EIT 2 - Solution Manual & Test Bank · Web viewChapter 2: Fundamentals of Electrical Circuits – Instructor Notes Chapter 2 develops the foundations for the first part of the book

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Chapter 2: Fundamentals of Electrical Circuits – Instructor Notes

Chapter 2 develops the foundations for the first part of the book. Coverage of the entire chapter would be typical in an introductory course. The first four sections provide the basic definitions (Section 2.1) and cover Kirchoff’s Laws and the passive sign convention (Sections 2.2, 2.3 And 2.4); A special feature, Focus on Methodology: The Passive Sign Convention (p. 39) and two examples illustrate this very important topic. A second feature, that will recur throughout the first six chapters, is presented in the form of sidebars. Make The Connection: Mechanical Analog of Voltage Sources (p. 24) and Make The Connection: Hydraulic Analog of Current Sources (p. 26) present the concept of analogies between electrical and other physical domains.

Sections 2.5and 2.6 introduce the i-v characteristic and the resistance element. Tables 2.1 and 2.2 on p. 45 summarize the resistivity of common materials and standard resistor values; Table 2.3 on p. 48 provides the resistance of copper wire for various gauges. The sidebar Make The Connection: Electric Circuit Analog of Hydraulic Systems – Fluid Resistance (p. 44) continues the electric-hydraulic system analogy.

Finally, Sections 2.7 and 2.8 introduce some basic but important concepts related to ideal and non-ideal current sources, and measuring instruments. In the context of measuring instruments, Chapter 2 introduces a third feature of this book, the Focus on Measurements boxes, referenced in the next paragraph.

The Instructor will find that although the material in Chapter 2 is quite basic, it is possible to give an applied flavor to the subject matter by emphasizing a few selected topics in the examples presented in class. In particular, a lecture could be devoted to resistance devices, including the resistive displacement transducer of Focus on Measurements: Resistive throttle position sensor (pp. 56-59), the resistance strain gauges of Focus on Measurements: Resistance strain gauges (pp. 58-59),and Focus on Measurements: The Wheatstone bridge and force measurements (pp. 59-60). The instructor wishing to gain a more in-depth understanding of resistance strain gauges will find a detailed analysis in the references1.

Early motivation for the application of circuit analysis to problems of practical interest to the non-electrical engineer can be found in the Focus on Measurements: The Wheatstone bridge and force measurements. The Wheatstone bridge material can also serve as an introduction to a laboratory experiment on strain gauges and the measurement of force (see, for example2). Finally, the material on practical measuring instruments in Section2.8b can also motivate a number of useful examples.

The homework problems include a variety of practical examples, with emphasis on instrumentation. Problem 2.51 illustrates analysis related to fuses; problems 2.65 relates to wire gauges; problem 2.70 discusses the thermistor; problems 2.71 discusses moving coil meters; problems 2.79 and 2.80 illustrate calculations related to strain gauge bridges; a variety of problems related to practical measuring devices are also included in the last section. The 5th Edition of this book includes 26 new problems; some of the 4th Edition problems were removed, increasing the end-of-chapter problem count from 66 to 80.

It has been the author's experience that providing the students with an early introduction to practical applications of electrical engineering to their own disciplines can increase the interest level in the course significantly.

1 E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York, 1990.2 G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3rd Edition, Kendall-Hunt, 1998.

2.1PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Learning Objectives for Chapter 21. Identify the principal elements of electrical circuits: nodes, loops, meshes, branches, and voltage

and current sources.2. Apply Kirchhoff’s Laws to simple electrical circuits and derive the basic circuit equations.3. Apply the passive sign convention and compute power dissipated by circuit elements. 4. Apply the voltage and current divider laws to calculate unknown variables in simple series, parallel

and series-parallel circuits.5. Understand the rules for connecting electrical measuring instruments to electrical circuits for the

measurement of voltage, current, and power.


Section 2.1: Definitions

Problem 2.1 An isolated free electron is traveling through an electric field from some initial point where its coulombic potential energy per unit charge (voltage) is 17 kJ/C and velocity = 93 Mm/s to some final point where its coulombic potential energy per unit charge is 6 kJ/C. Determine the change in velocity of the electron. Neglect gravitational forces.

Solution:Known quantities:

Initial Coulombic potential energy, V i=17 kJ /C ; initial velocity,U i=93 M m

s ; final Coulombic potential energy, V f=6 kJ /C .

Find:The change in velocity of the electron.

Assumptions:D PEg << D PE cD KE=- D PEc

12

me(U f2−U i

2 )=−qe(V f−V i )

U f2=U i

2−2qe

me(V f−V i )

=(93 M ms )

2−

2 (−1.6×10−19 C )9 . 11×1031 kg

(6 kV−17 kV )

=8 .649×1015 m2

s2 −3 .864×1015 m2

s2

U f=6 . 917 ´107 ms

|U f−U i|=93 M ms−69 .17 M m

s=23 .83 M m

s.



Problem 2.2 The unit used for voltage is the volt, for current the ampere, and for resistance the ohm. Using the definitions of voltage, current, and resistance, express each quantity in fundamental MKS units.

Solution:Known quantities:MKSQ units.

Find:Equivalent units of volt, ampere and ohm.

Analysis:

Voltage=Volt=JouleCoulomb

V =JC

Current=Ampere=Coulombsecond

A=Cs

Resistance=Ohm=VoltAmpere

=Joule×secondCoulomb2

=J⋅sC2

Conductance=Siemens =AmpereVolt

=C2

J⋅s

Problem 2.3 A battery can deliver 2.7 · 106 coulombs of charge. Find the capacity and delivered electrons of the battery.

Solution:Known quantities:qBattery = 2.7 · 106 C.

Find:a) The current capacity of the battery in ampere-hoursb) The number of electrons that can be delivered.

Analysis:a) There are 3600 seconds in one hour. Amperage is defined as 1 Coulomb per second and is directly

proportional to ampere-hours.

2.7 ∙106C ∙ 1hr3600 s

=750 Ah

b) The charge of a single electron is -1.602·10-19 C. The negative sign indicates that the particle delivered will be an electron. Simple division gives the solution:



2.7 ∙106 C1.602 ∙10−19 C

1 electron

=1.685 ∙1025 electrons

Problem 2.4 A three rate charge cycle is used to charge a battery. The current to the battery is held at 30 mA for 6 h and then switched to 20 mA for the next 3 h.

Solution:Known quantities:See Figure P2.4

Find:a) The total charge transferred to the battery.b) The energy transferred to the battery.

Analysis:

a) Current is equal to Coulombs

Second, therefore given the current and a duration of that current, the transferred

charge can be calculated by the following equation:A ∙ t=C

The two durations should be calculated independently and then added together.0.030 A ∙21600 s=648 C0.020 A ∙10800 s=216 C

648 C+216 C=864 Cb) P=V·I, therefore, an equation for power can be found by multiplying the two graphs together.

First separate the voltage graph into three equations:0 h → 3 h :V=9.26 ·1 0−6 t+0.53 h → 6 h :V=5.55 ∙1 0−5 t6 h → 9 h :V=1.11 ∙ 10−4 t−1.9Next, multiply the first two equations by 0.03A and the third by 0.02A.0 h → 3 h :P=2.77 ∙ 10−7 t+0.0153 h → 6 h :P=1.66 ∙ 10−6 t6 h → 9 h :P=2.22 ∙ 10−6 t−0.038Finally, since Energy is equal to the integral of power, take the integral of each of the equations for their specified times and add them together.

0 h → 3 h :E=[ 2.77∙ 10−7 t 2

2+0.015 t ]∨10800

0= 178.2 J

3 h → 6 h :E=[ 1.66∙ 10−6 t 2

2 ]∨2160010800

= 290.43 J

6 h → 9 h :E=[ 2.22∙ 10−6 t2

2+0.038 t ]∨32400

21600 = 1057.8 J



ETotal=1526 J

Problem 2.5 Batteries (e.g., lead-acid batteries) store chemical energy and convert it to electric energy on demand. Batteries do not store electric charge or charge carriers. Charge carriers (electrons) enter one terminal of the battery, acquire electrical potential energy, and exit from the other terminal at a lower voltage. Remember the electron has a negative charge! It is convenient to think of positive carriers flowing in the opposite direction, that is, conventional current, and exiting at a higher voltage. All currents in this course, unless otherwise stated, are conventional current. (Benjamin Franklin caused this mess!) For a battery with a rated voltage = 12 V and a rated capacity = 350 A-h,determinea. The rated chemical energy stored in the battery.b. The total charge that can be supplied at the rated

Solution:Known quantities:Rated voltage of the battery; rated capacity of the battery.

Find:a) The rated chemical energy stored in the batteryb) The total charge that can be supplied at the rated voltage.

Analysis:a)

DVºD PEc

DQI=DQ

DtChemical energy=D PE c=DV⋅DQ=DV⋅(I⋅Dt )

=12V 350 A−hr 3600 shr=15. 12 MJ.

As the battery discharges, the voltage will decrease below the rated voltage. The remaining chemical energy stored in the battery is less useful or not useful.b) DQ is the total charge passing through the battery and gaining 12 J/C of electrical energy.

ΔQ=I⋅Δt=350 A hr=350 Cs

hr⋅3600 shr=1 .26 MC.

Problem 2.6What determines the following?a. How much current is supplied (at a constant voltage) by an ideal voltage source.b. How much voltage is supplied (at a constant current) by an ideal current source.

Solution:Known quantities:Resistance of external circuit.



Find:a) Current supplied by an ideal voltage sourceb) Voltage supplied by an ideal current source.

Assumptions:Ideal voltage and current sources.

Analysis:a) An ideal voltage source produces a constant voltage at or below its rated current. Current is determined by the power required by the external circuit (modeled as R).

I=V s

RP=V s⋅I

b) An ideal current source produces a constant current at or below its rated voltage. Voltage is determined by the power demanded by the external circuit (modeled as R).

V=I s⋅RP=V⋅I s

A real source will overheat and, perhaps, burn up if its rated power is exceeded.

