=0.75V
k=2mA/V
2
ENSC 324 HOMEWORK #2 Fall 2015
DUE: Monday October 19, 2015 at 2 PM (note new time!)
Please note that unless you show work in the derivations and solutions you will get no credit for the
answers. Obviously copied answers from study partners or other sources, etc., will also receive no credit.
Please do all parts of all eight problems. It is suggested that you make a copy of your homework before
turning it in in case it cannot be returned before Exam 2 (solution key will be provided).
Problem #1
In a particular sample of n-type sil ...
The document contains 10 problems involving electromagnetic induction and Maxwell's equations. Problem 10.1 involves calculating the voltage and current in a circuit with a changing magnetic flux. Problem 10.2 replaces a voltmeter with a resistor and calculates the resulting current. Problem 10.3 calculates the emf induced in closed paths with changing magnetic fluxes.
The document contains 10 problems involving electromagnetic induction and Maxwell's equations. Problem 10.1 involves calculating the voltage and current in a circuit with a changing magnetic flux. Problem 10.2 replaces a voltmeter with a resistor and calculates the resulting current. Problem 10.3 calculates the emf induced in two different rectangular paths with a given magnetic field.
This document contains solutions to tutorial exercises on electronics circuits involving capacitors and other circuit elements. The first exercise involves determining currents and equivalent capacitance in a circuit with two capacitors. The second involves calculating capacitor voltage over time when a switch changes positions. The third determines capacitor voltage over time for virtually parallel capacitors driven by a voltage source. Subsequent exercises involve determining voltages, currents, and impedances in various RC, RL, and RLC circuits.
SolutionsPlease see answer in bold letters.Note pi = 3.14.docxrafbolet0
Solution
s:
Please see answer in bold letters.
Note pi = 3.1415….
1. The voltage across a 15Ω is as indicated. Find the sinusoidal expression for the current. In addition, sketch the v and i waveform on the same axis.
Note: For the graph of a and b please see attached jpg photo with filename 1ab.jpg and for c and d please see attached photo with filename 1cd.jpg.
a. 15sin20t
v= 15sin20t
By ohms law,
i = v/r
i = 15sin20t / 15
i = sin20t A
Computation of period for graphing:
v= 15sin20t
i = sin20t
w = 20 = 2pi*f
f = 3.183 Hz
Period =1/f = 0.314 seconds
b. 300sin (377t+20)
v = 300sin (377t+20)
i = 300sin (377t+20) /15
i = 20 sin (377t+20) A
Computation of period for graphing:
v = 300sin (377t+20)
i = 20 sin (377t+20)
w = 377 = 2pi*f
f = 60 Hz
Period = 1/60 = 0.017 seconds
shift to the left by:
2pi/0.017 = (20/180*pi)/x
x = 9.44x10-4 seconds
c. 60cos (wt+10)
v = 60cos (wt+10)
i = 60cos (wt+10)/15
i = 4cos (wt+10) A
Computation of period for graphing:
let’s denote the period as w sifted to the left by:
10/180*pi = pi/18
d. -45sin (wt+45)
v = -45sin (wt+45)
i = -45sin (wt+45) / 15
i = -3 sin (wt+45) A
Computation of period for graphing:
let’s denote the period as w sifted to the left by:
45/180 * pi = 1/4*pi
2. Determine the inductive reactance (in ohms) of a 5mH coil for
a. dc
Note at dc, frequency (f) = 0
Formula: XL = 2*pi*fL
XL = 2*pi* (0) (5m)
XL = 0 Ω
b. 60 Hz
Formula: XL = 2*pi*fL
XL = 2 (60) (5m)
XL = 1.885 Ω
c. 4kHz
Formula: XL = 2*pi*fL
XL = = 2*pi* (4k)(5m)
XL = 125.664 Ω
d. 1.2 MHz
Formula: XL = 2*pi*fL
XL = 2*pi* (1.2 M) (5m)
XL = 37.7 kΩ
3. Determine the frequency at which a 10 mH inductance has the following inductive reactance.
a. XL = 10 Ω
Formula: XL = 2*pi*fL
Express in terms in f:
f = XL/2 pi*L
f = 10 / (2pi*10m)
f = 159.155 Hz
b. XL = 4 kΩ
f = XL/2pi*L
f = 4k / (2pi*10m)
f = 63.662 kHz
c. XL = 12 kΩ
f = XL/2piL
f = 12k / (2pi*10m)
f = 190.99 kHz
d. XL = 0.5 kΩ
f = XL/2piL
f = 0.5k / (2pi*10m)
f = 7.958 kHz
4. Determine the frequency at which a 1.3uF capacitor has the following capacitive reactance.
a. 10 Ω
Formula: XC = 1/ (2pifC)
Expressing in terms of f:
f = 1/ (2pi*XC*C)
f = 1/ (2pi*10*1.3u)
f = 12.243 kΩ
b. 1.2 kΩ
f = 1/ (2pi*XC*C)
f = 1/ (2pi*1.2k*1.3u)
f = 102.022 Ω
c. 0.1 Ω
f = 1/ (2pi*XC*C)
f = 1/ (2pi*0.1*1.3u)
f = 1.224 MΩ
d. 2000 Ω
f = 1/ (2pi*XC*C)
f = 1/ (2pi*2000*1.3u)
f = 61.213 Ω
5. For the following pairs of voltage and current, indicate whether the element is a capacitor, an inductor and a capacitor, an inductor, or a resistor and find the value of C, L, or R if insufficient data are given.
a. v = 55 sin (377t + 50)
i = 11 sin (377t -40)
Element is inductor
In this case voltage leads current (ELI) by exactly 90 degrees so that means the circuit is inductive and the element is inductor.
XL = 55/11 = 5 Ω
we know the w=2pif so
w= 377=2pif
f= 60 Hz
To compute for th.
This document provides answers to questions about electric current and resistance from Chapter 27. It includes calculations of current, resistance, resistivity, and other electrical properties. Key points addressed include:
- How geometry and temperature affect resistance
- Models for electrical conduction and how resistance changes with temperature
- Properties of superconductors and electrical power calculations
- Questions cover topics like current density, resistivity, Ohm's law, and resistances of different materials.
The document provides examples and problems related to real gases and the van der Waals equation of state. Some key points:
1) Example 1 calculates pressure, compression factor Z, and compares to the ideal gas result for CO gas using the virial equation of state.
2) Example 2 calculates derivatives of pressure with respect to volume and temperature for an equation of state.
3) Problems cover a range of calculations for real gases including determining pressure, molar volume, density, compression factor, and van der Waals parameters.
4) Some problems require determining if a gas has a critical point based on its equation of state and deriving critical constants.
This document summarizes Chapter 1 from the textbook "Electrical Engineering: Principles and Applications" by Allan R. Hambley. It includes sample exercises and problems from Chapter 1 on topics like charge, current, voltage, power, energy, circuits, and circuit analysis techniques. The chapter introduces fundamental concepts of electrical engineering.
The document provides examples and exercises on collision theory and gas kinetics. It defines key terms like collision frequency, collision density, collision flux, and mean free path. It gives examples of calculating these values for different gases in various conditions. The exercises ask the reader to calculate collision properties for gases like hydrogen, oxygen, nitrogen, and others at different temperatures and pressures.
The document describes an experiment to determine the thermal conductivity (k) of copper using a modified version of Searle's method. Thermal conductivity is calculated in two ways: (1) by measuring the rate of heat extracted from the copper bar by circulating water, and (2) by measuring the rate of heat supplied to the bar electrically. Measurements are taken at different water flow rates and power inputs. The values of k calculated by the two methods are compared to each other and literature values to evaluate accuracy and sources of error.
Este documento presenta un resumen de un capítulo sobre capacitancia y dieléctricos. Explica que los capacitores permiten almacenar energía eléctrica de forma mecánica sin reacciones químicas, y consisten en dos conductores cargados con cargas opuestas separados por un dieléctrico. También define la capacitancia eléctrica como la habilidad de un conductor para almacenar carga sin un cambio sustancial en su potencial, y explica cómo la capacitancia depende de la geometría del conductor.
