ﬁ b ¼ verify the proposed solution b ¼ %& v dr ¼ dr ðqiw4/academic/mems0031/chapter 5...

Comparing Eq. 5.10-5 to Eq. 5.10-2 shows that the amplifier gain b must satisfy

b 0:4162( 10#3% &¼ 5

Hence, the amplifier gain is

b ¼ 12,013

Verify the Proposed SolutionSubstituting b ¼ 12,013 into Eq. 5.10-5 gives

vo ¼ 12,013ð Þ 0:4162( 10#3% &DR ¼ 4:9998 DR ð5:10-6Þ

which agrees with Eq. 5.10-2.

50 Ω

100 kΩ+–

vi

Rt

vt b vi

vo

+

– + –

Voltmeter

+–

i = 0

FIGURE 5.10-3 Solution to the design problem.

5.11 SUMMARYSource transformations, summarized in Table 5.11-1, are used totransform a circuit into an equivalent circuit. A voltage source voc

in series with a resistor Rt can be transformed into a currentsource isc¼ voc/Rt and a parallel resistor Rt. Conversely, a currentsource isc in parallel with a resistor Rt can be transformed into avoltage source voc ¼ Rtisc in series with a resistor Rt. The circuitsin Table 5.11-1 are equivalent in the sense that the voltage andcurrent of all circuit elements in circuit B are unchanged by thesource transformation.The superposition theorem permits us to determine thetotal response of a linear circuit to several independent sourcesby finding the response to each independent source separatelyand then adding the separate responses algebraically.Th!evenin and Norton equivalent circuits, summarized inTable 5.11-2, are used to transform a circuit into a smaller,yet equivalent, circuit. First the circuit is separated into twoparts, circuit A and circuit B, in Table 5.11-2. Circuit A canbe replaced by either its Th!evenin equivalent circuit or its

Norton equivalent circuit. The circuits in Table 5.11-2 areequivalent in the sense that the voltage and current of allcircuit elements in circuit B are unchanged by replacingcircuit A with either its Th!evenin equivalent circuit or itsNorton equivalent circuit.Procedures for calculating the parameters voc, isc, and Rt ofthe Th!evenin and Norton equivalent circuits are summarizedin Figures 5.4-3 and 5.4-4.The goal of many electronic and communications circuits isto deliver maximum power to a load resistor RL. Maximumpower is attained when RL is set equal to the Th!eveninresistance Rt of the circuit connected to RL. This results inmaximum power at the load when the series resistance Rt

cannot be reduced.The computer programs MATLAB and SPICE can be used toreduce the computational burden of calculating the parame-ters voc, isc, and Rt of the Th!evenin and Norton equivalentcircuits.

Table 5.11-1 Source Transformations

TH!EVENIN CIRCUIT NORTON CIRCUIT

+–

CircuitB

+– voc

a

b

CircuitB

Rtisc

a

b

CircuitB

CircuitB

Rt

Summary 203

PROBLEMS

Section 5.2 Source Transformations

P 5.2-1 The circuit shown in Figure P 5.2-1a has beendivided into two parts. The circuit shown in Figure P 5.2-1b wasobtained by simplifying the part to the right of the terminals usingsource transformations. The part of the circuit to the left of theterminals was not changed.

(a) Determine the values of Rt and vt in Figure P 5.2-1b.(b) Determine the values of the current i and the voltage v in

Figure P 5.2-1b. The circuit in Figure P 5.2-1b is equiv-alent to the circuit in Figure P 5.2-1a. Consequently,the current i and the voltage v in Figure P 5.2-1a havethe same values as do the current i and the voltage v inFigure P 5.2-1b.

(c) Determine the value of the current ia in Figure P 5.2-1a.

9 V 0.5 A

i

+–

+–4 Ω 2 Ω 2 V

4 Ω 2 Ω

(a)

ia

v

+

–

9 V

i

+–

+–

4 Ω

(b)

ia

vt

Rt

v

+

–

Figure P 5.2-1

P 5.2-2 Consider the circuit of Figure P 5.2-2. Find ia bysimplifying the circuit (using source transformations) to asingle-loop circuit so that you need to write only one KVLequation to find ia.

+–10 V

ia2 A6 Ω

3 Ω

8 Ω

4 Ω

Figure P 5.2-2

P 5.2-3 Find vo using source transformations if i ¼ 5=2 Ain the circuit shown in Figure P 5.2-3.

Hint: Reduce the circuit to a single mesh that contains thevoltage source labeled vo.

Answer: vo ¼ 28 V

6 Ω

20 Ω

3 Ω 10 Ω8 V

16 Ω

3 A

2 A 12 Ω 7 Ω

+ –

+ –i

v0

Figure P 5.2-3

P 5.2-4 Determine the value of the current ia in the circuitshown in Figure P 5.2-4.

4 kΩ10 V4 kΩ6 kΩ

3 kΩ12 V 6 V4 kΩ

+ –

+– +

–ia

Figure P 5.2-4

Table 5.11-2 Th!evenin and Norton Equivalent Circuits

ORIGINAL CIRCUIT TH!EVENIN CIRCUIT NORTON EQUIVALENT CIRCUIT

a

b

CircuitB

CircuitA

+– voc

Rta

b

CircuitB

isc Rt

a

b

CircuitB

Problem available in WileyPLUS at instructor’s discretion.