Problem 2.7An automotive battery is rated at 120 A-h. This means that under certain test conditions it can output1 A at 12 V for 120 h (under other test conditions, the battery may have other ratings).a. How much total energy is stored in the battery?b. If the headlights are left on overnight (8 h), how much energy will still be stored in the battery in themorning? (Assume a 150-W total power rating for both headlights together.)

Solution:Known quantities:Rated discharge current of the battery; rated voltage of the battery; rated discharge time of the battery.

Find:a) Energy stored in the battery when fully chargedb) Energy stored in the battery after discharging

Analysis:

a) Energy=Power ´ time=(1 A ) (12V ) (120 hr )(60 min

hr )(60 secmin )

w = 5. 184 × 106 Jb) Assume that 150 W is the combined power rating of both lights; then,

wused=(150 W )(8 hrs )(3600 sechr )= 4 .32 × 106 J

w stored=w−wused=864×103 J

Problem 2.8A car battery kept in storage in the basement needs recharging. If the voltage and the current provided by the charger during a charge cycle are shown in FigureP2.8,a. Find the total charge transferred to the battery.b. Find the total energy transferred to the battery.



Solution:Known quantities:Recharging current and recharging voltage

Find:a) Total transferred chargeb) Total transferred energy

Analysis:

a)

Q = area under the current-time curve=∫ Idt

=12(4 )(30 )(60)+6 (30 )(60 )+1

2(2)(90 )(60)+ 4 (90)(60 )+1

2( 4 )(60)(60 )=48,600 C

Q = 48,600 C

b)

dwdt

=p

w=∫ pdt=∫ vidt

v = 9 + 310800

t V, 0≤t≤10800 s

i1 = 10- 41800

t A, 0≤t≤1800 s

i2 = 6 - 25400

t A, 1800≤t≤72 00 s

i3 = 12-43600

t A, 7200≤t≤10800 s

wherei=i1+i2+i3

Therefore,



w=∫0

1800vi1dt+∫1800

7200vi2 dt+∫7200

10800vi3 dt

¿ (90 t+t2

720−t2

100−t3

4 .86×106 )|01800

+(54 t+ t2

1200−t 2

600−t3

29 .16×106 )|18007200

+(108 t+t2

600−t 2

200−t3

9 .72×106 )|720010800

¿132 .9×103+332.8×103−94 .3×103+648×103−566 .4×103

Energy = 453 kJ

Problem 2.9Suppose the current flowing through a wire is givenby the curve shown in Figure P2.9.a. Find the amount of charge, q, that flows throughthe wire between t1 = 0 and t2 = 1 s.b. Repeat part a for t2 = 2, 3, 4, 5, 6, 7, 8, 9, and 10 s.c. Sketch q(t) for 0 ≤ t ≤ 10 s.

Solution:Known quantities:Current-time curve

Find:a) Amount of charge during 1st secondb) Amount of charge for 2 to 10 secondsc) Sketch charge-time curve

Analysis:

a) i= 4×10−3 t

1

Q1=∫0

1idt=∫0

14×10−3 tdt=4×10−3 t 2

2|01=2×10−3 A⋅s=2×10−3 Coulombs

b) The charge transferred from t=1 to t=2 is the same as from t=0 tot=1 . Q2=4 ´ 10−3C oulombs

The charge transferred from to is the same in magnitude and opposite in direction to that from t=1 tot=2 . Q3=2 ´ 10−3 Coulombst=4

Q4=2´ 10−3−∫3

44 ´ 10−3dt=2´ 10−3−4 ´ 10−3 =-2 ´ 10−3 Coulombs



t=5,6,7

Q5 =-2 ´ 10−3+∫4

52´ 10−3 dt=0

Q6=0+∫5

62 ´ 10−3 dt=2 ´ 10−3 Coulombs

Q7=2 ´ 10−3+∫6

72 ´ 10−3 dt=4 ´ 10−3 Coulombs

t=8,9 ,10 sQ=4 ´ 10−3 Coulombs

In Figure P2.10 v1(t) appears twice and should be v2(t) for the second hour of the voltage graph

Problem 2.10 A two rate charge cycle is used to charge a battery. The current to the battery is held at 70 mA for 1 hour and then switched to 60 mA for the next hour.

Solution:Known quantities:See Figure P2.10

Find:a) The total charge transferred to the battery.b) The energy transferred to the battery.

Analysis:

a) Current is equal to Coulombs

Second, therefore given the current and a duration of that current, the transferred

charge can be calculated by the following equation:A ∙ t=C

The two durations should be calculated independently and then added together.0.070 A ∙3600 s=252C0.060 A ∙3600 s=216 C252 C+216 C=468C

b) P=V·I, therefore, an equation for power can be found by multiplying the two graphs together.First separate the voltage graph into two equations:

0 h → 1 h :V=5+et

5194.8 V

1 h → 2 h :V=(6− 4e1−1 )+ 4

e2−e1∗e t

Next, multiply the first equation by 0.07A and the second by 0.06A.

0 h → 1 h :P=0.35+0.07 et

5194.8

1 h → 2 h :P=0.06(6− 4e1h−1 )+0.06 4

e2 h−e1h∗e t /3600 V

Finally, since Energy is equal to the integral of power, take the integral of each of the equations for their specified times and add them together.



0 h → 1 h :E=[0.35t+363.64 et

5194.8 ]∨36000

= 1623.53 J

1 h → 2 h :E=[0.22t+184.98 e t /3600 ]∨72003600= 1656 J

ETota l=3279.53 J

Problem 2.11The charging scheme used in Figure P2.11 is an example of a constant-current charge cycle. Thecharger voltage is controlled such that the current into the battery is held constant at 40 mA, as shown inFigure P2.11. The battery is charged for 6 h. Find:a. The total charge delivered to the battery.b. The energy transferred to the battery during thecharging cycle.Hint: Recall that the energy, w, is the integral of power, or P = dw/dt.

Solution:Known quantities:Current-time curve and voltage-time curve of battery recharging


Analysis:

a) 40 mA=0 . 04 A

Q =area under the current-time curve=∫ Idt= (0. 04 ) (6 ) (3600 )=864 C

Q = 864 C

b)

dwdt

=P

Energy =1 ,167 J

Problem 2.12The charging scheme used in Figure P2.12 is calleda tapered-current charge cycle. The current starts atthe highest level and then decreases with time for theentire charge cycle, as shown. The battery is charged



for 12 h. Find:a. The total charge delivered to the battery.b. The energy transferred to the battery during thecharging cycle.Hint: Recall that the energy, w, is the integral of power, orP = dw/dt.

Solution:Known quantities:Current-time curve and voltage-time curve of battery recharging


Analysis:

a) Q = area under the current-time curve=∫ Idt=(3600 )∫

0

12

e-5t/12 dt=8,582 C

Q = 8,582 C

b) dwdt=P

w=∫Pdt=∫ vidt=(3600 )∫0

12

(12−3 e-5t/12) (e-5t/12)dt

¿90,022JEnergy = 90,022J

Problem 2.13Use KCL to determine the unknown currents in Figure P2.13.

Solution:Known quantities:io = 2 A, i2 = -7 A

Find:a) i1

b) i3

Analysis:

a) Use KCL at the node between Ro ,R1, and R2. i0−i1+i2=0

i1=i0+i2

i1=−5 A

b) Use KCL at the node between R2 ,R3, and the current source.2.11

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


6 A+i3−i2=0i3=i2−6 Ai3=−11 A


Solution:Known quantities:ia = 3 A, ib = -2 A, ic = 1 A, id = 6 A, ie = -4 A

Find:a) i1

b) i2

Analysis:

a) Use KCL at Node A.i1+ ib−ia−ic=0

i1=ia−ib+ic

i1=6 A

b) Use KCL at Node B.ie+id−i1−i2=0

i2=ie+id−i1

i2=−4 A


Solution:Known quantities:ia = 2 mA, ib = 7 mA, ic = 4 mA

Find:a) i1

b) i2

c) i3



Analysis:

a) Use KCL at Node A.ib−ia−i1=0

i1=ib−ia

i1=5 mA

b) Use KCL at Node C.ic−i2−ib=0

i2=ic−ib

i2=−3 mA

c) Use KCL at Node D.i3+ia−ic=0

i3=ic−ia

i3=2mA

Problem 2.16Use KVL to determine the unknown voltages in Figure P2.16.

Solution:Known quantities:Va = 2 V, Vb = 4 V, Vc = 5 V

Find:a) V1

b) V2

c) V3

Analysis:

a) Use KVL at the third loop.V 3−V c=0

V 3=V c

V 3=5 V

b) Use KVL at the second loop.V 2−V 3−V b=0

V 2=V 3+V b

V 2=9V

c) Use KCL at the first loop.V a−V 1−V 2=0



V 1=V a−V 2

V 1=−7V


Solution:Known quantities:ia = -2 A, ib = 6 A, ic = 1 A, id = -4 A

Find:a) i1

b) i2

c) i3

d) i4

Analysis:

a) Use KCL at Node A.i1−ia−ic=0

i1=ia+ic

i1=−1 A

b) Use KCL at Node B.i2−i1−ib=0

i2=i1+ ibi2=5 A

c) Use KCL at Node C.i3−i2−id=0

i3=i2+id

i3=1 A

d) Use KCL at Node D.ic+ i4−i3=0

i4=i3−ic

i4=0 A



Section 2.4 Sign Convention

Problem 2.18In the circuits of Figure P2.18, the directions ofcurrent and polarities of voltage have already beendefined. Find the actual values of the indicated currentsand voltages.

Solution:Known quantities:Circuit shown in Figure P2.18.

Find:Voltages and currents in every figure.

Analysis:

(a) Using I=15

30+20(clockwise current )

:i1=−0 .3 A;i2=0 .3 A;v1=6 V )

(b) The voltage across the 20 resistor is

204=5V

; since the current flows from

top to bottom, the polarity of this voltage is positive on top. Then it follows that

v1=5 V and i2=

530

=-0 .167A

(the negative sign follows from the direction of i2 in the drawing).