Download luận văn đồ án tốt nghiệp với đề tài: Tìm hiểu nội dung và phương pháp dạy học phép nhân, phép chia các số tự nhiên theo sách giáo khoa Toán lớp 3
Linear circuit analysis - solution manuel (R. A. DeCarlo and P. Lin) (z-lib.o...Ansal Valappil
This document provides solutions to multiple physics problems related to electricity and circuits. Solution 1.1 calculates the net charge on a sphere with 10^13 electrons. Solution 1.2 calculates the number of atoms and total charge in a sample of copper. Solution 1.3 calculates current flowing based on the charge and time.
The document advertises free solution manuals and textbooks for many university-level books. It states that the solution manuals contain fully solved and clearly explained solutions to all the exercises in the textbooks. It encourages visiting the website to download the files for free.
PROBLEMAS RESUELTOS (87) DEL CAPÍTULO I DE LABORATORIO DE FÍSICA II - SEARSLUIS POWELL
This document contains 15 examples that demonstrate concepts related to current, resistance, and electromotive force. The examples calculate values like charge, current, drift velocity, electric field, resistivity, and temperature coefficient of resistance using formulas like I=Q/t, J=I/A, vd=J/nq, E=ρJ, and R=R0[1+α(T-T0)]. The examples use materials like copper, silver, tungsten, and aluminum to illustrate how properties vary between conductors.
ECE 342 Problem Set #9 Due 5 P.M. Wednesday, October 28.docxjacksnathalie
ECE 342
Problem Set #9
Due: 5 P.M. Wednesday, October 28, 2015
Fall Semester 2015 Prof. E. Rosenbaum
Prof. T. Trick
Reading Assignment: Sections 6.6.3, 6.6.4, 6.6.6, 5.6.6
1. Consider the circuit shown below,
(a) Find Gv (vout/vsig) for IBIAS values of 0.1, 0.2, 0.5, 1 and 1.25 mA.
(b) Why isn’t Gv a linear function of IBIAS?
(c) For IBIAS = 1mA, what is the maximum allowed value of Vsig, where vsig = Vsig*sin( t)? The circuit
must provide linear amplification.
vsig
∞
VCC
vout
∞
IBIA S
RB=480kΩ
-VEE
RL=100kΩ
∞
Rsig=10kΩ
RC=10kΩ
ß = 100
VA=25V
2. Consider the circuit shown below,
(a) Choose the value of Re such that Gv is maximized, subject to the constraint that vbe ≤ 10mV and the
transistor stays in the active mode. What is the value of Gv?
(b) If β drops by 10%, what is the new value of Gv? All other design variables are unchanged.
vsig=0.05sin(ωt) ∞
+5V
vout
∞
0.2mA
100kΩ
20kΩ
∞
20kΩ
20kΩ
ß = 100
VA=∞
Re
-5V
10V
-10V
3. Find Rin, Rout, Gv, and the overall current gain io/isig (“Gi”). You are given that β=100 and VA=∞.
VBE,ON = 0.7V.
100kΩ
vsig
∞
5V
-5V
3.3kΩ 2kΩ
vo
io
ii
4. Choose the value of IBIAS such that vout is a sinusoidal signal with amplitude 200 mV or larger.
5V
10kΩ
vout
75Ω
∞
vsig=0.5sin(ωt)
VIN=3V
IBIAS
ß = 100
VA=∞
5. Consider the circuit shown below,
(a) Find the dc bias point and small-signal model parameters. Assume λ = 0.
(b) Find Rin, Rout, Avo and Gv.
(c) Repeat part (b) for the case that λ = 0.03. You will have to recalculate ro.
(d) Finally, you will consider the body effect; i.e., you will no longer assume that V B = VS but, instead,
that VB = -VSS, which is -5V in this circuit. You are given γ = 0.4V
1/2
and 2ϕF = 0.6V. You may
assume λ = 0 for simplicity. First, you need to recalculate the dc bias point. You may iteratively
solve for VS and Vt, using
( )
and (√ √ ).
Do not iterate more than 2 or 3 times, as this should be sufficient to obtain the value of V t with less
than 10mV error. Then, redraw the small-signal model, now including the gmb current source shown
in Fig. 5.62 of the textbook. The value of gmb can be found using equations (5.110) and (5.111). You
are to calculate the value of Gv.
1MΩ
vsig
∞
5V
5kΩ
vout
4.7MΩ
-5V
∞
0.5mA
Vto=0.75V
k=2mA/V
2
ENSC 324 HOMEWORK #2 Fall 2015
DUE: Monday October 19, 2015 at 2 PM (note new time!)
Please note that unless you show work in the derivations and solutions you will get no credit for the
answers. Obviously copied answers from study partners or other sources, etc., will also receive no credit.
Please do all parts of all eight problems. It is suggested that you make a copy of your homework before
turning it in in case it cannot be returned before Exam 2 (solution key will be provided).
Problem #1
In a particular sample of n-type sil ...
The document contains 10 problems involving electromagnetic induction and Maxwell's equations. Problem 10.1 involves calculating the voltage and current in a circuit with a changing magnetic flux. Problem 10.2 replaces a voltmeter with a resistor and calculates the resulting current. Problem 10.3 calculates the emf induced in closed paths with changing magnetic fluxes.
The document contains 10 problems involving electromagnetic induction and Maxwell's equations. Problem 10.1 involves calculating the voltage and current in a circuit with a changing magnetic flux. Problem 10.2 replaces a voltmeter with a resistor and calculates the resulting current. Problem 10.3 calculates the emf induced in two different rectangular paths with a given magnetic field.
This document contains solutions to tutorial exercises on electronics circuits involving capacitors and other circuit elements. The first exercise involves determining currents and equivalent capacitance in a circuit with two capacitors. The second involves calculating capacitor voltage over time when a switch changes positions. The third determines capacitor voltage over time for virtually parallel capacitors driven by a voltage source. Subsequent exercises involve determining voltages, currents, and impedances in various RC, RL, and RLC circuits.
SolutionsPlease see answer in bold letters.Note pi = 3.14.docxrafbolet0
Solution
s:
Please see answer in bold letters.
Note pi = 3.1415….
1. The voltage across a 15Ω is as indicated. Find the sinusoidal expression for the current. In addition, sketch the v and i waveform on the same axis.
Note: For the graph of a and b please see attached jpg photo with filename 1ab.jpg and for c and d please see attached photo with filename 1cd.jpg.
a. 15sin20t
v= 15sin20t
By ohms law,
i = v/r
i = 15sin20t / 15
i = sin20t A
Computation of period for graphing:
v= 15sin20t
i = sin20t
w = 20 = 2pi*f
f = 3.183 Hz
Period =1/f = 0.314 seconds
b. 300sin (377t+20)
v = 300sin (377t+20)
i = 300sin (377t+20) /15
i = 20 sin (377t+20) A
Computation of period for graphing:
v = 300sin (377t+20)
i = 20 sin (377t+20)
w = 377 = 2pi*f
f = 60 Hz
Period = 1/60 = 0.017 seconds
shift to the left by:
2pi/0.017 = (20/180*pi)/x
x = 9.44x10-4 seconds
c. 60cos (wt+10)
v = 60cos (wt+10)
i = 60cos (wt+10)/15
i = 4cos (wt+10) A
Computation of period for graphing:
let’s denote the period as w sifted to the left by:
10/180*pi = pi/18
d. -45sin (wt+45)
v = -45sin (wt+45)
i = -45sin (wt+45) / 15
i = -3 sin (wt+45) A
Computation of period for graphing:
let’s denote the period as w sifted to the left by:
45/180 * pi = 1/4*pi
2. Determine the inductive reactance (in ohms) of a 5mH coil for
a. dc
Note at dc, frequency (f) = 0
Formula: XL = 2*pi*fL
XL = 2*pi* (0) (5m)
XL = 0 Ω
b. 60 Hz
Formula: XL = 2*pi*fL
XL = 2 (60) (5m)
XL = 1.885 Ω
c. 4kHz
Formula: XL = 2*pi*fL
XL = = 2*pi* (4k)(5m)
XL = 125.664 Ω
d. 1.2 MHz
Formula: XL = 2*pi*fL
XL = 2*pi* (1.2 M) (5m)
XL = 37.7 kΩ
3. Determine the frequency at which a 10 mH inductance has the following inductive reactance.
a. XL = 10 Ω
Formula: XL = 2*pi*fL
Express in terms in f:
f = XL/2 pi*L
f = 10 / (2pi*10m)
f = 159.155 Hz
b. XL = 4 kΩ
f = XL/2pi*L
f = 4k / (2pi*10m)
f = 63.662 kHz
c. XL = 12 kΩ
f = XL/2piL
f = 12k / (2pi*10m)
f = 190.99 kHz
d. XL = 0.5 kΩ
f = XL/2piL
f = 0.5k / (2pi*10m)
f = 7.958 kHz
4. Determine the frequency at which a 1.3uF capacitor has the following capacitive reactance.
a. 10 Ω
Formula: XC = 1/ (2pifC)
Expressing in terms of f:
f = 1/ (2pi*XC*C)
f = 1/ (2pi*10*1.3u)
f = 12.243 kΩ
b. 1.2 kΩ
f = 1/ (2pi*XC*C)
f = 1/ (2pi*1.2k*1.3u)
f = 102.022 Ω
c. 0.1 Ω
f = 1/ (2pi*XC*C)
f = 1/ (2pi*0.1*1.3u)
f = 1.224 MΩ
d. 2000 Ω
f = 1/ (2pi*XC*C)
f = 1/ (2pi*2000*1.3u)
f = 61.213 Ω
5. For the following pairs of voltage and current, indicate whether the element is a capacitor, an inductor and a capacitor, an inductor, or a resistor and find the value of C, L, or R if insufficient data are given.