204 5. Circuit Theorems

P 5.2-5 Use source transformations to find the current ia inthe circuit shown in Figure P 5.2-5.

Answer: ia ¼ 1 A

6 Ω 6 V

4 A

1 A12 V 3 Ω

+–

+– ia

Figure P 5.2-5

P 5.2-6 Use source transformations to find the value of thevoltage va in Figure P 5.2-6.

Answer: va ¼ 7 V

+–

+ –

va

+

–

100 Ω

100 Ω 100 Ω 30 mA

8 V

10 V

Figure P 5.2-6

P 5.2-7 Theequivalent circuit inFigureP5.2-7 isobtained fromthe original circuit using source transformations and equivalentresistances. (The lower case letters a and b identify the nodes ofthe capacitor in both the original and equivalent circuits.)Determine the values of Ra, Va, Rb, and Ib in the equivalent circuit

2.2 A 36 V2.5 A

18 Ω9 Ω10 Ω

18 Ω

32 V

Ca b

original circuit

+–

+–

+–

equivalent circuit

R a

I bR bVa

Ca b

Figure P 5.2-7

P 5.2-8 The circuit shown in Figure P 5.2-8 contains anunspecified resistance R.

(a) Determine the value of the current i when R ¼ 4V.(b) Determine the value of the voltage v when R ¼ 8V.(c) Determine the value of R that will cause i ¼ 1 A.(d) Determine the value of R that will cause v ¼ 16 V.

18 Ω

24 Ω

24 Ω

12 Ω12 V 2 A

i R

v

+–

+ –

Figure P 5.2-8

P 5.2-9 Determine the value of the power supplied by thecurrent source in the circuit shown in Figure P 5.2-9.

15 Ω

24 Ω

12 Ω

25 Ω24 V

32 V

2 A+–

+–

Figure P 5.2-9

Section 5.3 Superposition

P 5.3-1 The inputs to the circuit shown in Figure P 5.3-1are the voltage source voltages v1 and v2. The output of thecircuit is the voltage vo. The output is related to the inputs by

vo ¼ av1 þ bv2

where a and b are constants. Determine the values of a and b.

20 Ω 5 Ω

20 Ω+–

+–vo v2v1

+

–

Figure P 5.3-1

P 5.3-2 A particular linear circuit has two inputs, v1 and v2,and one output, vo. Three measurements are made. The firstmeasurement shows that the output is vo ¼ 4 V when the inputsare v1 ¼ 2 V and v2 ¼ 0. The second measurement shows that theoutput is vo ¼ 10 V when the inputs are v1 ¼ 0 and v2 ¼#2.5 V.In the third measurement, the inputs are v1 ¼ 3 V and v2 ¼ 3 V.What is the value of the output in the third measurement?

P 5.3-3 The circuit shown in Figure P 5.3-3 has twoinputs, vs and is, and one output, io. The output is related to theinputs by the equation

io ¼ ais þ bvs

Problems 205

Given the following two facts:

The output is io ¼ 0:45Awhen the inputs are is ¼ 0:25 Aand vs ¼ 15 V

and

The output is io ¼ 0:30Awhen the inputs are is ¼ 0:50 Aand vs ¼ 0 V

Determine the values of the constants a and b and the values ofthe resistances are R1 and R2.

Answers:a¼0.6 A/A, b¼0.02 A/V, R1¼ 30V, and R2¼20V.

+– vs is

io R2

R1

Figure P 5.3-3

P 5.3-4 Use superposition to find v for the circuit ofFigure P 5.3-4.

20 Ω15 Ω6 A9 A10 Ω

v+ –

Figure P 5.3-4

P 5.3-5 Determine v(t), the voltage across the vertical resistorin the circuit in Figure P 5.3-5.

12 cos(5t ) V40 Ω

10 Ω

12 V

+

–

v (t )+–

+–

40 Ω

Figure P 5.3-5

P 5.3-6 Use superposition to find i for the circuit ofFigure P 5.3-6.

Answer: i ¼ 3.5 mA

4 kΩ 15 V

15 mA

12 kΩ 6 kΩ30 mA2 kΩi

+ –

Figure P 5.3-6

P 5.3-7 Determine v(t), the voltage across the 40 Ω resistor inthe circuit in Figure P 5.3-7.

1+ sin(5t ) A40 Ω10 Ω

12+15cos(8t ) V

+ –+

–

v (t )

Figure P 5.3-7

P 5.3-8 Use superposition to find the value of the currentix in Figure P 5.3-8.

Answer: ix ¼ 1=6 A

ix

3ix

6 Ω 3 Ω

2 A8 V+–

+–

Figure P 5.3-8

*P 5.3-9 The input to the circuit shown in Figure P 5.3-9 is thevoltage source voltage vs. The output is the voltage vo. Thecurrent source current ia is used to adjust the relationshipbetween the input and output. Design the circuit so that inputand output are related by the equation vo ¼ 2vs þ 9.

ix

A ix

vs ia12 Ω

6 Ω

12 Ω+

−vo

+ –

+–

Figure P 5.3-9

Hint: Determine the required values of A and ia.

P 5.3-10 The circuit shown in Figure P 5.3-10 has threeinputs: v1, v2, and i3. The output of the circuit is vo. The outputis related to the inputs by

vo ¼ av1 þ bv2 þ ci3

where a, b, and c are constants. Determine the values of a, b, and c.

i3v1

v2

10 Ω40 Ω

8 Ω

vo+–

+ –+

–

Figure P 5.3-10

P 5.3-11 Determine the voltage vo(t) for the circuit shown inFigure P 5.3-11.