(c) Since -0.5A pointing upward is the same current as 0.5A pointing downward, the

voltage across the 30 resistor is

V 30=15V (positive on top); and i2=

1520

=0 .75A,

sinceV 30 is also the voltage across the 20 resistor. Finally,i1=−( i2+0 . 5)=−1. 25 A ,and v1=−30 i1+15=52.5 V

Problem 2.19In the circuits of Figure P2.19, find the power delivered by each source.


a) Power delivered by the 3A Current Sourceb) Power delivered by the -9V Voltage Source



Analysis:

a) Follow the counterclockwise current:

P=+vi=(+3 A )∙(+10 V )P=+30W delivered

b) Follow the counterclockwise current:

P=−vi=−(+5 A)∙(−9V )P=+45 W delivered

Problem 2.20Determine which elements in the circuit of FigureP2.20 are supplying power and which are dissipatingpower. Also determine the amount of power dissipatedand supplied.


Find:Determine power dissipated or supplied for each power source.

Analysis:

Element A:P=-vi=−(−12 V )(25 A )=300 W (dissipating)

Element B:P=vi=(15V )(25 A )=375W (dissipating)

Element C:P=−vi=−(27 V )(25 A )=−675 W (supplying)

Problem 2.21In the circuits of Figure P2.21, find the power absorbed by R4 and the power delivered by the current source.


Find:a) Power absorbed by R4

b) Power delivered by the current source



Analysis:

a) Follow the current in the rightmost loop through R4:

P=(2 A )∙(15 V )P=+30W dissipated

b) Use KVL at the leftmost loop to find V3:10 V−2V−1V−v3=0

v3=7 V

Use KVL at the rightmost loop to find V5:7 V−15 V−v5=0

v5=−8V

The current source has a -8V drop across it. Use this to calculate the power dissipated using the proper sign

convention.

P=−vi=− (+2 A ) ∙ (−8V )=+16 W dissipated

Problem 2.22For the circuit shown in Figure P2.22:a. Determine which components are absorbing powerand which are delivering power.b. Is conservation of power satisfied? Explain youranswer.


Find:a) Determine power absorbed or power deliveredb) Testify power conservation

Analysis:

By KCL, the current through element B is 5A, to the right.

By KVL on the outer loop, −va−3+10+5=0 .

Therefore, the voltage across element A is

(positive at the top).

A supplies(12 V ) (5 A )=60 W



B supplies (3 V ) (5 A )=15 W

C absorbs (5V ) (5 A )=25 W

D absorbs (10V ) (3 A )=30W

E absorbs (10 V ) (2 A )=20 W

Total power supplied = 60 W +15 W =75W

Total power absorbed = 25 W +30 W +20 W=75W

Tot. power supplied = Tot. power absorbed

conservation of power is satisfied.

Problem 2.23For the circuit shown in Figure P2.23, determinethe power absorbed by the 5 Ω resistor.


Find:Power absorbed by the5 resistance.

Analysis:

The current flowing clockwise in the series circuit is i=20 V

20=1 A

The voltage across the 5 resistor, positive on the left, is v5=(1 A ) (5)=5 V

Therefore, P5=(5 V ) (1 A )=5 W

Problem 2.24For the circuit shown in Figure P2.24, determinewhich components are supplying power and which aredissipating power. Also determine the amount ofpower dissipated and supplied.


Find:Determine power absorbed or power delivered and corresponding amount.

Analysis:

If current direction is out of power source, then power source is supplying, otherwise it is absorbing.

A supplies(100 V ) (4 A )=400 W2.18



B absorbs (10 V ) (4 A )=40 W

C supplies (100 V ) (1 A )=100W

D supplies (−10 V ) (1 A )=−10 W , i.e absorbs 10 W

E absorbs (90 V ) (5 A )=450W

Problem 2.25For the circuit shown in Figure P2.25.determinewhich components are supplying power and which aredissipating power. Also determine the amount ofpower dissipated and supplied.


Find:Determine power absorbed or power delivered and corresponding amount.

Analysis:

If current direction is out of the positive terminal, then this is a power source which is supplying, otherwise it is

absorbing.

A absorbs (5 V ) (4 A )=20 W

B supplies (2 V ) (6 A )=12 W

D supplies (3 V ) (4 A )=12W

Since conservation of power is satisfied, Tot. power supplied = Tot. power absorbed

Total power supplied = 12W+12W=24W

Cabsorbs24W−20W=4 W



Problem 2.26If an electric heater requires 23 A at 110 V, determinea. The power it dissipates as heat or other losses.b. The energy dissipated by the heater in a 24-hperiod.c. The cost of the energy if the power company charges at the rate 6 cents/kWh.

Solution:Known quantities:Current absorbed by the heater; voltage at which the current is supplied; cost of the energy.

Find:a) Power consumptionb) Energy dissipated in 24 hr.c) Cost of the Energy

Assumptions:The heater works for 24 hours continuously.

Analysis:

a) P=VI=110 V (23 A )=2 .53×103 W=2 .53 KW

b) W=Pt=2 . 53×103 J

s×24 hr×3600 s

hr=218 . 6 MJ=60. 72 KW-hr

c) Cost=(Rate )×W=0 .06 $

kW−hr (2 .53 kW ) (24 hr )=$ 3.64

Problem 2.27In the circuit shown in Figure P2.27, determine theterminal voltage of the source, the power supplied tothe circuit (or load), and the efficiency of the circuit.Assume that the only loss is due to the internalresistance of the source. Efficiency is defined as theratio of load power to source power.


Circuit shown in Figure P2.27 with voltage source, vs=12V ; internal resistance of the source, R s=5 k ; and

resistance of the load, R0=7 k .

Find:The terminal voltage of the source; the power supplied to the circuit, the efficiency of the circuit.

Assumptions:Assume that the only loss is due to the internal resistance of the source.



Analysis:KVL : −v S+iT RS+vT=0OL : vT= iT R0

∴ iT=vT

RL

−vS+vT

R0RS+vT=0

vT=vS

1+RS

R0

=12 V

1+5 k7 k

=7 V or

VD : vT=vS R0

RS+R0=12 V⋅7 k

5 k+7 k=7 V .

PL=vT

2

R0=

(7 V )2

7×103 VA

=7 mW

η=Pout

Pin=

PL

PRS+PL

=iT

2 R0

iT2 RS+iT

2 R0

= 7 k5 k+7 k

=0. 5833 or 58 .33 %.



Problem 2.28A 24-volt automotive battery is connected to twoheadlights, such that the two loads are in parallel; eachof the headlights is intended to be a 75-W load,however, a 100-W headlight is mistakenly installed.What is the resistance of each headlight, and what isthe total resistance seen by the battery?

Solution:Known quantities:Headlights connected in parallel to a 24-V automotive battery; power absorbed by each headlight.

Find:Resistance of each headlight; total resistance seen by the battery.

Analysis:

Headlight no. 1:

P=v ´ i=100W=v2

R or

R=v2

100=576

100=5 .76

Headlight no. 2:

P=v ´ i=75W=v2

R or

R=v2

75=576

75=7 .68

The total resistance is given by the parallel combination:1

RTOTAL= 1

5 .76+ 1

7 .68 or RTOTAL = 3.29

Problem 2.29What is the equivalent resistance seen by thebattery of Problem 2.28 if two 15-W taillights areadded (in parallel) to the two 75-W (each)headlights?

Solution:Known quantities:Headlights and 24-V automotive battery of problem 2.13 with 2 15-W taillights added in parallel; power absorbed by each headlight; power absorbed by each taillight.

Find:Equivalent resistance seen by the battery.

Analysis:

The resistance corresponding to a 75-W headlight is:



R75 W=v2

75=576

75=7 . 68

For each 15-W tail light we compute the resistance:

R15 W=v2

15=576

15=38 . 4

Therefore, the total resistance is computed as:1

RTOTAL= 1

7 . 68+ 1

7 .68+ 1

38 . 4+ 1

38. 4 orRTOTAL=3.2

Problem 2.30Using the circuit in Figure P2.30, determine the powerabsorbed by the potentiometer and plot it as a function of R.

Solution:Known quantities:vs=15V, Rs=10 Ohms, and the circuit in Figure P2.30.

Find:RAnalysis:Use ohms law to find an equation for P as a function of R:

PR=V R∗I R

The voltage across R is equal to the source voltage minus the voltage across Rs:V R=15 V−V Rs

VRS is determined by the current through the loop which can be found by adding the resistors in series:

I R=15 V

(Rs+R)V RS=10Ω∗I R

Simplify:

PR=[15 V−(10 Ω∗15 V10Ω+R )]∗[ 15 V

10 Ω+R ]Plot:

PR



R

Problem 2.31Using the circuit in Figure P2.27 to determine the followingvalues relating to the power supply and consumption of the circuit.

Solution:Known quantities:vs=15V, Rs=100 Ohms, iT= 0, 10, 20, 30, 80, 100 mA.The circuit in Figure P2.27.

Find:a) The total power supplied by the ideal sourceb) The power dissipated within the non-ideal sourcec) How much power is supplied to the load resistord) Plot vT and power supplied to R0 as a function of iT.

Analysis:a) The power supplied by the ideal source is equal to the current through the loop times the 15V of the supply.

From current lowest to highest the power supplied would be:0W 0.15 W 0.3 W 0.45W 1.2 W 1.5W

b) The power dissipated within the non-ideal source is the power dissipated by Rs which can be found using P=i2*r. From current lowest to highest the power dissipated would be:

0 W 0.01 W 0.04 W 0.09 W 0.64 W 1Wc) The power supplied to the load resistor is equal to the total power supplied minus the power dissipated by

the non-ideal source. From current lowest to highest the power supplied would be:0 W 0.14 W 0.26 W 0.36 W 0.56 W 0.5 W

d) For the vT plot Ohm’s Law can be used to find the voltage drop across Rs which is vT = 15 – 100iT. For the power plot, the data from part c can be used directly.



Vs across R2 in figure P2.32 should be replaced with V2

Problem 2.32In the circuit in Figure P2.32, assume v2 = vs/6 and the powerdelivered by the source is 150 mW. Find R, vs, v2, and i.

Solution:Known quantities:R1=8kΩ, R2=10kΩ, R3=12kΩ, Ps = 150 mW and the circuit in Figure P2.32.