a. v = 55 sin (377t + 50)
i = 11 sin (377t -40)
Element is inductor
In this case voltage leads current (ELI) by exactly 90 degrees so that means the circuit is inductive and the element is inductor.
XL = 55/11 = 5 Ω
we know the w=2pif so
w= 377=2pif
f= 60 Hz
To compute for th.
This document provides answers to questions about electric current and resistance from Chapter 27. It includes calculations of current, resistance, resistivity, and other electrical properties. Key points addressed include:
- How geometry and temperature affect resistance
- Models for electrical conduction and how resistance changes with temperature
- Properties of superconductors and electrical power calculations
- Questions cover topics like current density, resistivity, Ohm's law, and resistances of different materials.
The document provides examples and problems related to real gases and the van der Waals equation of state. Some key points:
1) Example 1 calculates pressure, compression factor Z, and compares to the ideal gas result for CO gas using the virial equation of state.
2) Example 2 calculates derivatives of pressure with respect to volume and temperature for an equation of state.
3) Problems cover a range of calculations for real gases including determining pressure, molar volume, density, compression factor, and van der Waals parameters.
4) Some problems require determining if a gas has a critical point based on its equation of state and deriving critical constants.
This document summarizes Chapter 1 from the textbook "Electrical Engineering: Principles and Applications" by Allan R. Hambley. It includes sample exercises and problems from Chapter 1 on topics like charge, current, voltage, power, energy, circuits, and circuit analysis techniques. The chapter introduces fundamental concepts of electrical engineering.
The document provides examples and exercises on collision theory and gas kinetics. It defines key terms like collision frequency, collision density, collision flux, and mean free path. It gives examples of calculating these values for different gases in various conditions. The exercises ask the reader to calculate collision properties for gases like hydrogen, oxygen, nitrogen, and others at different temperatures and pressures.
Transient analysis examines how circuit voltages and currents change during a transition between steady state conditions. When a change occurs, such as a switch opening or closing, the circuit moves from one steady state to another.
The summary discusses transient analysis of RC circuits. In a source-free RC circuit, the capacitor voltage decays exponentially with a time constant of RC after a switch change. In a driven RC circuit with a voltage source, the capacitor voltage reaches a final value that is the combination of an exponentially decaying complementary solution and a steady-state particular solution.
Lecture slides Ist & 2nd Order Circuits[282].pdfsami717280
- The document discusses first-order differential circuits that contain a single storage element like a capacitor or inductor. It describes how to analyze such circuits by examining their behavior over time after a switch opens or closes.
- The time constant, represented by tau (τ), is defined as the time required for the storing element in a circuit to charge. Common time constants include L/R for inductors and RC for capacitors.
- Differential equations can be used to model first-order circuits and solutions involve finding the particular integral and complementary solutions based on initial conditions.
TIME-VARYING FIELDS AND MAXWELL's EQUATIONS -Unit4- problemsDr.SHANTHI K.G
1) A 200-turn coil has an induced emf of -3.8 volts at t=10 seconds when the flux through each turn is t(t-1) mwb.
2) A conductor moving at 100 m/s perpendicular to a 1 Tesla magnetic field induces an emf of 100 volts.
3) A parallel plate capacitor with a 10 cm^2 area, 5 mm separation, and 10 sin(100πt) voltage applied has a displacement current of 5.56×10^-9 cos(100πt) amps.
1) This document discusses alternating current (AC) circuits and includes problems involving capacitors and inductors in AC circuits.
2) It covers key concepts like reactance, time constants, and calculating current, voltage, charge and inductance in circuits containing resistors, capacitors, inductors and AC sources.
3) Many problems involve calculating values after a certain time or frequency using the equations for RC circuits, RL circuits and reactance of capacitors and inductors.
This document contains solved problems related to electrostatics and dielectric materials. Key points include:
- The dielectric constant of a composite material is the weighted average of the dielectric constants of its constituent materials.
- Boundary conditions require the tangential electric field to be continuous across material interfaces, while the normal component is scaled by the relative permittivities of the materials.
- Energy density and stored electrostatic energy depend on the dielectric constant and electric field strength within each material.
- High dielectric constants and breakdown field strengths are desirable for capacitors to maximize the CVmax product.
Similar to [E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda] (20)
[E book] introduction to electric circuits 6th ed [r. c. dorf and j. a. svoboda]
1. Solution Manual
to accompany
Introduction to Electric Circuits, 6e
By R. C. Dorf and J. A. Svoboda
1
2. Table of Contents
Chapter 1 Electric Circuit Variables
Chapter 2 Circuit Elements
Chapter 3 Resistive Circuits
Chapter 4 Methods of Analysis of Resistive Circuits
Chapter 5 Circuit Theorems
Chapter 6 The Operational Amplifier
Chapter 7 Energy Storage Elements
Chapter 8 The Complete Response of RL and RC Circuits
Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements
Chapter 10 Sinusoidal Steady-State Analysis
Chapter 11 AC Steady-State Power
Chapter 12 Three-Phase Circuits
Chapter 13 Frequency Response
Chapter 14 The Laplace Transform
Chapter 15 Fourier Series and Fourier Transform
Chapter 16 Filter Circuits
Chapter 17 Two-Port and Three-Port Networks
2
3. Errata for Introduction to Electric Circuits, 6th Edition
Errata for Introduction to Electric Circuits, 6th Edition
Page 18, voltage reference direction should be + on the right in part B:
Page 28, caption for Figure 2.3-1: "current" instead of "cuurent"
Page 41, line 2: "voltage or current" instead of "voltage or circuit"
Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit.
Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..."
Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources,
then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read:
"Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab.
Then Rt = Vab/1."
Page 340, Problem P8.3-5: The answer should be .
Page 340, Problem P8.3-6: The answer should be .
Page 341, Problem P.8.4-1: The answer should be
Page 546, line 4: The angle is instead of .
Page 554, Problem 12.4.1 Missing parenthesis:
Page 687, Equation 15.5-2: Partial t in exponent:
http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM
4. Errata for Introduction to Electric Circuits, 6th Edition
Page 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2
(s) and Hc(s) = V1(s) / Vs(s).
http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM
5. Chapter 1 – Electric Circuit Variables
Exercises
Ex. 1.3-1
i (t ) = 8 t 2 − 4 t A
t t 8 t 8
q(t ) = ∫ 0
i dτ + q(0) = ∫ 0
(8τ 2 − 4τ ) dτ + 0 = τ 3 −2τ 2 = t 3 − 2 t 2 C
3 0 3
Ex. 1.3-3
t t 4 4 4
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 3τ dτ + 0 = −
t
cos 3τ 0 = − cos 3 t + C
0 0 3 3 3
Ex. 1.3-4
0 t <0
dq ( t )
i (t ) = i (t ) = 2 0< t < 2
dt −2( t − 2 )
−2e t >2
Ex. 1.4-1
i1 = 45 µA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A
Ex. 1.4-2
∆ q = i∆ t = ( 4000 A )( 0.001 s ) = 4 C
Ex. 1.4-3
∆ q 45 × 10−9
i= = −3
= 9 × 10−6 = 9 µA
∆t 5 × 10
Ex. 1.4-4
electron −19 C 9 electron −19 C
i = 10 billion
s 1.602 ×10 electron =
10×10
s 1.602 × 10 electron
electron C
= 1010 × 1.602 ×10−19
s electron
C
= 1.602 × 10−9 = 1.602 nA
s
1-1
6. Ex. 1.6-1
(a) The element voltage and current do not adhere to the passive convention in
Figures 1.6-1B and 1.6-1C so the product of the element voltage and current
is the power supplied by these elements.