4 ix

ix

10 Ω

12 cos 2t V

40 Ω 10 Ω2 V 5 Ω vo(t)

+

–

+–

+–

Figure P 5.3-11

P 5.3-12 Determine the value of the voltage vo in thecircuit shown in Figure P 5.3-12.

0.3 A

20 V96 Ω 32 Ω

30 Ω120 Ω +vo–

+ –

Figure P 5.3-12

P 5.3-13 The input to the circuit shown in Figure P 5.3-13is the current i1. The output is the voltage vo. The current i2 isused to adjust the relationship between the input and output.Determine values of the current i2 and the resistance R, thatcause the output to be related to the input by the equation

vo ¼ #0:5i1 þ 4

i1 i28 Ω

2 Ω

8 Ω4 ΩR

a bvo +–

Figure P 5.3-13

P 5.3-14 Determine values of the current ia and theresistance R for the circuit shown in Figure P 5.3-14.

7 mA

20 kΩ

8 V

4 kΩ

5 kΩ

ia

R

2 mA

+ –

Figure P 5.3-14

P 5.3-15 The circuit shown in Figure P 5.3-15 has threeinputs: v1, i2, and v3. The output of the circuit is the current io.The output of the circuit is related to the inputs by

i1 ¼ avo þ bv2 þ ci3

where a, b, and c are constants. Determine the values ofa, b, and c.

20 Ω

12 Ω

10 Ω

40 Ω

v1

v3

i2io

+–

+–

Figure P 5.3-15

P 5.3-16 Using the superposition principle, find the valueof the current measured by the ammeter in Figure P 5.3-16a.

Hint: Figure P 5.3-16b shows the circuit after the idealammeter has been replaced by the equivalent short circuitand a label has been added to indicate the current measuredby the ammeter, im.

Answer: im ¼ 253 þ 2

# 32 þ 3

5 ¼ 5 # 3 ¼ 2 A

5 A

25 V

2 Ω3 Ω

Ammeter+–

(a)

2 Ω

im5 A

25 V

3 Ω

+–

(b)

Figure P 5.3-16 (a) A circuit containing two independentsources. (b) The circuit after the ideal ammeter has been replacedby the equivalent short circuit and a label has been added toindicate the current measured by the ammeter, im.

Section 5.4 Th!evenin’s Theorem

P 5.4-1 Determine values of Rt and voc that cause thecircuit shown in Figure P 5.4-1b to be the Th!evenin equivalentcircuit of the circuit in Figure P 5.4-1a.

Hint: Use source transformations and equivalent resistancesto reduce the circuit in Figure P 5.4-1a until it is the circuit inFigure P 5.4-1b.

Problems 207

Answer: Rt ¼ 5V and voc ¼ 2 V

3 A voc

Rt

12 V

3 Ω 3 Ω

6 Ω +–

+–

a

b

a

b

(a) (b)

Figure P 5.4-1

P 5.4-2 The circuit shown in Figure P 5.4-2b is the Th!eveninequivalent circuit of the circuit shown in Figure P 5.4-2a.Find the value of the open-circuit voltage voc and Th!eveninresistance Rt.

Answer: voc ¼ #12 V and Rt ¼ 16V

+–

(a) (b)

10 Ω

40 Ω15 V

8 Ω

+– voc

Rt

Figure P 5.4-2

P 5.4-3 The circuit shown in Figure P 5.4-3b is the Th!eveninequivalent circuit of the circuit shown in Figure P 5.4-3a. Find thevalue of the open-circuit voltage voc and Th!evenin resistance Rt.

Answer: voc ¼ 2 V and Rt ¼ 4V

+–

(a) (b)

6 Ω

6 Ω

6 Ω

12 V

1 A +– voc

Rt

Figure P 5.4-3

P 5.4-4 Find the Th!evenin equivalent circuit for the circuitshown in Figure P 5.4-4.

+–

12 Ω

6 Ω

3 Ω

a

b

18 V

10 Ω

Figure P 5.4-4


Answer: voc ¼ #2 V and Rt ¼ #8=3V

+–

8 Ω

4 Ω

a

b

6 V va

0.75va

–

+

Figure P 5.4-5


+–

va2va 3 A

a

b

+

–

3 Ω

6 Ω

3 Ω

Figure P 5.4-6

P 5.4-7 The equivalent circuit in Figure P 5.4-7 is obtained byreplacing part of the original circuit by its Th!evenin equivalentcircuit. The values of the parameters of the Th!evenin equivalentcircuit are

voc ¼ 15 V and Rt ¼ 60V

Determine the following:

(a) The values of Vs and Ra. (Four resistors in the originalcircuit have equal resistance, Ra.)

(b) The value of Rb required to cause i ¼ 0.2 A.(c) The value of Rb required to cause v ¼ 12 V.

voc

+–

Ra

Ra

Vs Ra Rbv

+

–

i

+–

R t

Rb

original circuit

equivalent circuit

Ra

Figure P 5.4-7


P 5.4-8 Aresistor,R,wasconnectedtoacircuitboxasshownin Figure P 5.4-8. The voltage v was measured. The resistance waschanged, and the voltage was measured again. The results areshown in the table. Determine the Th!evenin equivalent of thecircuit within the box and predict the voltage v when R ¼ 8 kV.