Find:R,vs, v2, and i.Analysis:Use ohms law to find vs and i:

v2=vs

6=R2∗i

Also:vs∗i=150 mW

So:



i= 2√ 150mWR2∗6

=1.58 mA

Since we know i, vs can easily be found:vs∗i=150mW

vs=94.94 V

v2=vs

6=15.82V

Use Ohm’s law to find Req:

vs

i=Req=60 k Ω

Use Req to find R:Req=R1+R+R2+R3

R=Req−(R¿¿1+R2+R3)=30 k Ω¿

Problem 2.33A GE SoftWhite Longlife lightbulb is rated as follows:PR = rated power = 60 WPOR = rated optical power = 820 lumens (lm) (average)1 lumen = 1/680WOperating life = 1,500 h (average)VR = rated operating voltage = 115 VThe resistance of the filament of the bulb, measured with a standard multimeter, is 16.7 Ω. When the bulb is connected into a circuit and is operating at the rated values given above, determinea. The resistance of the filament.b. The efficiency of the bulb.

Solution:Known quantities:Rated power; rated optical power; operating life; rated operating voltage; open-circuit resistance of the filament.

Find:a) The resistance of the filament in operationb) The efficiency of the bulb.

Analysis:a)

P=VI ∴ I=PR

V R=60 W

115 V=521 .7 mA



OL: R=V

I=

V R

I=115V

521 .7 mA=220 . 4

b)Efficiency is defined as the ratio of the useful power dissipated by or supplied by the load to the total power supplied by the source. In this case, the useful power supplied by the load is the optical power.

Po , out= Optical Power Out =820 lum W

680 lum=1 .206 W

η= efficiency

=Po ,out

PR=1. 206 W

60 W=0. 0201=2. 01 %

.

Problem 2.34An incandescent lightbulb rated at 100 W will dissipate 100 W as heat and light when connected across a 110-V ideal voltage source. If three of these bulbs are connected in series across the same source, determine the power each bulb will dissipate.

Solution:Known quantities:Rated power; rated voltage of a light bulb.

Find:The power dissipated by a series of three light bulbs connected to the nominal voltage.

Assumptions:The resistance of each bulb doesn’t vary when connected in series.

Analysis:When connected in series, the voltage of the source will divide equally across the three bulbs. The voltage across each bulb will be 1/3 what it was when the bulbs were connected individually across the source. Power dissipated in a resistance is a function of the voltage squared, so the power dissipated in each bulb when connected in series will be 1/9 what it was when the bulbs were connected individually, or 11.11 W:

Ohm’s Law: P=IV B=I 2 RB=

V B2

RB V B=V S=110VRB=

V B2

P=

(110 V )2

100 W=121

Connected in series and assuming the resistance of each bulb remains the same as when connected individually:

KVL: −V S+V B 1+V B2+V B 3=0

OL: −V S+ IRB 3+ IRB 2+ IRB 1=0

I=V S

RB 1+RB 2+RB 3=110 V

(121+121+121 )=303 mA

PB 1=I 2 RB 1=(303 mA )2 (121 )=11. 11 W=19

100 W .



Problem 2.35An incandescent lightbulb rated at 60 W willdissipate 60 W as heat and light when connectedacross a 100-V ideal voltage source. A 100-W bulbwill dissipate 100 W when connected across the samesource. If the bulbs are connected in series across thesame source, determine the power that either one of thetwo bulbs will dissipate.

Solution:Known quantities:Rated power and rated voltage of the two light bulbs.

Find:The power dissipated by the series of the two light bulbs.

Assumptions:The resistance of each bulb doesn’t vary when connected in series.

Analysis:

Ohm’s Law: P=IV B=I 2 RB=

V B2

RB

V B=V S=110V

R60=V B

2

P60=

(110 V )2

60 W=201 .7

R100=V B

2

P100=

(110 V )2

100 W=121

When connected in series and assuming the resistance of each bulb remains the same as when connected individually:

KVL: −V S+V B 60+V B 100=0

OL: −V S+ IRB 60+ IRB 100=0

I=V S

RB 60+RB100=110V

201.7+121 VA

=340.9 mA

PB 60=I 2 RB 60=(340 . 9 mA )2 (201 .7 )=23 .44 WPB 100=I 2 RB 100= (340. 9 mA )2 (121 )=14 . 06 W

Notes: 1.It’s strange but it’s true that a 60 W bulb connected in series with a 100 W bulb will dissipate more power than the 100 W bulb. 2. If the power dissipated by the filament in a bulb decreases, the temperature at which the filament operates and therefore its resistance will decrease. This made the assumption about the resistance necessary.

Problem 2.36For the circuit shown in Figure P2.36, find the voltagevab and the power dissipated in R2.



Solution:Known quantities:Circuit of Figure P2.36. R1=5Ω, R2=3Ω, R3=4Ω,

R4=5Ω, vs=12V.

Find:vab and the power dissipated in R2.

Analysis:Find the current through each resistor pair by combining them in series:

R1+R2=8ΩR3+R4=9 Ω

Since there is 12V across both of them Ohm’s law is used to find the currents:12V=8 Ω∗I a

I a=1.5 A12 V=9 Ω∗Ib

I b=1.33 ANow find the voltage drop across R1 and R3:

V a=12V −( I a∗R1)=4.5VV b=12V −( I b∗R3)=6.67 V

V ab=−2.17 VPR2 is equal to Ib times Va:

PR 2=4.5V∗1.5 A=6.75 W

Problem 2.37For the circuit shown in Figure P2.37, find the currentsi2 and i1 and the power supplied by the voltage source.

Solution:Known quantities:Circuit of Figure P2.37. R1=20Ω, R2=12Ω, R3=10Ω,

Is=3A, vs=7V.

Find:i1, i2, and Pv.

Analysis:Use KVL of the leftmost loop:

vs−R1∗i1−R2∗i2−R3∗i1=0vs=R1∗i1+R2∗i2+R3∗i1

Use KCL where i1 meets is:i1+ is−i2=0



i1+ is=i2

Combine the two equations to solve for i1:

vs=R1∗i1+R2∗(i1+is)+R3∗i1

i1=−0.69 A

Solve for i2:i1+ is=i2

i2=2.31 A

Power delivered equals the vs times i1:Pv=vs∗i1=−4.83 W

Problem 2.38For the circuit shown in Figure P2.38, find the currentsi2 and i1 and the power supplied by the voltage sources.

Solution:Known quantities:Circuit of Figure P2.38. R1=18Ω, R2=10Ω, v1=15V, v2=6V.

Find:i1, i2, Pv1,Pv2.

Analysis:Use KVL of the rightmost loop:

v2+R1∗(−i1)−R2∗i2=0v2=R1∗i1+R2∗i2

Use KCL where i1 meets is:i−i1−i2=0

i1+ i2=i

The assumption can be made that v1 is equal to the voltage across R2.v1=R2∗i2

i2=1.5 A

The assumption can be made that the voltage drop across R1 is equal to the difference between v1 and v2.v1−v2=R1∗i1

i1=0.5 A

Solve for i:i1+ i2=i

i=2.0 A2.30



Power delivered equals the voltage source times the current through it:P1=v1∗i=30.0 W

P2=v2∗(−i1)=−3W

Problem 2.39Consider NiMH hobbyist batteries shown in thecircuit of Figure P2.39.a. If V1 = 12.0 V, R1 = 0.15 Ω and RL = 2.55Ωfind the load current IL and the power dissipated bythe load.b. If we connect a second battery in parallel withbattery 1 that has voltage V2 = 12 V andR2 = 0.28 Ω, will the load current IL increase ordecrease? Will the power dissipated by the loadincrease or decrease? By how much?

Solution:Known quantities:Schematic of the circuit in Figure P2.39.

Find:a) If V 1=12. 0V , R1=0.15 , RL=2.55 , the load current and the power dissipated by the loadb) If a second battery is connected in parallel with battery 1with V 2=12. 0 V , R2=0. 28 , determine the

variations in the load current and in the power dissipated by the load due to the parallel connection with a second battery.

Analysis:

a)I L=

V 1

R1+RL=12

0 . 15+2. 55=12

2. 7 = 4.444 APLoad=I L

2 RL = 50.4 W.

b) with another source in the circuit we must find the new power

dissipated by the load. To do so, we write KVL twice using mesh

currents to obtain 2 equations in 2 unknowns:

−v2+i2 R2+(i2−iL ) R1+v1=0 ¿¿¿¿

0 .43⋅i2−0 . 15⋅iL=0 ¿ ¿¿¿Solving the above equations gives us:

I 2=1 . 59 A I L=4 . 55 A PLoad=I L2 RL=52.79W



This is an increase of 5%.

Problem 2.40With no load attached, the voltage at the terminalsof a particular power supply is 50.8 V. When a 10-Wload is attached, the voltage drops to 49 V.a. Determine vS and RS for this non ideal source.b. What voltage would be measured at the terminalsin the presence of a 15-Ωload resistor?c. How much current could be drawn from this powersupply under short-circuit conditions?

Solution:Known quantities:Open-circuit voltage at the terminals of the power source is 50.8 V; voltage drop with a 10-W load attached is 49 V.

Find:a) The voltage and the internal resistance of the sourceb) The voltage at its terminals with a 15- load resistor attachedc) The current that can be derived from the source under short-circuit conditions.

Analysis:

(a)

(49 V )2

RL=10W ⇒RL=240

vs=50.8 V , the open-circuit voltage isRL

RS+RLv S=49⇒240

RS+24050. 8=49

RS=

(240)(50 . 8)49

−240=8 .82

(b)v=

RL

RS+RLv S=

158 .82+15

50 .8=31 . 99 V

(c)iCC (RL=0 )=

v S

RS=50 .8

8.82=5 .76

A

Problem 2.41A 220-V electric heater has two heating coilswhich can be switched such that either coil can be usedindependently or the two can be connected in series orparallel, yielding a total of four possible configurations.If the warmest setting corresponds to 2,000-W powerdissipation and the coolest corresponds to 300 W, finda. The resistance of each of the two coils.b. The power dissipation for each of the other twopossible arrangements.