(b) The element voltage and current adhere to the passive convention in Figures
1.6-1A and 1.6-1D so the product of the element voltage and current is the
power delivered to, or absorbed by these elements.
(c) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
delivered by this element: (2 V)(6 A) = 12 W. The power received by the
element is the negative of the power delivered by the element, -12 W.
(d) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
supplied by this element: (2 V)(6 A) = 12 W.
(e) The element voltage and current adhere to the passive convention in Figure
1.6-1D, so the product of the element voltage and current is the power
delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the
element is the negative of the power delivered to the element, -12 W.
Problems
Section 1-3 Electric Circuits and Current Flow
P1.3-1
d
i (t ) =
dt
( )
4 1 − e −5t = 20 e −5t A
P1.3-2
4 4
( )
t t t t
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 1 − e −5τ dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C
0 0 0 0 5 5
P1.3-3
t t
q ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0.
−∞ −∞
t t
q ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2 τ 2 = 2 t − 4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C.
t
2 2
t t
q ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8 − t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C.
t
4 4
t t
q ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t .
8 8
1-2
7. P1.3-4
C
i = 600 A = 600
s
C s mg
Silver deposited = 600 ×20 min×60 ×1.118 = 8.05×105 mg=805 g
s min C
Section 1-6 Power and Energy
P1.6-1
a.) q = ∫ i dt = i∆t = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10
4
C
b.) P = v i = (110 V )(10 A ) = 1100 W
0.06$
c.) Cost = × 1.1kW × 2 hrs = 0.132 $
kWhr
P1.6-2
P = ( 6 V )(10 mA ) = 0.06 W
∆w 200 W⋅s
∆t = = = 3.33×103 s
P 0.06 W
P1.6-3
30
for 0 ≤ t ≤ 10 s: v = 30 V and i = t = 2t A ∴ P = 30(2t ) = 60t W
15
25
for 10 ≤ t ≤ 15 s: v ( t ) = − t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V
5
v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W
30
for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = − t +b A
10
i (25) = 0 ⇒ b = 75 ⇒ i (t ) = −3t + 75 A
∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W
1-3
8. 60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt
10 15 25
Energy = ∫ P dt = ∫0
15 25
+ 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J
10
= 30t 2
0 3 10 2 15
P1.6-4
a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully
charged).
5( 3600 )
t 5 ( 3600 ) 0.5 τ 0.5 2
w = ∫ Pdt = ∫0 vi dτ = ∫0 2 11 + dτ = 22 t + 3600 τ
3600 0
= 441× 103 J = 441 kJ
1 hr 10¢
b.) Cost = 441kJ × × = 1.23¢
3600s kWhr
P1.6-5
1 1
p (t ) = ( cos 3 t )( sin 3 t ) = sin 6 t
3 6
1
p ( 0.5 ) = sin 3 = 0.0235 W
6
1
p (1) = sin 6 = −0.0466 W
6
1-4
9. Here is a MATLAB program to plot p(t):
clear
t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time
v=4*cos(3*t); % device voltage
i=(1/12)*sin(3*t); % device current
for k=1:length(t)
p(k)=v(k)*i(k); % power
end
plot(t,p)
xlabel('time, s');
ylabel('power, W')
P1.6-6
p ( t ) = 16 ( sin 3 t )( sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t W
Here is a MATLAB program to plot p(t):
clear
t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time
v=8*sin(3*t); % device voltage
i=2*sin(3*t); % device current
for k=1:length(t)
p(k)=v(k)*i(k); % power
end
plot(t,p)
xlabel('time, s');
ylabel('power, W')
1-5
10. P1.6-7
( ) ( )
p ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e −2t
Here is a MATLAB program to plot p(t):
clear
t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time
v=4*(1-exp(-2*t)); % device voltage
i=2*exp(-2*t); % device current
for k=1:length(t)
p(k)=v(k)*i(k); % power
end
plot(t,p)
xlabel('time, s');
ylabel('power, W')
P1.6-8
P = V I =3 × 0.2=0.6 W
w = P ⋅ t = 0.6 × 5 × 60=180 J
1-6
11. Verification Problems
VP 1-1
Notice that the element voltage and current of each branch adhere to the passive convention. The
sum of the powers absorbed by each branch are:
(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W
=0W
The element voltages and currents satisfy conservation of energy and may be correct.
VP 1-2
Notice that the element voltage and current of some branches do not adhere to the passive
convention. The sum of the powers absorbed by each branch are:
-(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A)
= -9 W + 6 W + 6 W + 12 W + 9 W -12 W
≠0W
The element voltages and currents do not satisfy conservation of energy and cannot be correct.
Design Problems
DP 1-1
The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25)
= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A
Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust
the estimates of the maximum voltage and current and a Grade A device otherwise.
1-7
12. DP1-2
( ) ( )
p ( t ) = 20 1 − e −8 t × 0.03 e −8 t = 0.6 1 − e−8t e−8t
Here is a MATLAB program to plot p(t):
clear
t0=0; % initial time
tf=1; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time
v=20*(1-exp(-8*t)); % device voltage
i=.030*exp(-8*t); % device current
for k=1:length(t)
p(k)=v(k)*i(k); % power
end
plot(t,p)
xlabel('time, s');
ylabel('power, W')
Here is the plot:
The circuit element must be able to absorb 0.15 W.
1-8
13. Chapter 2 - Circuit Elements
Exercises
Ex. 2.3-1
m ( i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied
m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied
Therefore the element is linear.
Ex. 2.3-2
m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied
Therefore the element is not linear.
Ex. 2.5-1
v 2 (10 )
2
P= = =1 W
R 100
Ex. 2.5-2
v 2 (10 cos t ) 2
P= = = 10 cos 2 t W
R 10
Ex. 2.8-1
ic = − 1.2 A, v d = 24 V
id = 4 ( − 1.2) = − 4.8 A
id and vd adhere to the passive convention so
P = vd id = (24) (−4.8) = −115.2 W
is the power received by the dependent source
2-1
14. Ex. 2.8-2
vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V
id and vd adhere to the passive convention so
P = vd id = (2.2) (−8) = −17.6 W
is the power received by the dependent source. The power supplied by the
dependent source is 17.6 W.
Ex. 2.8-3
ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A
id and vd adhere to the passive convention so
P = vd id = (2.5) (1.75) = 4.375 W
is the power received by the dependent source.
2-2
15. Ex. 2.9-1
θ = 45° , I = 2 mA, R p = 20 kΩ
θ 45
a= ⇒ aR = (20 kΩ) = 2.5 kΩ
360 p 360
vm = (2 ×10−3 )(2.5 ×103 ) = 5 V
Ex. 2.9-2
µA
v = 10 V, i = 280 µA, k = 1 for AD590
°K
i °K
i = kT ⇒ T = = (280µA)1 = 280° K
k µA
Ex. 2.10-1
At t = 4 s both switches are open, so i = 0 A.
Ex. 2.10.2
At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V .
At t = 6 s the switch is in the down position, so v = 0 V.
Problems
Section 2-3 Engineering and Linear Models
P2.3-1
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the
voltage. Hence, the property of homogeneity is not satisfied.
P2.3-2
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and
the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed
linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
4
(c) When v = 4 V, i = = 33 A = 33 A.
0.12
2-3
16. P2.3-3
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and
the line passes through the origin so the equation of the line is v = 256.5i . The element is indeed
linear.
(b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V
12
(c) When v = 12 V, i = = 0.04678 A = 46.78 mA.
256.5
P2.3-4
Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,
the property of homogeneity is not satisfied. The element is not linear.
Section 2-5 Resistors
P2.5-1
i = is = 3 A and v = Ri = 7 × 3 = 21 V
v and i adhere to the passive convention
∴ P = v i = 21 × 3 = 63 W
is the power absorbed by the resistor.