2 kΩ4 kΩ

6 V2 V

Circuit v+

–R

R v

i

Figure P 5.4-8

P 5.4-9 A resistor, R, was connected to a circuit box asshown in Figure P 5.4-9. The current i was measured. Theresistance was changed, and the current was measured again.The results are shown in the table.

(a) Specify the value of R required to cause i ¼ 2 mA.(b) Given that R > 0, determine the maximum possible value

of the current i.

Hint: Use the data in the table to represent the circuit by aTh!evenin equivalent.

2 kΩ4 kΩ

4 mA3 mA

Circuit v

+

–R

R i

i

Figure P 5.4-9

P 5.4-10 For the circuit of Figure P 5.4-10, specify theresistance R that will cause current ib to be 2 mA. The current iahas units of amps.

Hint: Find the Th!evenin equivalent circuit of the circuitconnected to R.

+–12 V R

6 kΩ

1 kΩ

2000ia

+ –

ia ib

Figure P 5.4-10

P 5.4-11 For the circuit of Figure P 5.4-11, specify thevalue of the resistance RL that will cause current iL to be #2 A.

Answer: RL ¼ 12V

10 A 2 Ω

4ia

b

+ –

iiL RL

Figure P 5.4-11

P 5.4-12 The circuit shown in Figure P 5.4-12 contains anadjustable resistor. The resistance R can be set to any value inthe range 0 - R - 100 kV.

(a) Determine the maximum value of the current ia that can beobtained by adjusting R. Determine the correspondingvalue of R.

(b) Determine the maximum value of the voltage va that can beobtained by adjusting R. Determine the correspondingvalue of R.

(c) Determine the maximum value of the power supplied to theadjustable resistor that can be obtained by adjusting R.Determine the corresponding value of R.

12 V

R

2 mA

24 kΩ

18 kΩ

12 kΩia

+ −va

+–

Figure P 5.4-12

P 5.4-13 The circuit shown in Figure P 5.4-13 consists oftwo parts, the source (to the left of the terminals) and the load.The load consists of a single adjustable resistor having resist-ance 0 - RL - 20V. The resistance R is fixed but unspecified.When RL ¼ 4V, the load current is measured to be io ¼ 0.375 A.When RL ¼ 8V, the value of the load current is io ¼ 0.300 A.

(a) Determine the value of the load current when RL ¼ 10V.(b) Determine the value of R.

24 V R

48 Ω

source load

RL

io+–

Figure P 5.4-13

P 5.4-14 The circuit shown in Figure P 5.4-14 contains anunspecified resistance, R. Determine the value of R in each ofthe following two ways.

(a) Write and solve mesh equations.(b) Replace the part of the circuit connected to the resistor R by

a Th!evenin equivalent circuit. Analyze the resulting circuit.

R

0.25 A

40 V

20 Ω 40 Ω

10 Ω20 Ω

+–

Figure P 5.4-14

Problems 209

P 5.4-15 Consider the circuit shown in Figure P 5.4-15.Replace the part of the circuit to the left of terminals a–b byits Th!evenin equivalent circuit. Determine the value of thecurrent io.

96 Ω 32 Ω

20 V32 Ω

30 Ω120 Ω

a

b

io

vo

+

–

+ –

Figure P 5.4-15

P 5.4-16 An ideal voltmeter is modeled as an open circuit. Amore realistic model of a voltmeter is a large resistance. FigureP 5.4-16a shows a circuit with a voltmeter that measures thevoltage vm. In Figure P 5.4-16b, the voltmeter is replaced by themodel of an ideal voltmeter, an open circuit. The voltmetermeasures vmi, the ideal value of vm.

Voltmeter

200 Ω 10 Ω

50 Ω25 V Rm

200 Ω 10 Ω

50 Ω25 V

200 Ω 10 Ω

50 Ω25 V

(a)

(b)

(c)

+–

+–

+– vm

+

–

vm

+

–

vmi

+

–

Figure P 5.4-16

As Rm!1, the voltmeter becomes an ideal voltmeterand vm ! vmi. When Rm < 1, the voltmeter is not ideal andvm > vmi. The difference between vm and vmi is a measurementerror caused by the fact that the voltmeter is not ideal.

(a) Determine the value of vmi.(b) Express the measurement error that occurs when Rm ¼

1000V as a percentage of vmi.(c) Determine the minimum value of Rm required to ensure

that the measurement error is smaller than 2 percent of vmi.

P 5.4-17 Given that 0 - R -1 in the circuit shown in FigureP 5.4-17, consider these two observations:

Observation 1: When R ¼ 2V then vR ¼ 4 V and iR ¼ 2 A.Observation 1: When R ¼ 6V then vR ¼ 6 V and iR ¼ 1 A.


(a) The maximum value of iR and the value of R that causes iRto be maximal.

(b) The maximum value of vR and the value of R that causes vR

to be maximal.(c) The maximum value of pR ¼ iRvR and the value of R that

causes pR to be maximal.

vs+–

ia iR

Bia vR R

+

–

24 Ω

6 Ω

Figure P 5.4-17

P 5.4-18 Consider the circuit shown in Figure P 5.4-18.Determine

(a) The value of vR that occurs when R ¼ 9V.(b) The value of R that causes vR ¼ 5.4 V.(c) The value of R that causes iR ¼ 300 mA.