Solution:Known quantities:Voltage of the heater, maximum and minimum power dissipation; number of coils, schematics of the configurations.

Find:a) The resistance of each coilb) The power dissipation of each of the other two possible arrangements.



Analysis:

(a) For the parallel connection, P=2000 W. Therefore,

2000= (220 )2

R1+

(220 )2

R2=(220 )2( 1

R1+ 1

R2 )or,

1R1+ 1

R2= 5

121 .

For the series connection, P=300 W. Therefore, 300= (220 )2

R1+R2 or,

1R1+R2

= 3484 .

Solving, we find that R1=131. 7and R2=29 .6 .

(b) the power dissipated by R1 alone is:

PR1= (220 )2

R1=368 W

and the power dissipated by R2 alone is

PR2=

(220 )2

R2=1635 W

.

Section 2.5, 2.6 Resistance and Ohm’s Law

Problem 2.42For the circuits of Figure P2.42, determine theresistor values (including the power rating) necessaryto achieve the indicated voltages. Resistors areavailable in 1/8-, 1/4-, 1/2-, and 1-W ratings.

Solution:Known quantities:Circuits of Figure 2.42.

Find:Values of resistance and power rating

Analysis:

(a)20=

Ra

Ra+15 , 000(50 )

Ra (50 - 20 )=20 (15 )×103

Ra=10 k

Pa=I 2 R=(5025000 )

2

(10 , 000 )=40 mW

PRa=1

8W



P1=I2 R =60 mW

PR1=1

8W

(b)

2 .25=5×(10 , 00010 , 000+Rb )

Rb=12 .2 k

PRb= 1

8W

PR2=1

8W

c) 28.3=110×( 2.7×103

2 .7×103+1×103+RL)

RL=6 . 8kPRL

=1W

PR3=1

8W

PR4=1

2W

Problem 2.43A 1-hp motor is powered by a DC generator 150m away. Using theresistance of copper wire chart, determine the minimum AWG thatcan be used to power the motor.

Solution:Known quantities:Figure P2.43.VM-min = 105V -> IM-FL = 7.10AVM-max = 117V -> IM-FL = 6.37AVG = 110V

Find:The minimum AWG wire that will meet the specifications.

Assumptions:Assume an ideal DC source.

Analysis:The circuit that this setup is describing is essentially a voltage source with a resistor, an inductor, and another resistor all in series.

150m is equal to 492.13 feet. This will help with the resistance later.

Assume the voltage of the motor is 105V. Use Ohm’s Law to calculate the Current through the first conductor:

I M=110V−105V

2RIM has to be 7.1 A or greater in order for the motor to draw full power:



7.1 A=2.5 VR

R=0.352 ΩUse the value of R along with the chart to find the correct AWG:

0.49213× 103 ft

=0.715 Ω /1000 ft

This means that 8 AWG wire must be used.

Problem 2. 44Cheap resistors are fabricated by depositing a thin layer of carbon onto a non-conducting cylindrical substrate (see Figure P2.50). If such a cylinder has radius a and length d, determine the thickness of the film required for a resistance R ifa = 1 mm R = 33 kΩ σ = 1/ρ= 2.9 MS/m d = 9 mmNeglect the end surfaces of the cylinder and assume that the thickness is much smaller than the radius.

Solution:Known quantities:Figure P2.50. Diameter of the cylindrical substrate; length of the substrate; conductivity of the carbon.

Find:The thickness of the carbon film required for a resistance R of 33 k.

Assumptions:Assume the thickness of the film to be much smaller than the radiusNeglect the end surface of the cylinder.



Analysis:

R=dσ⋅A

≃dσ⋅2 πa⋅Δt

Δt=dR⋅2 πa⋅σ

=9⋅10−3 m

33⋅103 ⋅2 .9⋅106 Sm⋅2 π⋅1⋅10−3 m

=14 . 97⋅10−12 m=14 .97 pm

Problem 2.45The resistive elements of fuses, lightbulbs, heaters,etc., are significantly nonlinear (i.e., the resistance isdependent on the current through the element).Assume the resistance of a fuse (Figure P2.51) is givenby the expression R = R0[1 + A(T −T0)] withT −T0 = kP; T0 = 25C; A = 0.7[C]−1;k = 0.35C/W; R0 = 0.11 Ω; and P is the powerdissipated in the resistive element of the fuse.Determine the rated current at which the circuit willmelt and open, that is, “blow” (Hint: The fuse blowswhen R becomes infinite.)

Solution:Known quantities:Figure P2.51. The constants A and k; the open-circuit resistance.

Find:The rated current at which the fuse blows, showing that this happens at:

I= 1√AkR0 .

Assumptions:Here the resistance of the fuse is given by:

R=R0 [1+A(T−T 0 )]where T 0 , room temperature, is assumed to be 25°C.We assume that:

T−T 0=kPwhere P is the power dissipated by the resistor (fuse).

Analysis:R=R0(1+A⋅DT )=R0(1+AkP)=R0(1+AkI 2 R )R−R0 AkI 2 R=R0

R=R0

1−R0 AkI 2→¥when

1−R0 AkI 2→0

I= 1√AkR0

=( 0. 7 m° C

0 .35 °CVa

0 . 11 Va)−

12

=6.09 A.



Problem 2.46Use Kirchhoff’s current law and Ohm’s law todetermine the current in each of the resistors R4, R5,andR6 in the circuit of Figure P2.52. VS = 10 V,R1 = 20 Ω, R2 = 40 Ω, R3 = 10 Ω, R4 = R5 =R6 = 15 Ω.

Solution:Known quantities:Circuit shown in Figure P2.52 with voltage source, V s=10 V and resistors, R1=20 , R2=40 , R3=10 , R4=R5=R6=15 .

Find:The current in the 15- resistors.

Analysis:

Since the 3 resistors must have equal currents,

I31I15

and,

I=V S

R1+R2||R3+R4||R5||R6=10

20+8+5=10

33=303

mA

Therefore, I 15=

1099

=101 mA

Problem 2.47Use Kirchhoff’s current law and Ohm’s law todetermine the unknown currents.

Solution:Known quantities:Circuit shown in Figure P2.13 R0=1, R1=2, R2=3, R3=4, vs=10V

Find:a) i0 b) i1 c) i2 d) i3

Analysis:Use KCL at node A:

i0+i2−i1=0Use KCL at node B:

i3+6 A−i2=0Use Ohms Law for i0:

i0=10 V−vA

1ΩUse Ohms Law for i1:


A B


i1=v A

2Ω

Use Ohms Law for i2:

i2=vB−va

3ΩUse Ohms Law for i3:

i3=0−vB

4 ΩUse algebra to solve the six equations with six unknowns. All that the problem statement asks for are the currents. The two voltages come along for free. (N.B. This solution is simply using the theory behind Nodal Analysis to solve the problem):

i0=1.83 Ai1=4.09 Ai2=2.26 A

i3=−3.74 ANote that the negative value for i3 indicates that it is flowing in the opposite direction from that assumed.

Problem 2.48Use Kirchhoff’s current law and Ohm’s law todetermine the power delivered by the voltage source.

Solution:Known quantities:Circuit shown in Figure P2.48k=0.25A/A2

Find:Power supplied by the voltage source.

Analysis:The 7 Ohm and 5 Ohm resistor are in series so i can be found by using Ohm’s law:

24 V =12∗ii=2 A

The dependent current source current can therefore be found using i:id=0.25∗¿

id=1 AUse KCL to determine the current through the voltage source:

iv+id−i=0iv=1 A

Use P=v*i to determine the power supplied by the voltage source:P=24 V∗1 A

P=24 W



Problem 2.49Assuming R0 = 2 Ω, R1 = 1Ω, R2 = 4/3 Ω R3 = 6 Ωand VS = 12 V in the circuit of FigureP2.55, use Kirchhoff’s voltage law and Ohm’s law to find a. ia, ib, and icb. the current through each resistor.

Solution:Known quantities:Schematic of the circuit shown in Figure P2.55 with resistors R0=2 ,R1=1 ,R2=4 /3 , R3=6 and voltage source V S=12V.

Find:a) The mesh currents ia , ib , ic

b) The current through each resistor.

Analysis:Applying KVL to mesh (a), mesh (b) and mesh (c):

iaR0+( ia−ib) R1=0 ¿ (ia−ib )R1−ib R2+(ic−ib) R3=0¿ ¿¿¿

2ia+ (ia−ib)=0 ¿ ( ia−ib )−43

ib+6 (ic−ib)=0 ¿ ¿¿¿

Solving the system we obtain:

ia= 2 A ¿ ib=6 A ¿¿¿¿

IR0=ia=2 A (positive in the direction of ia )¿ I R 1

=ib−ia=4 A ( positive in the direction of ib) ¿ I R2=ib=6 A (positive in the direction of ib )¿ ¿¿¿

Problem 2.50Assuming R0 = 2Ω, R1 = 2 Ω, R2 = 5 Ω, R3 = 4 A, and VS = 24 V in the circuit of FigureP2.55, use Kirchhoff’s voltage law and Ohm’s law to finda. ia, ib, and ic.b. The voltage across each resistance.

Solution:Known quantities:Schematic of the circuit shown in Figure P2.55 with resistors R0=2 , R1=2 , R2=5 , R3=4 and voltage source V S=24 V.

Find:a) The mesh currents ia , ib , ic

b) The current through each resistor.





2ia+2 ( ia−ib )=0 ¿ 2( ia−ib)−5 ib+4 (ic−ib)=0 ¿¿¿¿Solving the system we obtain:

ia= 2 A ¿ ib=4 A ¿ ¿¿¿

V R0=R0 ia=4 V (⊕ down )¿ V R1

=R1 (ib−ia )=4 V (⊕ down ) ¿ V R2=R2 ib=20 V (⊕ up )¿ ¿¿¿

Problem 2.51Assume that the voltage source in the circuit ofFigure P2.55 is now replaced by a current source, andR0 = 1 Ω, R1 = 3 ΩR2= 2Ω, R3 = 4 Ω, andIS = 12 A. Use Kirchhoff’s voltage law and Ohm’slaw to determine the voltage across each resistance.