P2.5-2
i = is = 3 mA and v = 24 V
v 24
R = = = 8000 = 8 k Ω
i .003
P = (3×10 −3 )× 24 = 72×10 −3 = 72 mW
P2.5-3
v = vs =10 V and R = 5 Ω
v 10
i = = =2 A
R 5
v and i adhere to the passive convention
∴ p = v i = 2⋅10 = 20 W
is the power absorbed by the resistor
2-4
17. P2.5-4
v = vs = 24 V and i = 2 A
v 24
R= = = 12 Ω
i 2
p = vi = 24⋅2 = 48 W
P2.5-5
v1 = v 2 = vs = 150 V;
R1 = 50 Ω; R2 = 25 Ω
v 1 and i1 adhere to the passive convention so
v 1 150
i1 = = =3 A
R 1 50
v 150
v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = − = −6 A
R2 25
The power absorbed by R1 is P = v1 i1 = 150 ⋅ 3 = 450 W
1
The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W
P2.5-6
i1 = i 2 = is = 2 A ;
R1 =4 Ω and R2 = 8 Ω
v 1 and i 1 do not adhere to the passive convention so
v 1 =− R 1 i 1 =−4⋅2=−8 V.
The power absorbed by R 1 is
P1 =−v 1i 1 =−(−8)(2) = 16 W.
v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V .
The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W.
P2.5-7
Model the heater as a resistor, then
v2 v2 (250) 2
with a 250 V source: P = ⇒ R = = = 62.5 Ω
R P 1000
v 2 (210) 2
with a 210 V source: P = = = 705.6 W
R 62.5
2-5
18. P2.5-8
P 5000 125
The current required by the mine lights is: i = = = A
v 120 3
Power loss in the wire is : i 2 R
Thus the maximum resistance of the copper wire allowed is
0.05P 0.05×5000
R= = = 0.144 Ω
i2 (125/3) 2
now since the length of the wire is L = 2×100 = 200 m = 20,000 cm
thus R = ρ L / A with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1
ρL 1.7×10−6 ×20,000
A= = = 0.236 cm 2
R 0.144
Section 2-6 Independent Sources
P2.6-1
v s 15
= 3 A and P = R i 2 = 5 ( 3 ) = 45 W
2
(a) i = =
R 5
(b) i and P do not depend on is .
The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A.
P2.6-2
v 2 102
(a) v = R i s = 5 ⋅ 2 = 10 V and P = = = 20 W
R 5
(b) v and P do not depend on v s .
The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V
2-6
19. P2.6-3
Consider the current source:
i s and v s do not adhere to the passive convention,
so Pcs =i s v s =3⋅12 = 36 W
is the power supplied by the current source.
Consider the voltage source:
i s and v s do adhere to the passive convention,
so Pvs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the voltage source.
∴ The voltage source supplies −36 W.
P2.6-4
Consider the current source:
i s and vs adhere to the passive convention
so Pcs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the current source.
Current source supplies − 36 W.
Consider the voltage source:
i s and vs do not adhere to the passive convention
so Pvs = i s vs = 3 ⋅12 =36 W
is the power supplied by the voltage source.
P2.6-5
(a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW
1
1 1 1 1
(b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t = 10 + 5 sin 2 mJ
2
2 4 0
2-7
20. Section 2-7 Voltmeters and Ammeters
P2.7-1
v 5
(a) R = = = 10 Ω
i 0.5
(b) The voltage, 12 V, and the
current, 0.5 A, of the voltage
source adhere to the passive
convention so the power
P = 12 (0.5) = 6 W
is the power received by the
source. The voltage source
delivers -6 W.
P2.7-2
The voltmeter current is zero
so the ammeter current is
equal to the current source
current except for the
reference direction:
i = -2 A
The voltage v is the voltage of
the current source. The power
supplied by the current source
is 40 W so
40 = 2 v ⇒ v = 20 V
2-8
21. Section 2-8 Dependent Sources
P2.8-1
vb 8
r = = =4 Ω
ia 2
P2.8-2
ia 2 A
vb = 8 V ; g v b = i a = 2 A ; g = = = 0.25
vb 8 V
P2.8-3
i a 32 A
i b = 8 A ; d i b = i a = 32A ; d = = =4
ib 8 A
P2.8-4
vb 8 V
va = 2 V ; b va = vb = 8 V ; b = = =4
va 2 V
Section 2-9 Transducers
P2.9-1
θ 360 vm
a= , θ =
360 Rp I
(360)(23V)
θ = = 75.27°
(100 kΩ)(1.1 mA)
P2.9-2
µA
AD590 : k =1 °
,
K
v =20 V (voltage condition satisfied)
4 µ A < i < 13 µ A
i ⇒ 4 ° K< T <13° K
T =
k
2-9
22. Section 2-10 Switches
P2.10-1
At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 10 V.
v 10
i= = = 2 mA
R 5×103
At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
15 V.
v 15
i= = = 3 mA
R 5×103
P2.10-2
At t = 1 s the current in the resistor
is 3 mA so v = 15 V.
At t = 4 s the current in the resistor
is 0 A so v = 0 V.
Verification Problems
VP2-1
vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .)
vo 40 V
So = = 20
is 2 A
Your lab partner is wrong.
VP2-2
vs 12
We expect the resistor current to be i = = = 0.48 A. The power absorbed by
R 25
this resistor will be P = i vs = (0.48) (12) = 5.76 W.
A half watt resistor can't absorb this much power. You should not try another resistor.
2-10
23. Design Problems
DP2-1
10 10
1.) > 0.04 ⇒ R < = 250 Ω
R 0.04
102 1
2.) < ⇒ R > 200 Ω
R 2
Therefore 200 < R < 250 Ω. For example, R = 225 Ω.
DP2-2
1.) 2 R > 40 ⇒ R > 20 Ω
15
2.) 2 2 R < 15 ⇒ R < = 3.75 Ω
4
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.
DP2-3
P = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W
2 2
1
P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W
2 2
P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W
2 2
2-11
24. Chapter 3 – Resistive Circuits
Exercises
Ex 3.3-1
Apply KCL at node a to get 2 + 1 + i3 = 0 ⇒ i3 = -3 A
Apply KCL at node c to get 2 + 1 = i4 ⇒ i4 = 3 A
Apply KCL at node b to get i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 3 = 0 ⇒ v2 = -3 V
Apply KVL to the loop consisting of elements C, E, D, and A to get
3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V
Apply KVL to the loop consisting of elements E and F to get
v6 – 6 = 0 ⇒ v6 = 6 V
Check: The sum of the power supplied by all branches is
-(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0
3-1
25. Ex 3.3-2
Apply KCL at node a to
determine the current in the
horizontal resistor as shown.
Apply KVL to the loop
consisting of the voltages source
and the two resistors to get
-4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A
2
Ex 3.3-3 −18 + 0 − 12 − va = 0 ⇒ va = −30 V and im = va + 3 ⇒ im = 9 A
5
18
Ex 3.3-4 −va − 10 + 4va − 8 = 0 ⇒ va = = 6 V and vm = 4 va = 24 V
3
Ex 3.4-1
From voltage division
3
v3 = 12 = 3V
3+9
then
v
i = 3 = 1A
3
The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W
The power supplied by the source is (12)(1) = 12 W.
3-2
26. Ex 3.4-2
P = 6 W and R1 = 6 Ω
P 6
i2 = = = 1 or i =1 A
R1 6
v0 = i R1 =(1) (6)=6V
from KVL: − v+ i (2 + 4 + 6 + 2) = 0
s
⇒ v = 14 i = 14 V
s
25
Ex 3.4-3 From voltage division ⇒ v =
m 25+75
(8) = 2 V
25
Ex 3.4-4 From voltage division ⇒ v =
m 25+75
( −8 ) = −2 V
Ex. 3.5-1
1 1 1 1 1 4 103 1
= + 3+ 3+ 3= 3 ⇒ R = = kΩ
R 3
10 10 10 10 10 eq 4 4
eq
1 -3 1
By current division, the current in each resistor = (10 ) = mA
4 4
Ex 3.5-2
10
From current division ⇒ i =
m 10+40
( −5 ) = − 1 A
3-3
27. Problems
Section 3-3 Kirchoff’s Laws
P3.3-1
Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 A
The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is
power received by element B. The power supplied by element B is 12 W.
Apply KVL to the loop consisting of elements D, F, E, and C to get
4 + v + (-5) – 12 = 0 ⇒ v = 13 V
The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W
is the power supplied by element F.