+–

iR

vR R

+

–

20 Ω 6 Ω

30 Ω300 mA9 V

Figure P 5.4-18

P 5.4-19 The circuit shown in Figure P 5.4-19a can be reducedto the circuit shown in Figure P 5.4-19b using source transfor-mations and equivalent resistances. Determine the values of thesource voltage voc and the resistance R.

(a)

(b)

42 Ω

R

C18 V+– 84 Ω

voc+–

46 ΩC

Figure P 5.4-19


P 5.4-20 The equivalent circuit in Figure P 5.4-20 is obtainedby replacing part of the original circuit by its Th!eveninequivalent circuit. The values of the parameters of the Th!eveninequivalent circuit are

voc ¼ 15 V and Rt ¼ 60 V


(a) The values of Vs and Ra. (Three resistors in the originalcircuit have equal resistance, Ra.)

(b) The value of Rb required to cause i ¼ 0.2 A.(c) The value of Rb required to cause v ¼ 5 V.

voc

+–

Ra Ra

Vs Ra Rbv

+

–

i

+–

R t

Rb

original circuit

equivalent circuit

Figure P 5.4-20

Section 5.5 Norton’s Equivalent Circuit

P 5.5-1 The part of the circuit shown in Figure P 5.5-1ato the left of the terminals can be reduced to its Nortonequivalent circuit using source transformations and equi-valent resistance. The resulting Norton equivalent circuit,shown in Figure P 5.5-1b, will be characterized by theparameters:

isc ¼ 0:5A and Rt ¼ 20V

(a) Determine the values of vS and R1.(b) Given that 0 - R2 -1, determine the maximum values of

the voltage v and of the power p ¼ vi.

Answers: vs ¼ 37:5 V; R1 ¼ 25V; max v ¼ 10 V and maxp ¼ 1.25 W

+ –

i

vvs

R1 R2

+

–

50 Ω

50 Ω0.25 A

i

vRt R2

+

–

isc

(a)

(b)

Figure P 5.5-1

P 5.5-2 Two black boxes are shown in Figure P 5.5-2. BoxA contains the Th!evenin equivalent of some linear circuit, andbox B contains the Norton equivalent of the same circuit. Withaccess to just the outsides of the boxes and their terminals, howcan you determine which is which, using only one shorting wire?

1 Ω1 A

a

b

1 V

1 Ω a

b

Box A Box B

+–

Figure P 5.5-2 Black boxes problem.

P 5.5-3 The circuit shown in Figure P 5.5-3a can be reducedto the circuit shown in Figure P 5.5-3b using source transfor-mations and equivalent resistances. Determine the values of thesource current isc and the resistance R.

80 Ω

i sc

L

L

R160 Ω

48 Ω

4.8 A

(a)

(b)

Figure P 5.5-3

Problems 211

P 5.5-4 Find the Norton equivalent circuit for the circuitshown in Figure P 5.5-4.

4 A 5 A

5 Ω3 Ω

8 Ω

a

b

Figure P 5.5-4

P 5.5-5 The circuit shown in Figure P 5.5-5b is the Nortonequivalent circuit of the circuit shown in Figure P 5.5-5a.Find the value of the short-circuit current isc and Th!eveninresistance Rt.

Answer: isc ¼ 1.13 A and Rt ¼ 7.57V

+–

(a) (b)

3 Ω

6 Ω Rt10 V

5 Ω

isc2ia

+–

ia

Figure P 5.5-5

P 5.5-6 The circuit shown in Figure P 5.5-6b is the Nortonequivalentcircuitof thecircuit showninFigureP5.5-6a.Find thevalue of the short-circuit current isc and Th!evenin resistance Rt.

Answer: isc ¼ #24 A and Rt ¼ #3V

–+

(a) (b)

3 Ω 6 Ω

Rt1.33va24 V iscva

+

–

Figure P 5.5-6

P 5.5-7 Determine the value of the resistance R in the circuitshown in Figure P 5.5-7 by each of the following methods:

(a) Replace the part of the circuit to the left of terminals a–b byits Norton equivalent circuit. Use current division todetermine the value of R.

(b) Analyze the circuit shown Figure P 5.5-7 using mesh equa-tions. Solve the mesh equations to determine the value of R.

5 kΩ 10 kΩ

0.5 mA4 ibib

b

R25 V

a

+–

Figure P 5.5-7


6 Ω

4 Ω 1 Ω

3 Ω2.5 A

2 ixa

b

ix

+ –

Figure P 5.5-8


a

b

3 Ω5 Ω

4 Ω

2.5 v1

v1

+

–1 3 A

Figure P 5.5-9

P 5.5-10 An ideal ammeter is modeled as a short circuit. Amore realistic model of an ammeter is a small resistance. FigureP 5.5-10a shows a circuit with an ammeter that measures thecurrent im. In Figure P 5.5-10b, the ammeter is replaced by themodel of an ideal ammeter, a short circuit. The ammetermeasures imi, the ideal value of im.