Schematic of the circuit shown in Figure P2.55 with resistors R0=1 , R1=3 , R2=2 , R3=4 and current source I S=12 A.

Find:The voltage across each resistance.



ia+3 (ia−ib )=0 ¿ 3 (ia−ib )−2ib+4 ( ib−ic )=0¿ ¿¿¿Solving the system we obtain:

ia=163

A ¿ ib=649

A ¿ ¿¿¿

V R0=R0 ia=5. 33 V (⊕ down )¿ V R1

=R1 (ib−ia )=5 .33 V (⊕ down) ¿ V R 2=R2 ib=14 .22 V (⊕ up) ¿¿ ¿¿

Problem 2.52The voltage divider network of Figure P2.58 isexpected to provide 5 V at the output. The resistors,



however, may not be exactly the same; that is, theirtolerances are such that the resistances may not beexactly 5 kΩa. If the resistors have +/-10 percent tolerance, find the worst-case output voltages.b. Find these voltages for tolerances of +/-5 percent.Given: v = 10 V, R1 = 5 kΩ, R2 = 5 kΩ.

Solution:Known quantities:Schematic of voltage divider network shown of Figure P2.58.

Find:a) The worst-case output voltages for ±10 percent toleranceb) The worst-case output voltages for ±5percent tolerance

Analysis:

a) 10% worst case: low voltageR2 = 4500 , R1 = 5500

vOUT , MIN=45004500+5500

5=2.25 V

10% worst case: high voltageR2 = 5500 , R1 = 4500

vOUT , MAX=55004500+5500

5=2. 75 V

b) 5% worst case: low voltage

R2 = 4750 , R1 = 5250

vOUT , MIN=47504750+5250

5=2.375 V

5% worst case: high voltage

R2 = 5250 , R1 = 4750

vOUT , MAX=52505250+4750

5=2.625 V

Problem 2.53Find the equivalent resistance seen by the source and the currents i and i1and v2in the circuit of Figure P2.53.


Find:Equivalent resistance and current i and i1and v2

Analysis:Simplify resistors:

¿



Req=10

Find i:

i=9VR eq

=9V10

i=0.9 AFind i1 using current division:

i1=(0.9 A )∗72(72+9)

i1=0.8 A

The voltage drop across the 72Ω resistor (v2) is the same as the voltage drop across the 9Ω resistor.v2=0.8 A ×9 Ω=7.2V

Problem 2.54Find the equivalent resistance seen by the sourceand the current i in the circuit of Figure P2.45.


Find:Equivalent resistance and current i

Analysis:

Step1: ( 4||4 )+22=24

Step 2: 24||8=6

Therefore, the equivalent circuit is as shown in the figure:

Further, (4+6 )||90=9W

The new equivalent circuit is shown below.

Thus, Rtotal=10 .

This means the circuit is delivering 5A (50V/10Ω).



We can now find the current i by the current divider rule as follows:

i=(1010 + 90 )(5 )=0 .5 A

Problem 2.55In the circuit of Figure P2.55, the power absorbed

by the 15-Ω resistor is 15 W. Find R.

Solution:Known quantities:Circuits of Figure P2.55.

Find:Resistance R

Analysis:

Combining the elements to the right of the 15 resistor, we computeReq=( (4||4 )+6 )||24+4=10 .

The power dissipated by the 15- resistor is

P15=v15

2

15=15W

, therefore, v15= 15 V and i1 = 1 A.

Using the Ohms Law: i2=

15 V10

=1. 5 A.

Applying KCL, we can find iR: iR=i1+i2=2 .5 A .

Using KVL: −25+2 .5 R +15=0

Therefore, R = 4.



Problem 2.56Find the equivalent resistance between terminals aand b in the circuit of Figure P2.47.


Find:Equivalent resistance.

Analysis:2+2=46‖12=44‖4=24‖4=22+2=4Req=3+4‖4=5

Problem 2.57For the circuit shown in Figure P2.48, find theequivalent resistance seen by the source. How muchpower is delivered by the source?


Find:a) Equivalent resistanceb) Power delivered.

Analysis:

(a)



2+1=33‖3=1 .54+1.5+5=10 .510 .5‖6=3 .818Req=3 .818+7=10 .818

(b)

P=142

10 . 818=18.12 W

Problem 2.58For the circuit shown in Figure P2.58, find theequivalent resistance seen by the current source and the number of nodes.

Solution:Known quantities:Circuits of Figure P2.58.R1 = 2Ω, R2 = 3Ω, R3 = 85Ω,R4 = 2Ω, R5 = 4Ω

Find:The equivalent resistance seen by the current source.The number of nodes in the circuit.

Analysis:

Simplify left side:R1∨¿ R2+R3=R¿¿

R¿=86.2 Ω¿

Simplify right side:R4+R5=R¿

R¿=6ΩCombine:

Req=R¿∨¿R ¿¿

Req=5.61 Ω

Count the nodes:

Nodes=4Nodes A, B, C, & D as shown above

Problem 2.59For the circuit shown in Figure P2.59, find the number of


A B C

D


nodes, the equivalent resistance seen by the voltage source,and the current through R7.

Solution:Known quantities:Circuit of Figure P2.59. R1=10Ω, R2=5 Ω, R3=8 Ω,

R4=2 Ω, R5=4 Ω, R6=2 Ω, R7=1 Ω,R8=10 Ω vs=20V.

Find:The equivalent resistance seen by the voltage source.The current through R7.

Analysis:Count the nodes:

Nodes=4Nodes A, B, C, & D as shown above.

Combine R1 and R2:R1∨¿ R2=3.33 Ω

R4-6are in parallel with each other and in series with R1||2:R4||R5||R6+R1∨¿2=4.13 Ω

The previous resistance is in parallel with R3:R3∨¿4.13Ω=2.72Ω

R7 is in parallel with R8:R8∨¿R7=0.91 Ω

These last two resistances are in series and are equal to the equivalent resistance:Req=3.63 Ω

The current through R7||8 will be the same as the total current through the circuit. Find the total current then use

current division to find the current through R7:

I=vs

Req=5.51 A

I R7=

I∗R78

R7=5.01 A

Problem 2.60For the circuit shown in Figure P2.60, find theequivalent resistance seen by the current source


AB

C

D

A B C


and count the number of nodes in the circuit.

Solution:Known quantities:Circuit of Figure P2.60.R1 = 12Ω, R2 = 5 Ω, R3 = 8 Ω, R4 = 2 Ω,R5 = 4 Ω, R6 = 2 Ω, R7 = 1 Ω, R8 = 10 Ω,R9 = 10 Ω

Find:The equivalent resistance seen by the current source.The number of nodes in the circuit.

Analysis:

Simplify starting on the right side:

R6∨¿(R¿¿2+R7)=R ¿¿

R¿=1.5 ΩSimplify middle:

(R¿+R8 )∨¿R5=Rmiddle

Rmiddle=2.968 Ω

Finish with the left:

¿¿Req=5.05 Ω

Count the nodes:

Nodes=6See nodes A, B, C, D, E, & F above

Problem 2.61For the circuit shown in Figure P2.61, find thenumber of nodes in the circuit, the power deliveredby the voltage source and the equivalent resistanceseen by the voltage source.

Solution:Known quantities:Circuit of Figure P2.61.R1 = 9Ω, R2 = 4 Ω, R3 = 4 Ω,R4 = 5 Ω, R5 = 4 Ω, vs = 10V

Find:The number of nodes in the circuit.The power delivered by the voltage source.The equivalent resistance seen by the voltage source.


DEF

A B C

D


Analysis:

Count the nodes:

Nodes=4See nodes A, B, C, & D above

Simplify starting on the right side:R5∨¿R4=R¿

R¿=2.22 ΩSimplify middle:

(R¿+R3)∨¿R2=Rmiddle

Rmiddle=2.43 Ω

Finish with the left side:Rmiddle+R1=Req

Req=11.43 Ω

Find power delivered using watts law:

P=vs

2

Req

¿102

11. 43=8 .75

P=8.75W

Problem 2.62Determine the equivalent resistance of the infinitenetwork of resistors in the circuit of Figure P2.62.

Solution:Known quantities:Schematic of the circuit shown in Figure P2.62.

Find:

The equivalent resistance Req of the infinite network of resistors.

Analysis:We can see the infinite network of resistors as the equivalent to the circuit in the picture:

Therefore,



Req=R+(R||Req )+R=2 R+RR eq

R+Req

Req=(1+√3 ) R¿2 .73 R

Problem 2.63For the circuit shown in Figure P2.63, find thenumber of nodes in the circuit, the power deliveredby the current source and the equivalent resistanceseen by the current source.

Solution:Known quantities:Circuit of Figure P2.63.is = 5A, R1 = 10Ω, R2 = 7 Ω.R3 = 8 Ω, R4 = 4 Ω, R5 = 2 Ω

Find:The number of nodes in the circuit.The power delivered by the current source.The equivalent resistance seen by the current source.

Analysis:

Count the nodes:


Simplify the right side:R5+R4=R¿

R¿=6ΩSimplify the left side:

R1∨¿ R2+R3=R¿¿

R¿=12.12Ω¿

Combine both sides:R¿∨¿ R¿=Req ¿

Req=4.01Ω



A B C

D


P=i s2 Req

¿52⋅4 .01=P=100.325 W

Problem 2.64In the circuit of Figure P2.64, find the equivalentresistance looking in at terminals a and b if terminals cand d are open and again if terminals c and d areshorted together. Also, find the equivalent resistancelooking in at terminals c and d if terminals a and b areopen and if terminals a and b are shorted together.


Find:The equivalent resistance at terminals a,b in the case that terminals c,d are a) open b) shorted; the same for terminals c,d with respect to terminals a,b.

Analysis:

With terminals c-d open, Req= (360+540 )‖(180+540)=400 ,

with terminals c-d shorted, Req= (360‖180)+(540‖540)=390 ,

with terminals a-b open, Req= (540+540 )‖(360+180)=360 ,

with terminals a-b shorted, Req= (360‖540)+(180‖540 )=351 .

Problem 2.65In the circuit of Figure P2.64, find the equivalentresistance looking in at terminals a and b if terminals cis shorted to terminal a and terminal d is shorted to terminalb.