Check: The sum of the power supplied by all branches is
-(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0
3-4
28. P3.3-2
Apply KCL at node a to get 2 = i2 + 6 = 0 ⇒ i2 = -4 A
Apply KCL at node b to get 3 = i4 + 6 ⇒ i4 = -3 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 6 = 0 ⇒ v2 = -6 V
Apply KVL to the loop consisting of elements C, D, and A to get
-v3 – (-2) – 6 = 0 ⇒ v4 = -4 V
Apply KVL to the loop consisting of elements E, F and D to get
4 – v6 + (-2) = 0 ⇒ v6 = 2 V
Check: The sum of the power supplied by all branches is
-(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0
3-5
29. P3.3-3
KVL : −12 − R 2 (3) + v = 0 (outside loop)
v − 12
v = 12 + 3R 2 or R 2 =
3
12
KCL i+ − 3 = 0 (top node)
R1
12 12
i = 3− or R1 =
R1 3−i
(a)
v = 12 + 3 ( 3) = 21 V
12
i = 3− =1 A
6
(b)
2 − 12 10 12
R2 = = − Ω ; R1 = =8Ω
3 3 3 − 1.5
(checked using LNAP 8/16/02)
(c)
24 = − 12 i, because 12 and i adhere to the passive convention.
12
∴ i = − 2 A and R1 = = 2.4 Ω
3+ 2
9 = 3v, because 3 and v do not adhere to the passive convention
3 − 12
∴ v = 3V and R 2 = = −3 Ω
3
The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative.
3-6
30. P3.3-4
12
i = =2A
1 6
20
i = = 5A
2 4
i = 3−i = − 2 A
3 2
i = i +i = 3A
4 2 3
Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W
2
Power absorbed by the 6 Ω resistor = 6 ⋅ i 2 = 24 W
1
Power absorbed by the 8 Ω resistor = 8 ⋅ i 2 = 72 W
4 (checked using LNAP 8/16/02)
P3.3-5
v1 = 8 V
v2 = −8 + 8 + 12 = 12 V
v3 = 2⋅ 4 = 8 V
v2
4Ω : P = 3 = 16 W
4
2
v2
6Ω : P = = 24 W
6
v2
8Ω : P = 1 = 8 W
(checked using LNAP 8/16/02) 8
P3.3-6
P2 mA = − 3 × ( 2 ×10−3 ) = −6 × 10−3 = −6 mW
P1 mA = − −7 × (1× 10−3 ) = 7 × 10−3 = 7 mW
(checked using LNAP 8/16/02)
3-7
31. P3.3-7
P2 V = + 2 × (1× 10−3 ) = 2 × 10−3 = 2 mW
P3 V = + 3 × ( −2 × 10 ) = −6 × 10−3 = −6 mW
−3
(checked using LNAP 8/16/02)
P3.3-8
KCL: iR = 2 + 1 ⇒ iR = 3 A
KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V
vR 12
∴ R= = =4Ω
iR 3
(checked using LNAP 8/16/02)
P3.3-9
KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V
KCL: iR + 8 = 0 ⇒ iR = −8 A
vR −80
∴ R= = = 10 Ω
iR −8
(checked using LNAP 8/16/02)
3-8
33. Section 3-4 A Single-Loop Circuit – The Voltage Divider
P3.4-1
6 6
v = 12 = 12 = 4 V
1 6+3+5+ 4 18
3 5 10
v = 12 = 2 V ; v = 12 = V
2 18 3 18 3
4 8
v = 12 = V
4 18 3
(checked using LNAP 8/16/02)
P3.4-2
(a) R = 6 + 3 + 2 + 4 = 15 Ω
28 28
(b) i = = = 1.867 A
R 15
( c ) p = 28 ⋅ i =28(1.867)=52.27 W
(28 V and i do not adhere
to the passive convention.)
(checked using LNAP 8/16/02)
3-10
34. P3.4-3
i R2 = v = 8 V
12 = i R1 + v = i R1 + 8
⇒ 4 = i R1
8 8 4 4 ⋅ 100
(a) i= = ; R1 = = = 50 Ω
R 2 100 i 8
4 4 8 8 ⋅ 100
(b) i = = ; R2 = = = 200 Ω
R1 100 i 4
4 8
( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω
i i
(checked using LNAP 8/16/02)
P3.4-4
Voltage division
16
v1 = 12 = 8 V
16 + 8
4
v3 = 12 = 4 V
4+8
KVL: v3 − v − v1 = 0
v = −4 V
(checked using LNAP 8/16/02)
P3.4-5
100 v
using voltage divider: v = ⇒ R = 50 s − 1
v
0 100 + 2 R s
v
o
with v = 20 V and v > 9 V, R < 61.1 Ω
s 0
R = 60 Ω
with v = 28 V and v < 13 V, R > 57.7 Ω
s 0
3-11
35. P3.4-6
240
a.) 18 = 12 V
120 + 240
18
b.) 18 = 0.9 W
120 + 240
R
c.) 18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω
R + 120
R
d.) 0.2 = ⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω
R + 120
(checked using LNAP 8/16/02)
3-12
36. Section 3-5 Parallel Resistors and Current Division
P3.5-1
1
6 1 1
i = 4= 4= A
1 1 + 1 + 1 +1 1+ 2 + 3 + 6 3
6 3 2 1
1
3 2
i = 4 = A;
2 1 + 1 + 1 +1 3
6 3 2 1
1
i = 2 4 =1 A
3 1 + 1 + 1 +1
6 3 2 1
1
i = 4=2 A
4 1 + 1 + 1 +1
6 3 2
P3.5-2
1 1 1 1 1
(a) = + + = ⇒ R = 2Ω
R 6 12 4 2
(b) v = 6 ⋅ 2 = 12 V
(c) p = 6 ⋅12 = 72 W
P3.5-3
8 8
i= or R1 =
R1 i
8 8
8 = R 2 (2 − i ) ⇒ i = 2 − or R 2 =
R2 2−i
8 4 8
(a) i = 2− = A ; R1 =
4
=6Ω
12 3
3
8 2 8
(b) i = = A ; R2 =
2
=6Ω
12 3 2−
3
3-13
37. 1
( c ) R1 = R 2 will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A.
2
R1 R 2 1
2 ⋅ = 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω
R1 + R 2 2
P3.5-4
Current division:
8
1 16 + 8 ( )
i = −6 = −2 A
8
2 8+8( )
i = −6 = −3 A
i = i −i = +1 A
1 2
P3.5-5
R
current division: i = 1 i and
2 R + R s
1 2
Ohm's Law: v = i R yields
o 2 2
v R + R
i = o 1 2
s R R
2 1
plugging in R = 4Ω, v > 9 V gives i > 3.15 A
1 o s
and R = 6Ω, v < 13 V gives i < 3.47 A
1 o s
So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V.
s o
3-14
38. P3.5-6
24
a) 1.8 = 1.2 A
12 + 24
R
b) 2 = 1.6 ⇒ 2 R = 1.6 R + 1.6 (12 ) ⇒ R = 48 Ω
R + 12
R
c) 0.4 = ⇒ ( 0.4 )(12 ) = 0.6 R ⇒ R = 8 Ω
R + 12
Section 3-7 Circuit Analysis
P3.7-1
48 ⋅ 24
(a) R = 16 + = 32 Ω
48 + 24
32 ⋅ 32
(b) v = 32 + 32 24 = 16 V ;
32 ⋅ 32
8+
32 + 32
16 1
i= = A
32 2
48 1 1
(c) i2 = ⋅ = A
48 + 24 2 3
3-15
39. P3.7-2
3⋅ 6
(a) R1 = 4 + =6Ω
3+ 6
1 1 1 1
(b) = + + ⇒ R p = 2.4 Ω then R 2 = 8 + R p = 10.4 Ω
Rp 12 6 6
(c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0
⇒ −24+6 (i1 −2)+10.4i1 = 0
36
⇒ i1 = =2.2 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.88 V
16.4
1
(d ) i2 = 6 ( 2.2 ) = 0.878 A,
1 1 1
+ +
6 6 12
v2 = ( 0.878 ) (6) = 5.3 V
6 2
(e) i3 = i2 = 0.585 A ⇒ P = 3 i3 = 1.03 W
3+ 6
3-16
40. P3.7-3
Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with
a 2 Ω resistor by the equivalent 1 Ω resistor
This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is
equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors:
1+1
i1 = (1.5 ) = 0.75 A
2 + (1 + 1)
3-17
41. P3.7-4
(a) 1 1 1 1 (10 + 8) ⋅ 9
= + + ⇒ R2 = 4 Ω and R1 = = 6Ω
R2 24 12 8 b g
10 + 8 + 9
(b)
First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next,
apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A .