Ammeter

im

imi

3 mA

4 kΩ

4 kΩ 2 kΩ

im

3 mA 4 kΩ

4 kΩ

4 kΩ

2 kΩ

3 mA 4 kΩ 2 kΩ

Rm

(a)

(b)

(c)

Figure P 5.5-10


As Rm! 0, the ammeter becomes an ideal ammeter andim! imi. When Rm > 0, the ammeter is not ideal and im < imi.The difference between im and imi is a measurement errorcaused by the fact that the ammeter is not ideal.

(a) Determine the value of imi.(b) Express the measurement error that occurs when Rm ¼

20V as a percentage of imi.(c) Determine the maximum value of Rm required to ensure

that the measurement error is smaller than 2 percent of imi.

P 5.5-11 Determine values of Rt and isc that cause the circuitshown in Figure P 5.5-11b to be the Norton equivalent circuitof the circuit in Figure P 5.5-11a.

Answer: Rt ¼ 3V and isc ¼ #2 A

isc Rt12 V

6 Ω 3 Ω

+–

a

b

a

b

+–

ia

2ia

(a) (b)

Figure P 5.5-11

P 5.5-12 Use Norton’s theorem to formulate a generalexpression for the current i in terms of the variable resistance Rshown in Figure P 5.5-12.

Answer: i ¼ 20=(8 þ R) A

30 V +– i

12 Ω 8 Ω

16 ΩR

a

b

Figure P 5.5-12

Section 5.6 Maximum Power Transfer

P 5.6-1 The circuit shown in Figure P 5.6-1 consists of twoparts separated by a pair of terminals. Consider the part of thecircuit to the left of the terminals. The open circuit voltage isvoc ¼ 8 V, and short-circuit current is isc ¼ 2 A. Determine thevalues of (a) the voltage source voltage vs and the resistance R2,and (b) the resistance R that maximizes the power delivered tothe resistor to the right of the terminals, and the correspondingmaximum power.

vs

8 Ω R2

R+–

+–

ia

4 iai

v

+

–

Figure P 5.6-1

P 5.6-2 The circuit model for a photovoltaic cell is givenin Figure P 5.6-2 (Edelson, 1992). The current is is proportionalto the solar insolation (kW/m2).

(a) Find the load resistance, RL, for maximum power transfer.(b) Find the maximum power transferred when is ¼ 1 A.

is RL

1 Ω

100 Ω

Figure P 5.6-2 Circuit model of a photovoltaic cell.

P 5.6-3 For the circuit in Figure P 5.6-3, (a) find R suchthat maximum power is dissipated in R, and (b) calculate thevalue of maximum power.

Answer: R ¼ 60V and Pmax ¼ 54 mW

R

100 Ω150 Ω

6 V 2 V+–

+–

Figure P 5.6-3

P 5.6-4 For the circuit in Figure P 5.6-4, prove that for Rs

variable and RL fixed, the power dissipated in RL is maximumwhen Rs ¼ 0.

vs RL

Rs

+–

sourcenetwork

load

Figure P 5.6-4

P 5.6-5 Determine the maximum power that can beabsorbed by a resistor, R, connected to terminals a–b of thecircuit shown in Figure P 5.6-5. Specify the required value of R.

20 A

20 Ω

8 Ω a

10 Ω

120 Ω 50 Ω

b

Figure P 5.6-5 Bridge circuit.

Problems 213

P 5.6-6 Figure P 5.6-6 shows a source connected to a loadthrough an amplifier. The load can safely receive up to 15 W ofpower. Consider three cases:

(a) A ¼ 20 V/V and Ro ¼ 10V. Determine the value of RL thatmaximizes the power delivered to the load and the corre-sponding maximum load power.

(b) A ¼ 20 V/V and RL ¼ 8V. Determine the value of Ro thatmaximizes the power delivered to the load and the corre-sponding maximum load power.

(c) Ro ¼ 10V and RL ¼ 8V. Determine the value of A thatmaximizes the power delivered to the load and the corre-sponding maximum load power.

500 mV 100 kΩ va

source amplifier load

RL

Ro

Ava+–

+–

+

–

Figure P 5.6-6

P 5.6-7 The circuit in Figure P 5.6-7 contains a variableresistance, R, implemented using a potentiometer. The resistanceof the variable resistor varies over the range 0 - R - 1000V.The variable resistor can safely receive 1=4 W power. Determinethe maximum power received by the variable resistor. Is the circuitsafe?

180 Ω

150 Ω 470 Ω

120 Ω

10 V 20 V

R

+–

+–

Figure P 5.6-7

P 5.6-8 For the circuit of Figure P 5.6-8, find the powerdelivered to the load when RL is fixed and Rt may be variedbetween 1V and 5V. Select Rt so that maximum power isdelivered to RL.

Answer: 13.9 W

10 V +– RL = 5 Ω

Rt

Figure P 5.6-8

P 5.6-9 A resistive circuit was connected to a variable resistor,and the power delivered to the resistor was measured as shown inFigure P 5.6-9. Determine the Th!evenin equivalent circuit.

Answer: Rt ¼ 20V and voc ¼ 20 V

10 20 30

5

Power(W)

400 R (ohms)

Figure P 5.6-9

P 5.6-10 The part circuit shown in Figure P 5.6-10a to left ofthe terminals can be reduced to its Norton equivalent circuitusing source transformations and equivalent resistance. Theresulting Norton equivalent circuit, shown in Figure P 5.6-10b,will be characterized by the parameters:

isc ¼ 1:5A and Rt ¼ 80V

(a) Determine the values of is and R1.(b) Given that 0 - R2 -1, determine the maximum value of

p = vi, the power delivered to R2.