Find:The equivalent resistance seen by terminals a and b.



Analysis:Shorting a to c eliminates the 360Ω resistor. Shorting b to d eliminates the rightmost 540Ω resistor. This leaves the remaining two resistors in parallel with each other.

540 Ω∨¿180 Ω=Req

Req=135 Ω



Problem 2.66For the circuit shown in Figure P2.66, find thenumber of nodes in the circuit and the equivalentresistance seen by the voltage source.

Solution:Known quantities:Circuit of Figure P2.66.R1 = 12Ω, R2 = 5 Ω, R3 = 8 Ω,R4 = 2 Ω, R5 = 4 Ω

Find:The equivalent resistance seen by the voltage source.The number of nodes in the circuit.

Analysis:

Count the nodes:


Simplify R2, R3, and R5 using a wye-delta transformation (N.B. The nodes are labeled A, B, & C to match the nodes

in figure 2.69a. Node D is the common node of the wye:Rx=wye=R2=5R y−wye=R5=4R z−wye=R3=8

Now using equations 2.37 through 2.39

R1−delta=Rx R y+Rx R z+R y R z

R y=92

4=23( this goes from A to C in parallel with the R1 of the circuit

R2−delta=92R z

=11. 5( this goes directly from A to B )

R3−delta=92Rx

=18 . 4( this goes from C to B in parallel with the R4 of the circuit

RAC=R1‖R1−delta=7 .89RCB=R4‖R3−delta=1 . 80Req=RAB=(R AC+RCB )‖R2−delta=5 .25

Problem 2.672.52


A

B

C D


Determine the voltage between nodes A and B inthe circuit shown in Figure P2.67.VS = 12 VR1 = 11 kΩ R3 = 6.8 kΩR2 = 220 kΩ R4 = 0.22 MΩ

Solution:Known quantities:Schematic of the circuit shown in Figure P2.67 with source voltage, V s=12 V ;

and resistances, R1=11k , R2=220 k , R3=6 .8 k , R4=0 .22 M

Find:The voltage between nodes A and B.

Analysis:The same current flows through R1 and R3. Therefore, they are connected in series. Similarly, R2 and R4 are connected in series.SPECIFY THE ASSUMED POLARITY OF THE VOLTAGE BETWEEN NODES A AND B. THIS WILL HAVE TO BE A GUESS AT THIS POINT.Specify the polarities of the voltage across R3 and R4 which will be determined using voltage division. The actual polarities are not difficult to determine. Do so.

VD : V R 3=V S R3

R1+R3=

[12 V ] [6 .8 k ]11+6 . 8 k

=4 .584 V

VD : V R 4=V S R4

R2+R4=

[12 V ] [0 . 22×103 k ](220+0. 22×103) k

=6 V

KVL : −V R 3+V AB+V R 4=0∴V AB=V R 3−V R 4=−1 . 416 V

The voltage is negative indicating that the polarity of V AB is opposite of that specified.A solution is not complete unless the assumed positive direction of a current or assumed positive polarity of a voltage IS SPECIFIED ON THE CIRCUIT.

Problem 2.68For the circuit shown in Figure P2.68, find thenumber of nodes in the circuit, the power deliveredby the voltage source and the equivalent resistanceseen by the voltage source.

Solution:Known quantities:Circuit of Figure P2.68.R1=12Ω, R2=5 Ω, R3=8 Ω,R4=2 Ω, R5=4 Ω, R6=2 Ω,R7=1 Ω, R8=R9=10 Ω, vs=15V

Find:The number of nodes in the circuit.The power delivered by the voltage source.The equivalent resistance seen by the voltage source.


A B

C

D

E

F


Analysis:

Count the nodes:

Nodes=6See nodes A, B, C, D, E, & F above

R9 is eliminated due to the short.

Start at the right and work left:

R8∨¿(R¿¿6+R7)=R¿¿

R¿=2.3 Ω(R¿¿¿+R4)∨¿R5=Rmiddle¿

Rmiddle=2.1 ΩRmiddle+(R3∨¿ (R1+R2 ))=Req

Req=7.51 Ω


P=vs

2

Req=152

7 .51=29 .94 W

P=29.9 W

Problem 2.69For the circuit shown in Figure P2.69, find thevoltages vac and vbd. Also find the power delivered by the voltage source.

Solution:Known quantities:Circuit of Figure 2.69. R1=12Ω, R2=5Ω, R3=8Ω, R4=2Ω, R5=4Ω, R6=2Ω, R7=1Ω,vs=15V.

Find:vac

vbd

The power delivered by the voltage source.

Analysis:

Find Req by adding all the resistors in series:Req=R1+R2+R3+R4+R5+R6+R7

Req=34 Ω

Use the equivalent resistance to find the current through the loop:15V=Req∗I



I=0.44 AFind voltage across R1, R2, and R3 using the current:

vac=(R1+R2+R3 )∗Ivac=11.03 V

Do the same with R2-6:vbd=(R2+R3+R4+R5+R6 )∗I

vbd=9.26 V

Find power delivered:

P=vs

2

Req=152

34=6 . 618

P=6.62W

Sections 2.7, 2.8: Practical Sources and Measuring Devices

Problem 2.70A thermistor is a nonlinear device which changes its terminal resistance value as its surroundingtemperature changes. The resistance and temperature generally have a relation in the form ofRth(T ) = R0e−β(T−T0)

where Rth = resistance at temperature T, ΩR0 = resistance at temperature T0 = 298 K, Ωβ = material constant, K−1T, T0 = absolute temperature, Ka. If R0 = 300Ω and β = −0.01 K−1, plot Rth(T ) asa function of the surrounding temperature T for350 ≤ T ≤ 750.b. If the thermistor is in parallel with a 250-Ω resistor, find the expression for the equivalent resistance andplot Rth(T ) on the same graph for part a.


Parameters R0=300 (resistance at temperature T 0 = 298 K), andβ =- 0 .01 K-1, value of the second resistor =

250Ω.

Find:a) Plot Rth (T ) versus T in the range 350≤T≤750 [K]b) The equivalent resistance of the parallel connection with the 250- resistor; plot Req (T ) versus T in the

range 350≤T≤750 [K] for this case on the same plot as part a.



Assumptions:Rth (T )=R0 e

−β (T−T0 ).

Analysis:

a) Rth (T )=300 e−0. 01⋅(T−298 )

b) Req (T )=Rth (T )||250=1500 e−0 . 01 (T−298 )

5+6e−0. 01 (T−298)

The two plots are shown below.

In the above plot, the solid line is for the thermistor alone; the dashed line is for the thermistor-resistor combination.

Problem 2.71A moving-coil meter movement has a meterresistance rM= 200 Ω, and full-scale deflection iscaused by a meter current Im= 10 μA. The movementmust be used to indicate pressure measured by thesensor up to a maximum of 100 kPa. See Figure P2.71.a. Draw a circuit required to do this, showing allappropriate connections between the terminals ofthe sensor and meter movement.b. Determine the value of each component in thecircuit.c. What is the linear range, that is, the minimum andmaximum pressure that can accurately bemeasured?

Solution:Known quantities:Meter resistance of the coil; meter current for full scale deflection; max measurable pressure.

Find:a) The circuit required to indicate the pressure measured by a sensorb) The value of each component of the circuit; the linear rangec) The maximum pressure that can accurately be measured.

Assumptions:Sensor characteristics follow what is shown in Figure P2.71

Analysis:a) The upper and lower terminals of the sensor is attached to the upper and lower terminals of the meter, respectively. A series resistor to drop excess voltage is required.b) At full scale, meter:

I mFS=10 μArm=2000 . L . : V m FS=I m FS r m=2 mV .

at full scale, sensor (from characteristics):PFS=100 kPa



V TFS=10 mVKVL : −V TFS+V RFS+V m FS=0V RFS=V TFS−V mFS=10 mV−2 mV=8 mVI RFS=I TFS=I m FS=10 mA

Ohm law: R=

V RFS

I RFS= 8 mV

10 μA=800

.c) from sensor characteristic: 30 kPa –100 kPa.

Problem 2.72The circuit of Figure P2.72 is used to measure theinternal impedance of a battery. The battery beingtested is a NiMH battery cell.a. A fresh battery is being tested, and it is found thatthe voltage Vout , is 2.28 V with the switch open and2.27 V with the switch closed. Find the internalresistance of the battery.b. The same battery is tested one year later, and Vout isfound to be 2.2 V with the switch open but 0.31 Vwith the switch closed. Find the internal resistanceof the battery.

Solution:Known quantities:Schematic of the circuit shown in Figure P2.72; voltage at terminals with switch open and closed for fresh battery; same voltages for the same battery after 1 year.

Find:The internal resistance of the battery in each case.

Analysis:

a)

V out=(1010+r B )V oc

r B=10( V oc

V out−1)=10(2.28

2.27−1)=0 .044

b)

r B=10( V oc

V out−1)=10( 2.2

0 .31−1)=60 .97



Problem 2.73Consider the practical ammeter, described inFigure P2.73, consisting of an ideal ammeter in serieswith a 1-kΩ resistor. The meter sees a full-scaledeflection when the current through it is 30 μA. If wedesire to construct a multirange ammeter readingfull-scale values of 10 mA, 100 mA, and 1 A,depending on the setting of a rotary switch, determineappropriate values of R1, R2, and R3.

Solution:Known quantities:Ammeter shown in Figure P2.73; Current for full-scale deflection; desired full scale values.

Find:Value of the resistors required for the given full scale ranges.

Analysis:We desire R1,R2, R3 such that Ia= 30 A for I = 10 mA, 100 mA, and 1

A, respectively. We use conductances to simplify the arithmetic:

Ga =

1Ra

=

11000

S

G1,2,3 =

1R1,2,3

By the current divider rule:

Ia =

Ga

Ga + Gx

I

or:

G x=Ga( II a )−Ga

or

1Gx

= 1Ga

( I a

I−I a)

Rx=Ra( I a

I−I a).

We can construct the following table:

x I Rx (Approx.)