(c)
1
i2 = 8 2.25 = 1125 A
. b gLM b10 +98g + 9 3OP = −10 V
and v1 = − 10
1 1 1
+ +
24 8 12
N Q
3-18
42. P3.7-5
30
v1 = 6 ⇒ v1 = 8 V
10 + 30
R2
12 = 8 ⇒ R2 = 20 Ω
R2 + 10
20 =
b
R1 10 + 30 g ⇒ R1 = 40 Ω
b
R1 + 10 + 30 g
Alternate values that can be used to change the numbers in this problem:
meter reading, V Right-most resistor, Ω R1, Ω
6 30 40
4 30 10
4 20 15
4.8 20 30
3-19
43. P3.7-6
P3.7-7
24
1× 10−3 = ⇒ R p = 12 ×103 = 12 kΩ
12 ×103 + R p
12 × 10 = R p
3
=
( 21×10 ) R3
⇒ R = 28 kΩ
( 21×10 ) + R
3
P3.7-8
130 500
Voltage division ⇒ v = 50 = 15.963 V
130 500 + 200 + 20
100 10
∴v = v
h = (15.963) = 12.279 V
100 + 30 13
v
∴ i = h = .12279 A
h 100
3-20
45. P3.7-10
15 ( 20 + 10 )
Req = = 10 Ω
15 + ( 20 + 10 )
60 30 60 20
ia = − = −6 A, ib = R
= 4 A, vc = ( −60 ) = −40 V
Req 30 + 15 eq 20 + 10
P3.7-11 a)
(24)(12)
Req = 24 12 = =8Ω
24 + 12
b) from voltage division:
100
20 100 5
v = 40 = V∴ i = 3 = A
x 20 + 4 3 x 20 3
8 5
from current division: i = i = A
x 8+8
6
3-22
46. P3.7-12
9 + 10 + 17 = 36 Ω
36 (18 )
a.) = 12 Ω
36+18
36 R
b.) = 18 ⇒ 18 R = (18 )( 36 ) ⇒ R = 36 Ω
36+R
P3.7-13
2 R( R ) 2
Req = = R
2R + R 3
v 2 240
Pdeliv. = = =1920 W
to ckt Req 2 R
3
Thus R =45 Ω
P3.7-14
R = 2 + 1 + ( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω
eq
40 40
∴i = = =5 A
Req 8
6
i1 = i
6 + 12
( )
= ( 5) 3 = 3 A
1 5 from current division
i2 = i
2
2+2
( )
= ( 5) 2 = 2 A
1 5
3-23
47. Verification Problems
VP3-1
KCL at node a: i = i + i
3 1 2
− 1.167 = − 0.833 + ( −0.333)
− 1.167= − 1.166 OK
KVL loop consisting of the vertical
6 Ω resistor, the 3 Ω and4Ω resistors,
and the voltage source:
6i + 3i + v + 12 = 0
3 2
yields v = −4.0 V not v = −2.0 V
VP3-2
reduce circuit: 5+5=10 in parallel with 20 Ω gives 6.67Ω
6.67
by current division: i = 5 = 1.25 A
20 + 6.67
∴Reported value was correct.
VP3-3
320
v =
o 320 + 650 + 230
( 24 ) = 6.4 V ∴Reported value was incorrect.
3-24
48. VP3-4
KVL bottom loop: − 14 + 0.1iA + 1.2iH = 0
KVL right loop: − 12 + 0.05iB + 1.2iH = 0
KCL at left node: iA + iB = iH
This alone shows the reported results were incorrect.
Solving the three above equations yields:
iA = 16.8 A iH = 10.3 A
iB = −6.49 A
∴ Reported values were incorrect.
VP3-5
1
Top mesh: 0 = 4 i a + 4 i a + 2 i a + − i b = 10 ( −0.5 ) + 1 − 2 ( −2 )
2
Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V
Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V
The KVL equations are satisfied so the analysis is correct.
3-25
49. VP3-6
Apply KCL at nodes b and c to get:
KCL equations:
Node e: −1 + 6 = 0.5 + 4.5
Node a: 0.5 + i c = −1 ⇒ i c = −1.5 mA
Node d: i c + 4 = 4.5 ⇒ i c = 0.5 mA
That's a contradiction. The given values of ia
and ib are not correct.
Design Problems
DP3-1
Using voltage division:
R 2 + aR p R 2 + aR p
vm = 24 = 24
R1 + (1 − a ) R p + R 2 + aR p R1 + R 2 + R p
vm = 8 V when a = 0 ⇒
R2 1
=
R1 + R 2 + R p 3
vm = 12 V when a = 1 ⇒
R2 + R p 1
=
R1 + R 2 + R p 2
The specification on the power of the voltage source indicates
242 1
≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω
R1 + R 2 + R p 2
Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives
3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives
R1 = 6000 Ω and R 2 = 4000 Ω .
With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate
24 mW, 16 mW and 8 mW respectively. Therefore the design is complete.
3-26
50. DP3-2
Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage
200
division, 12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2
R 2 + 200
82 1
is = 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the
400 8
series combination of two 200 Ω resistors. The power required to be dissipated by each of these
42 1
resistors is = 0.08 W < W .
200 8
Now let’s check the voltage:
190 210
11.88 < v < 12.12
190 + 420 0 210 + 380
3.700 < v0 < 4.314
4 − 7.5% < v0 < 4 + 7.85%
Hence, vo = 4 V ± 8% and the design is complete.
DP3-3
Vab ≅ 200 mV
10 10
v= 120 Vab = (120) (0.2)
10 + R 10 + R
240
let v = 16 = ⇒ R=5Ω
10 + R
162
∴P= = 25.6W
10
DP3-4
N N 1 1
i = G v = v where G = ∑ = N
T R T
n = 1 Rn R
iR ( 9 )(12 )
∴N= = = 18 bulbs
v 6
3-27
52. Chapter 4 – Methods of Analysis of Resistive Circuits
Exercises
Ex. 4.3-1
v v −v
a a b
KCL at a: + + 3 = 0 ⇒ 5 v − 3 v = −18
3 2 a b
v −v
b a
KCL at b: − 3 −1 = 0 ⇒ v − v = 8
2 b a
Solving these equations gives:
va = 3 V and vb = 11 V
Ex. 4.3-2
KCL at a:
v v −v
a a b
+ + 3 = 0 ⇒ 3 v − 2 v = −12
4 2 a b
v v −v
b a b
− −4=0
KCL at a: 3 2
⇒ − 3 v + 5 v = 24
a b
Solving:
va = −4/3 V and vb = 4 V
Ex. 4.4-1
Apply KCL to the supernode to get
v + 10 v
2+ b + b =5
20 30
Solving:
v = 30 V and v = v + 10 = 40 V
b a b
4-1
53. Ex. 4.4-2
( vb + 8) − ( −12) + vb = 3 ⇒ v = 8 V and v = 16 V
10 40 b a
Ex. 4.5-1
Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into
vb = 4 ia and solve for vb .
6 vb 9 + vb
+ =i ⇒ v = 4i = 4 ⇒ v = 4.5 V
8 12 a b a 12 b
Ex. 4.5-2
The controlling voltage of the dependent source is a node voltage so it is already expressed as a
function of the node voltages. Apply KCL at node a.
v −6 v −4v
a + a a = 0 ⇒ v = −2 V
20 15 a
Ex. 4.6-1
Mesh equations:
−12 + 6 i + 3 i − i − 8 = 0 ⇒ 9 i − 3 i = 20
1 2
1 1 2
8 − 3 i − i + 6 i = 0 ⇒ − 3 i + 9 i = −8
1 2
2 1 2
Solving these equations gives:
13 1
i = A and i = − A
1 6 2 6
The voltage measured by the meter is 6 i2 = −1 V.
4-2
54. Ex. 4.7-1
3 −12
Mesh equation: 9 + 3 i + 2 i + 4 i + = 0 ⇒ ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i= A
4 9
The voltmeter measures 3 i = −4 V
Ex. 4.7-2
−33 2
Mesh equation: 15 + 3 i + 6 ( i + 3) = 0 ⇒ ( 3 + 6 ) i = −15 − 6 ( 3) ⇒ i= = −3 A
9 3
Ex. 4.7-3
3 3
Express the current source current in terms of the mesh currents: = i1 − i 2 ⇒ i1 = + i 2 .