50 Ω +

–

i s vR1

i

R225 V +

–

50 Ω(a)

+

–

v

i

R2Rti sc

(b)

Figure P 5.6-10

P 5.6-11 Given that 0 - R -1 in the circuit shown in FigureP 5.6-11, determine (a) maximum value of ia, (b) the maximumvalue of va, and (c) the maximum value of pa = ia va.

+– v a

+

−8 Ω R

4 Ω

12 V

i a

Figure P 5.6-11

P 5.6-12 Given that 0 - R -1 in the circuit shown inFigure P 5.6-12, determine value of R that maximizes thepower pa = ia va and the corresponding maximum value of pa.


+– v a

+

−

20 Ω R

8 Ω

6 V

i a 2 Ω

Figure P 5.6-12

Section 5.8 Using PSpice to Determine the Th!eveninEquivalent Circuit

P 5.8-1 The circuit shown in Figure P 5.8-1 is separated into twoparts by a pair of terminals. Call the part of the circuit to the left ofthe terminals circuit A and the part of the circuit to the right of theterminal circuit B. Use PSpice to do the following:

(a) Determine the node voltages for the entire circuit.(b) Determine the Th!evenin equivalent circuit of circuit A.(c) Replace circuit A by its Th!evenin equivalent and determine

the node voltages of the modified circuit.(d) Compare the node voltages of circuit B before and after

replacing circuit A by its Th!evenin equivalent.

+–

60 Ω

10 Ω 10 Ω

15 Ω10 Ω20 Ω

10 Ω

12 Ω

40 Ω

250 mA15 V

Figure P 5.8-1

Section 5.9 How CanWe Check . . . ?

P 5.9-1 For the circuit of Figure P 5.9-1, the current i has beenmeasured for three different values of R and is listed in thetable. Are the data consistent?

4 kΩ

4 kΩ1 kΩ

R

10 V

ix

i

5000ix+–+ –

R(Ω) i(mA)

5000500

0

16.543.897.2

Figure P 5.9-1

P 5.9-2 Your lab partner built the circuit shown inFigure P 5.9-2 and measured the current i and voltage vcorresponding to several values of the resistance R. The resultsare shown in the table in Figure P 5.9-2. Your lab partner says

that RL ¼ 8000V is required to cause i ¼ 1 mA. Do you agree?Justify your answer.

6 kΩ

18 kΩ2 mA 12 kΩ

24 kΩ

+–

i

v

R

R vi

+ –

12 V

open10 kΩshort

0 mA0.857 mA

3 mA

12 V8.57 V

0 V

Figure P 5.9-2

P 5.9-3 In preparation for lab, your lab partner determined theTh!evenin equivalent of the circuit connected to RL in FigureP 5.9-3. She says that the Th!evenin resistance is Rt ¼ 6

11 R andthe open-circuit voltage is voc ¼ 60

11 V. In lab, you built the circuitusing R ¼ 110V and RL ¼ 40V and measured that i ¼ 54.5 mA.Is this measurement consistent with the prelab calculations? Justifyyour answers.

+–

+–

+–

Load

3R

2R

R

RL

30 V

20 V

10 V

i

Figure P 5.9-3

P 5.9-4 Your lab partner claims that the current i in FigureP 5.9-4 will be no greater than 12.0 mA, regardless of the valueof the resistance R. Do you agree?

+–12 V 3 kΩ 6 kΩ

500 Ω

1500 Ω

i

R

Figure P 5.9-4

Problems 215

P 5.9-5 Figure P 5.9-5 shows a circuit and some correspond-ing data. Two resistances, R1 and R, and the current sourcecurrent are unspecified. The tabulated data provide values ofthe current i and voltage v corresponding to several values ofthe resistance R.

(a) Consider replacing the part of the circuit connected to theresistor R by a Th!evenin equivalent circuit. Use the data inrows 2 and 3 of the table to find the values of Rt and voc, theTh!evenin resistance, and the open-circuit voltage.

18 Ω

12 Ω

+–

i

R

R1

is12 V 24 Ω

v

+

–

(a)

R, Ω v, Vi, A

010204080

31.3330.857

0.5?

013.3317.14

?21.82

(b)

Figure P 5.9-5

(b)Use the results of part (a) to verify that the tabulated dataare consistent.

(c) Fill in the blanks in the table.(d) Determine the values of R1 and is.

PSpice ProblemsSP 5-1 The circuit in Figure SP 5-1 has three inputs: v1, v2,and i3. The circuit has one output, vo. The equation

vo ¼ a v1 þ b v2 þ c i3

expresses the output as a function of the inputs. Thecoefficients a, b, and c are real constants.

(a) Use PSpice and the principle of superposition to determinethe values of a, b, and c.

(b) Suppose v1 ¼ 10 V and v2 ¼ 8 V, and we want the output tobe vo ¼ 7 V. What is the required value of i3?

Hint: The output is given by vo¼ a when v1¼1 V, v2¼0 V, andi3 ¼ 0 A.