1 10-2 A 3

2 10-1 A 0.3

3 100 A 0.03



Problem 2.74A circuit that measures the internal resistance of apractical ammeter is shown in Figure P2.74, whereRS = 50,000 Ω, VS = 12 V, and Rp is a variableresistor that can be adjusted at will.a. Assume that ra<<50,000 Ω. Estimate the current i.b. If the meter displays a current of 150 μA whenRp= 15 Ω, find the internal resistance of the meter ra.

Solution:Known quantities:Schematic of the circuit shown in Figure P2.74; for part b: value of Rp and current displayed on the ammeter.

Find:The current i; the internal resistance of the meter.

Assumptions:ra <<50 k

Analysis:

a) Assuming that ra <<50 k

i

V s

R s=12

50000

= 240 A

b) With the same assumption as in part a)

imeter = 150(10)-6 =

R p

ra +R pi

or:

150(10)-6=

15ra +15

240⋅10-6

.

Therefore, ra = 9 .

Problem 2.75A practical voltmeter has an internal resistance rm.What is the value of rmif the meter reads 11.81 V when connected as shown in Figure P2.75.

Solution:Known quantities:Voltage read at the meter; schematic of the circuit shown in Figure P2.75 with source voltage, V s=12V and source resistance,R s=25 k .

Find:The internal resistance of the voltmeter.



Analysis:

Using the voltage divider rule:

V = 11.81 =

rm

rm + Rs(12 )

Therefore, r m = 1.55 M.

Problem 2.76Using the circuit of Figure P2.75, find the voltagethat the meter reads if VS = 24 V and RS has thefollowing values:RS = 0.2rm, 0.4rm, 0.6rm, 1.2rm, 4rm, 6rm, and 10rm.How large (or small) should the internal resistance ofthe meter be relative to RS?

Solution:Known quantities:Circuit shown in Figure P2.75 with source voltage, V s=24 V ; and ratios

between sR and rm .

Find:The meter reads in the various cases.

Analysis:

By voltage division:

V =

rm

rm + Rs(24 )

Rs V

0.2rm20 V

0.4rm17.14 V

0.6rm15 V

1.2rm10.91 V

4rm4.8 V

6rm3.43 V

10 rm2.18 V

For a voltmeter, we always desire rm >> R s .

Problem 2.77A voltmeter is used to determine the voltage acrossa resistive element in the circuit of Figure P2.77. Theinstrument is modeled by an ideal voltmeter in parallelwith a 120-kΩresistor, as shown. The meter is placed



to measure the voltage across R4. Assume R1 = 8 kΩR2 = 22 kΩR3 = 50 kΩ, RS = 125 kΩ, andIS = 120 mA. Find the voltage across R4 with andwithout the voltmeter in the circuit for the followingvalues:a. R4 = 100 Ωb. R4 = 1 kΩc. R4 = 10 kΩd. R4 = 100 kΩ

Solution:Known quantities:Schematic of the circuit shown in Figure P2.77, values of the components.

Find:The voltage across R4 with and without the voltmeter for the following values:a) R4=100

b) R4=1 k

c) R4=10 k

d) R4=100 k .

Assumptions:The voltmeter behavior is modeled as that of an ideal voltmeter in parallel with a 120- k resistor.

Analysis:

We develop first an expression for V R4 in terms of 4R

. Next, using current division:

432

2

4321

14

1 ||

RRRR

II

RRRRRR

II

RR

S

SSR

Therefore,

432

2

4321 ||4 RRRR

RRRRRR

IIS

SSR

V R4=I R4

R4

=I S (RS R4

RS+R1+R2‖(R3+R4 ) )⋅(R2

R2+R3+R4 )¿

2 .129×103⋅R4

R4+68 . 877×103

Without the voltmeter:

a) V R4 = 3.08 V

b) V R4 = 30.47 V



c) V R4 = 269.91 V

d) V R4 = 1260.7 V.

Now we must find the voltage drop across R4 with a 120-k resistor across R4 . This is the voltage that the

voltmeter will read.

IR4=I S( RS

RS+R1+R2||(R3+(R4||120k) )¿( R2

R2+R3+(R4||120 k))V R4

=I R4R4

=I S (RS R4

RS+R1+R2||( R3+(R4||120 k ) )⋅(R2

R2+R3+(R4||120 k) )¿11.272×10−3⋅

(120000+R4)⋅R4

R4+43. 760×103

With the voltmeter:

a) V R4 = 3.09 V b)

V R4 = 30.47 V

c) V R4 = 272.58 V d)

V R4 = 1724.99 V.

Problem 2.78An ammeter is used as shown in Figure P2.78. Theammeter model consists of an ideal ammeter in serieswith a resistance. The ammeter model is placed in thebranch as shown in the figure. Find the current through R5both with and without the ammeter in the circuit for thefollowing values, assuming that RS = 20 Ω, R1 =800 Ω, R2 = 600 ΩR3 = 1.2 kΩR4 = 150 Ω, andVS = 24 V.a. R5 = 1 kΩb. R5 = 100 Ωc. R5 = 10 Ωd. R5 = 1 Ω

Solution:Known quantities:Schematic of the circuit shown in Figure P2.78, value of the components.

Find:The current through R5 both with and without the ammeter, for the following values of the resistor R5 :a) R5=1 k

b) R5=100

c) R5=10

d) R5=1 .



Analysis:

First we should find an expression for the current through R5

in terms of R5

and the meter resistance, Rm

. By the

voltage divider rule we have:

V R3 =

R3‖(R4+R5+Rm)⋅V R1

R3‖(R4+R5+Rm )+R2

V R1=

R1‖( R3‖(R4+R5+Rm )+R2 )¿V s

R1‖( R3‖(R4+R5+Rm )+R2 )+Rs

AndI R5 =

V R3

R4+R5+Rm

Therefore,

I R5=

R1‖( R3‖(R4+R5+Rm )+R2 )¿V s

R1‖( R3‖(R4+R5+Rm )+R2 )+RS

¿R3‖(R4+R5+Rm )R3‖(R4+R5+Rm )+R2

¿1R4+R5+Rm

=553 . 846 (R5+Rm+550 )⋅24

573 . 846 (R5+Rm+558 .579 )⋅1200 (R5+Rm+150 )1800 (R5+Rm+550 )

⋅1(R5+Rm+150 )

¿15 . 4424R5+Rm+558. 579

Using the above equation will give us the following table:with meter in circuit

without meter in circuit

a 9.75 mA 9.91 mA b 22.59 mA 23.45 mA c 26.02 mA 27.16 mA d 26.42 mA 27.60 mA

Problem 2.79Shown in Figure P2.79 is an aluminumcantilevered beam loaded by the force F. Strain gaugesR1, R2, R3, and R4 are attached to the beam as shownin Figure P2.79 and connected into the circuit shown.The force causes a tension stress on the top of thebeam that causes the length (and therefore theresistance) of R1 and R4 to increase and a compressionstress on the bottom of the beam that causes the length(and therefore the resistance) of R2 and R3 to decrease.This causes a voltage of 50 mV at node B with respectto node A. Determine the force ifRo = 1 kΩ VS = 12 V L = 0.3 mw = 25 mm h = 100 mm Y = 69 GN/m2

Solution:Known quantities:Schematic of the circuit and geometry of the beam shown in Figure P2.79, characteristics of the material, reads on the bridge.



Find:The force applied on the beam.

Assumptions:Gage Factor for Strain gauge is 2

Analysis:R1 and

R2 are in series; R3 and

R4 are in series.

Voltage Division: V R2

=V S R2

R1+R2=

V S(R0−DR)R0+DR+R0−DR

=V S(R0−DR )

2 R0

Voltage Division: V R4

=V S R4

R3+R4=

V S(R0+ΔR )R0−ΔR+R0+ΔR

=V S(R0+ΔR )

2 R0

KVL: −V R2

−V BA+V R4=0

V BA=V R4−V R2

=V S(R0+ΔR )

2 R0−

V S (R0−ΔR )2 R0

=V S 2 ΔR

2 R0=V SGF ε=

V S 2∗6 LFwh2 Y

assuming GF=2 for aluminum.

F=V BA wh2Y

V s 12 L=

0 .050 V (0 .025 m)(0 .100 m )2 69×109 Nm2

12 V (12 ) 0.3 m=19 .97 kN .

Problem 2.80Shown in Figure P2.79 is a structural steelcantilevered beam loaded by a force F. Strain gaugesR1, R2, R3, and R4 are attached to the beam as shownand connected into the circuit shown. The force causesa tension stress on the top of the beam that causes thelength (and therefore the resistance) of R1 and R4 toincrease and a compression stress on the bottom of thebeam that causes the length (and therefore theresistance) of R2 and R3 to decrease. This generates avoltage between nodes B and A. Determine thisvoltage if F = 1.3 MN andRo = 1 kΩVS = 24 V L = 1.7 mw = 3 cm h = 7 cm Y = 200 GN/m2

Solution:Known quantities:Schematic of the circuit and geometry of the beam shown in Figure P2.79, characteristics of the material, reads on the bridge.

Find:The force applied on the beam.

Assumptions:Gage Factor for Strain gauge is 2



Analysis:R1 and

R2 are in series; R3 and

R4 are in series.

VD: V R2

=V S R2

R1+R2=

V S(R0−DR)R0+DR+R0−DR

=V S(R0−DR )

2R0

VD: V R4

=V S R4

R3+R4=

V S(R0−DR )R0−DR+R0+DR

=V S(R0+DR )

2R0

KVL: −V R2

−V BA+V R4=0

V BA=V R4−V R2

=V S(R0+ΔR )

2R0−

V S (R0−ΔR )2 R0

=V S 2 ΔR

2R0=V SGF ε=

V S ¿GF⋅6 LFwh2Y

Assuming GF=2 for aluminum.

V BA=24 V⋅2⋅6⋅1 .7 m⋅1. 3×106 N

(0. 03 m)(0 . 07 m)2 200×109 Nm2

=21. 6 mV


Documents

EIT 2 - Solution Manual & Test Bank · Web viewChapter 2: Fundamentals of Electrical Circuits – Instructor Notes Chapter 2 develops the foundations for the first part of the book