4 4
3
Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4 + i 2 + 5 i 2 = 9 ⇒ 9 i 2 = 6
4
2 4
so i 2 = A and the voltmeter reading is 2 i 2 = V
3 3
4-3
55. Ex. 4.7-4
Express the current source current in terms of the mesh currents: 3 = i1 − i 2 ⇒ i1 = 3 + i 2 .
Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3
1
Finally, i 2 = − A is the current measured by the ammeter.
3
Problems
Section 4-3 Node Voltage Analysis of Circuits with Current Sources
P4.3-1
KCL at node 1:
v v −v
1 1 2 −4 − 4 − 2
0= + +i = + + i = −1.5 + i ⇒ i = 1.5 A
8 6 8 6
(checked using LNAP 8/13/02)
4-4
56. P4.3-2
KCL at node 1:
v −v v
1 2 1
+ + 1 = 0 ⇒ 5 v − v = −20
20 5 1 2
KCL at node 2:
v −v v −v
1 2 2 3
+2= ⇒ − v + 3 v − 2 v = 40
20 10 1 2 3
KCL at node 3:
v −v v
2 3 3
+1 = ⇒ − 3 v + 5 v = 30
10 15 2 3
Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.
(checked using LNAP 8/13/02)
P4.3-3
KCL at node 1:
v −v v
1 2 1 4 − 15 4
+ =i ⇒ i = + = −2 A
5 20 1 1 5 20
KCL at node 2:
v −v v −v
1 2 2 3
+i =
5 2 15
4 − 15 15 − 18
⇒ i = − + =2A
2 5 15
(checked using LNAP 8/13/02)
4-5
58. Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage Sources
P4.4-1
Express the branch voltage of the voltage source in terms of its node voltages:
0 − va = 6 ⇒ va = −6 V
KCL at node b:
va − vb v −v −6 − vb v −v vb v −v
+2= b c ⇒ +2= b c ⇒ −1− +2= b c ⇒ 30 = 8 vb − 3 vc
6 10 6 10 6 10
vb − vc vc 9
KCL at node c: = ⇒ 4 vb − 4 vc = 5 vc ⇒ vb = vc
10 8 4
9
Finally: 30 = 8 vc − 3 vc ⇒ vc = 2 V
4
(checked using LNAP 8/13/02)
P4.4-2
Express the branch voltage of each voltage source in terms of its node voltages to get:
va = −12 V, vb = vc = vd + 8
4-7
59. KCL at node b:
vb − va vb − ( −12 )
= 0.002 + i ⇒ = 0.002 + i ⇒ vb + 12 = 8 + 4000 i
4000 4000
KCL at the supernode corresponding to the 8 V source:
v
0.001 = d + i ⇒ 4 = vd + 4000 i
4000
so vb + 4 = 4 − vd ⇒ ( vd + 8 ) + 4 = 4 − vd ⇒ vd = −4 V
4 − vd
Consequently vb = vc = vd + 8 = 4 V and i = = 2 mA
4000
(checked using LNAP 8/13/02)
P4.4-3
Apply KCL to the supernode:
va − 10 va va − 8
+ + − .03 = 0 ⇒ va = 7 V
100 100 100
(checked using LNAP 8/13/02)
P4.4-4
Apply KCL to the supernode:
va + 8 ( va + 8 ) − 12 va − 12 va
+ + + =0
500 125 250 500
Solving yields
va = 4 V
(checked using LNAP 8/13/02)
4-8
60. P4.4-5
The power supplied by the voltage source is
v −v v −v 12 − 9.882 12 − 5.294
va ( i1 + i 2 ) = va a b + a c = 12 +
4 6 4 6
= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W
(checked using LNAP 8/13/02)
P4.4-6
Label the voltage measured by the meter. Notice that this is a node voltage.
Write a node equation at the node at which
the node voltage is measured.
12 − v m v m v −8
− + + 0.002 + m =0
6000 R 3000
That is
6000 6000
3 + v m = 16 ⇒ R = 16
R −3
vm
(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ.
(b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.
4-9
61. Section 4-5 Node Voltage Analysis with Dependent Sources
P4.5-1
Express the resistor currents in terms of the
node voltages:
va − vc
i 1= = 8.667 − 10 = −1.333 A and
1
v −v 2 − 10
i 2= b c = = −4 A
2 2
Apply KCL at node c:
i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333)
−5.333
⇒ A= =4
−1.333
(checked using LNAP 8/13/02)
P4.5-2
Write and solve a node equation:
va − 6 v v − 4va
+ a + a = 0 ⇒ va = 12 V
1000 2000 3000
va − 4va
ib = = −12 mA
3000
(checked using LNAP 8/13/02)
P4.5-3
First express the controlling current in terms of
the node voltages:
2 − vb
i =
a 4000
Write and solve a node equation:
2 − vb v 2 − vb
− + b − 5 = 0 ⇒ vb = 1.5 V
4000 2000 4000
(checked using LNAP 8/14/02)
4-10
62. P4.5-4
Apply KCL to the supernode of the CCVS to get
12 − 10 14 − 10 1
+ − + i b = 0 ⇒ i b = −2 A
4 2 2
Next
10 − 12 1
ia = =− −2 V
4 2 ⇒ r = =4
1 A
r i a = 12 − 14
−
2
(checked using LNAP 8/14/02)
P4.5-5
First, express the controlling current of the CCVS in
v2
terms of the node voltages: i x =
2
Next, express the controlled voltage in terms of the
node voltages:
v2 24
12 − v 2 = 3 i x = 3 ⇒ v2 = V
2 5
so ix = 12/5 A = 2.4 A.
(checked using ELab 9/5/02)
4-11
63. Section 4-6 Mesh Current Analysis with Independent Voltage Sources
P 4.6-1
2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0
15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0
−6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0
or
14 i1 − 3 i 2 − 9 i 3 = 0
−3 i1 + 9 i 2 − 6 i 3 = −15
−9 i1 − 6 i 2 + 15 i 3 = 21
so
i1 = 3 A, i2 = 2 A and i3 = 4 A.
(checked using LNAP 8/14/02)
P 4.6-2
Top mesh:
4 (2 − 3) + R(2) + 10 (2 − 4) = 0
so R = 12 Ω.
Bottom, right mesh:
8 (4 − 3) + 10 (4 − 2) + v 2 = 0
so v2 = −28 V.
Bottom left mesh
−v1 + 4 (3 − 2) + 8 (3 − 4) = 0
so v1 = −4 V.
(checked using LNAP 8/14/02)
4-12
64. P 4.6-3
−6
Ohm’s Law: i 2 = = −0.75 A
8
KVL for loop 1:
R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0
KVL for loop 2
+ (−6) − 3 − 4 ( i1 − i 2 ) = 0
⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0
⇒ i 1 = −3 A
R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω
(checked using LNAP 8/14/02)
P4.6-4
KVL loop 1:
25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0
450 ia −100 ib = −2
KVL loop 2:
−100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0
−100 ia + 500 ib = − 4
⇒ ia = − 6.5 mA , ib = − 9.3 mA
(checked using LNAP 8/14/02)
P4.6-5
Mesh Equations:
mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0
mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0
mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0
Solving:
5
i = i2 ⇒ i = − = −0.294 A
17
(checked using LNAP 8/14/02)
4-13
65. Section 4-7 Mesh Current Analysis with Voltage and Current Sources
P4.7-1
1
mesh 1: i1 = A
2
mesh 2: 75 i2 + 10 + 25 i2 = 0
⇒ i2 = − 0.1 A
ib = i1 − i2 = 0.6 A
(checked using LNAP 8/14/02)
P4.7-2
mesh a: ia = − 0.25 A
mesh b: ib = − 0.4 A
vc = 100(ia − ib ) = 100(0.15) =15 V
(checked using LNAP 8/14/02)
P4.7-3
Express the current source current as a function of the mesh currents:
i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5
Apply KVL to the supermesh:
30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10
5
50 i2 − 15 = − 10 ⇒ i2 = = .1 A
50
i1 =−.4 A and v2 = 20 i2 = 2 V
(checked using LNAP 8/14/02)
4-14