+–

+ –

vo

v2

i3v1

+

–

100 Ω

100 Ω 100 Ω

Figure SP 5-1

Answer: (a) vo ¼ 0.3333v1 þ 0.3333v2 þ 33.33i3, (b) i3 ¼ 30mA

SP 5-2 The pair of terminals a–b partitions the circuit inFigure SP 5-2 into two parts. Denote the node voltages atnodes 1 and 2 as v1 and v2. Use PSpice to demonstrate thatperforming a source transformation on the part of the circuit tothe left of the terminal does not change anything to the right ofthe terminals. In particular, show that the current io and the nodevoltages v1 and v2 have the same values after the sourcetransformation as before the source transformation.

+–

+ –100 Ω 8 V

a 1 2

b

10 V 100 Ω 100 Ω 30 mAio

Figure SP 5-2


SP 5-3 Use PSpice to find the Th!evenin equivalent circuitfor the circuit shown in Figure SP 5-3.

Answer: voc ¼ #2 V and Rt ¼ #8=3V

+–

8 Ω

4 Ω

a

b

6 V va

0.75va

–

+

Figure SP 5-3

SP 5-4 The circuit shown in Figure SP 5-4b is the Nortonequivalent circuit of the circuit shown in Figure SP 5-4a.Find the value of the short-circuit current isc and Th!eveninresistance Rt.

Answer: isc ¼ 1.13 V and Rt ¼ 7.57V

+–

(a) (b)

3 Ω

6 Ω Rt10 V

5 Ω

isc2ia

+–

ia

Figure SP 5-4

Design ProblemsDP 5-1 The circuit shown in Figure DP 5-1a has four un-specified circuit parameters: vs, R1, R2, and R3. To design thiscircuit, we must specify the values of these four parameters. Thegraph shown in Figure DP 5-1b describes a relationship betweenthe current i and the voltage v.

R2

R3

vs

R1 i

+

–

v+–

v, V

i, mA

–2

–4

–6

–8

2

4

6

2 4 6 8–2–4–6

(a)

(b)

Figure DP 5-1

Specify values of vs, R1, R2, and R3 that cause the current iand the voltage v in Figure DP 5-1a to satisfy the relationshipdescribed by the graph in Figure DP 5-1b.

First Hint: The equation representing the straight line in FigureDP 5-1b is

v ¼ #Rti þ vocThat is, the slope of the line is equal to #1 times the Th!eveninresistance, and the v-intercept is equal to the open-circuit voltage.

Second Hint: There is more than one correct answer to thisproblem. Try setting R1 ¼ R2.

DP 5-2 The circuit shown in Figure DP 5-2a has four un-specified circuit parameters: is, R1, R2, and R3. To design thiscircuit, we must specify the values of these four parameters. Thegraph shown in Figure DP 5-2b describes a relationship betweenthe current i and the voltage v.

Specify values of is, R1, R2, and R3 that cause the current iand the voltage v in Figure DP 5-2a to satisfy the relationshipdescribed by the graph in Figure DP 5-2b.

First Hint: Calculate the open-circuit voltage voc and theTh!evenin resistance Rt, of the part of the circuit to the left ofthe terminals in Figure DP 5-2a.

Second Hint: The equation representing the straight line inFigure DP 5-2b is

v ¼ #Rti þ vocThat is, the slope of the line is equal to #1 times the Th!eveninresistance, and the v-intercept is equal to the open-circuitvoltage.

Third Hint: There is more than one correct answer to thisproblem. Try setting both R3 and R1 þ R2 equal to twice theslope of the graph in Figure DP 5-2b.

R3R1

R2

is

i

+

–

v

(a)

Design Problems 217

v, V

i, mA

–2

–4

–6

–8

2

4

6

2 4 6 8–2–4–6

(b)

Figure DP 5-2

DP 5-3 The circuit shown in Figure DP 5-3a has four un-specified circuit parameters: vs, R1, R2, and R3. To design thiscircuit, we must specify the values of these four parameters. Thegraph shown in Figure DP 5-3b describes a relationship betweenthe current i and the voltage v.

R2

R3

vs

R1 i

+

–

v+–

v, V

i, mA

–2

–4

–6

–8

2

4

6

2 4 6 8–2–4–6

(a)

(b)

Figure DP 5-3

Is it possible to specify values of vs, R1, R2, and R3 thatcause the current i and the voltage v in Figure DP 5-1a to satisfythe relationship described by the graph in Figure DP 5-3b?Justify your answer.

DP 5-4 The circuit shown in Figure DP 5-4a has four un-specified circuit parameters: vs, R1, R2, and d, where d is the gainof the CCCS. To design this circuit, we must specify the valuesof these four parameters. The graph shown in Figure DP 5-4bdescribes a relationship between the current i and the voltage v.

Specify values of vs, R1, R2, and d that cause the current iand the voltage v in Figure DP 5-4a to satisfy the relationshipdescribed by the graph in Figure DP 5-4b.

First Hint: The equation representing the straight line in FigureDP 5-4b is

v ¼ #Rti þ voc

That is, the slope of the line is equal to #1 times the Th!eveninresistance and the v-intercept is equal to the open-circuitvoltage.

Second Hint: There is more than one correct answer to thisproblem. Try setting R1 ¼ R2.

dia R2

R1

(a)

vs+–

ia i

+

–

v

(b)

v, V

i, mA

–2

–4

–6

–8

2

4

6

2 4 6 8–2–4–6

Figure DP 5-4


ﬁ b ¼ verify the proposed solution b ¼ %& v dr ¼ dr ðqiw4/academic/mems0031/chapter 5...

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