Solution manual fundamentals of electric circuits, 5th edition www solutionmanual info

s eqv i R If i1 is the current through the 24-Ω resistor and io is the current through the 50-Ω resistor, using current division gives... b We note that the two 12k-ohm resistors are in

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Chapter 1, Problem 1

How many coulombs are represented by these amounts of electrons:

10 482

.

10 24

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= +

1600 900

e 10 q(0)

t 40 sin 10e q(t)

-30t 30t

-Chapter 1, Problem 4

A current of 3.2 A flows through a conductor Calculate how much charge passes

through any cross-section of the conductor in 20 seconds

25 C 0

t

q = ∫ idt = ∫ tdt = =

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6 t 2 25A,

-2 t 0 A,

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1 10

idt

q = ∫ = × + × =

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15

1 5 2

1 5 10 1 10 idt

0

= + +

=

× +

A lightning bolt with 8 kA strikes an object for 15 μ s How much charge is

deposited on the object?

Chapter 1, Solution 10

q = it = 8x103x15x10-6 = 120 mC

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A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h How much charge can it release at that rate? If its terminals voltage is 1.2 V, how much energy can the battery deliver?

15s

t 10 A, 12 -

10s

t 6 A, 18

6s

t 0 A,

3t t

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(a) Find the power delivered to the element at t = 0.3 s

(b) Calculate the energy delivered to the element between 0 and 0.6s

0 8

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The voltage v across a device and the current I through it are

( ) 5 cos 2 V, ( ) 10 ( 1 0 t 5 ) A

e t

i t t

Calculate:

(a) the total charge in the device at t = 1 s

(b) the power consumed by the device at t = 1 s

10

2e t 10 dt e - 1 10 idt

q

0.5 -

1 0 0.5t - 1

0

0.5t -

i = − and the voltage across the device is v ( ) t = 5 di / dt V

(a) Find the charge delivered to the device between t = 0 and t = 2 s

(b) Calculate the power absorbed

(c) Determine the energy absorbed in 3 s

5

.

1

e 2

3 dt 3e idt

q

4-

20

2t2

02t-

1.4725 C

(b)

W e 90

) (

t 4

e 6 dt

0 4t

4

90 dt

e -90 pdt

w

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Figure 1.27 shows the current through and the voltage across a device (a) Sketch the power delivered to the device for t >0 (b) Find the total energy absorbed by the device for the period of 0< t < 4s

i (mA)

60

0 2 4 t(s) v(V)

0

Figure 1.27 For Prob 1.16

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(a)

30 mA, 0 < t <2 ( )

0 J

W = ∫ pdt =

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Figure 1.28 shows a circuit with five elements If

W, 30 W,

45 W,

60 W,

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Find I in the network of Fig 1.30

I

1A + +

+

4 A 9V 9V

– – – + – 6 V

Figure 1.30 For Prob 1.19

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hr 60

4

kW

1.2

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A flashlight battery has a rating of 0.8 ampere-hours (Ah) and a lifetime of 10 hours (a) How much current can it deliver?

(b) How much power can it give if its terminal voltage is 6 V?

(c) How much energy is stored in the battery in kWh?

(a) = . ⋅ = 80 mA

10h

h A 8 0

i

(b) p = vi = 6 × 0.08 = 0.48 W

(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh

A constant current of 3 A for 4 hours is required to charge an automotive battery If the

terminal voltage is 10 + t/2 V, where t is in hours,

(a) how much charge is transported as a result of the charging?

(b) how much energy is expended?

(c) how much does the charging cost? Assume electricity costs 9 cents/kWh

3 T 3 3dt idt

q

3600 4

kJ 475.2

.

)

( (

=

×

× +

25 0 3600 40

3 3600

25 0 10

3

dt 3600

t 5 0 10 3 vidt

0 2 0

T 0

t t

T

cents 1.188

475.2

Cost

Ws) (J kWs, 475.2

W

c)

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A 30-W incandescent lamp is connected to a 120-V source and is left burning

continuously in an otherwise dark staircase Determine:

(a) the current through the lamp,

(b) the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh

A 0.25

24 365 30 pt

W

b)

An electric stove with four burners and an oven is used in preparing a meal as follows

Burner 1: 20 minutes Burner 2: 40 minutes

Burner 3: 15 minutes Burner 4: 45 minutes

+ + +

=

3.3 cents 12

Cost

kWh 3.3 0.9 2.4

hr 60

30 kW 1.8 hr 60

45) 15 40 (20 kW 2 1

pt

w

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Reliant Energy (the electric company in Houston, Texas) charges customers as

follows:

Monthly charge $6 First 250 kWh @ $0.02/kWh All additional kWh @ $0.07/kWh

If a customer uses 1,218 kWh in one month, how much will Reliant Energy charge?

Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 968 kWh @ $0.07/kWh= $67.76

In a household, a 120-W PC is run for 4 hours/day, while a 60-W bulb runs for 8

hours/day If the utility company charges $0.12/kWh, calculate how much the household pays per year on the PC and the bulb

Total energy consumed = 365(120x4 + 60x8) W

Cost = $0.12x365x960/1000 = $42.05

A telephone wire has a current of 20 μ A flowing through it How long does it take for a

charge of 15 C to pass through the wire?

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, (

A 4

h 000 160 0.001A

160

i

(a)

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A bar of silicon is 4 cm long with a circular cross section If the resistance of the bar is

240 Ω at room temperature, what is the cross-sectional radius of the bar?

6.4 10 4 10

0.033953 240

(a) Calculate current i in Fig 2.68 when the switch is in position 1

(b) Find the current when the switch is in position 2

(a) i = 3/100 = 30 mA

(b) i = 3/150 = 20 mA

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Determine the number of branches and nodes in the circuit of Fig 2.71

Figure 2.71 For Prob 2.7

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Find i1, i ,2 and i3 in Fig 2.73.

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In the circuit in Fig 2.67 decrease in R3 leads to a decrease of:

(a) current through R3

3A

-2A 2

3

At node 1, 4 + 3 = i1 i1 = 7A

At node 3, 3 + i2 = -2 i2 = -5A Chapter 2, Problem 11

In the circuit of Fig 2.75, calculate V1 and V2

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ix

+ _

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From the circuit in Fig 2.80, find I, the power dissipated by the resistor, and the power

supplied by each source

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2 Vx

_ +

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Find Vo in the circuit in Fig 2.85 and the power dissipated by the controlled source

4 Ω + V0 -

The current through the controlled source is

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The current through the 1.2- resistor is 0.5iΩ x = 1A The voltage across the 12-

resistor is 1 x 4.8 = 4.8 V Hence the power is

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For the circuit in Fig 2.86, find Vo / Vs in terms of α, R1, R2, R3, and R4 If R1 = R2 = R3 =

R4, what value of α will produce | Vo / Vs | = 10?

(a) I0 =

2

1 R R

4321

s

R R

R R R R

V

+

⋅ +

R R Vs

V

+ +

−

(b) If R1 = R2 = R3= R4 = R,

10 4 2

R R 2 V

+

= ( 0 01 50 ) 20

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4 16

o

+

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15 6 Ω

We now apply voltage division,

+ 6 ( 40 ) 14

14

28 V

v2 = v3 = =

+ 6 ( 40 ) 14

6

12 V Hence, v1 = 28 V, v2 = 12 V, vs = 12 V

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30 x

= 40

50

40 x

Using current division principle,

A 12 ) 20 ( 20

12 i i , A 8 ) 20 ( 12 8

8 i

+

= +

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= ( 9 )

2

1 1

1

i 6 A, v = 3(1) = 3 V

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20 20 40

eq

212

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30 x 70

At node a, KCL must be satisfied

i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 A

Hence v0 = 8 V and I0 = 0.2A

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s eq

v i R

If i1 is the current through the 24-Ω resistor and io is the current through the 50-Ω

resistor, using current division gives

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o eq

i R

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Req = [(1x2.667)/3.667]k = 727.3 Ω

(b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in parallel with the 4k-ohm resistor producing,

2 A

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1 12

1 R

1o

+ +

= Ro = 4

) R 14 ( 60 30 ) R R 10 ( 60 30

R 74

) R 14 ( 60 30 50

+

+ +

(b) Rab = 2 + 4 ( 5 + 3 ) 8 + 5 10 4 = 2 + 4 4 + 5 2 857 = 2 + 2 + 1 8181 = 5.818 Ω

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20 x 40 10 20

1 20

1 60

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Combining the resistors in parallel

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Find the equivalent resistance Rab in the circuit of Fig 2.111

= 20

25

20 x

6 3 = = 2 Ω

9

3 6

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R R R R R R

3

1 3 3 2 2

50 x 20 50 x 30 20 x 30

Ra

, 155 20

R R

cba

ca

R1 = R2 = R3 = 4 Ω

+ +

10 30 60

30 x 60

R3

R1 = 18Ω, R2 = 6Ω, R3 = 3Ω

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3 R 4

RxR

) R 4 /(

3 ) R 4 /(

) RxR 3 ( R

R R 2

3 R 3

R 2

3 Rx 3 R 2

3 R 3 R 4

3 R 4

3 R 3

= +

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10 x

+

50 x 10

100

50 x 40

We convert the balanced Δ s to Ts as shown below:

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20 x 10 10 x 40 40 x 20 R

R R R R R

R

3

1 3 3 2 2

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15 x 12 12 x 10 10

Req = 19.688||(12 + 16.667) = 11.672Ω

By voltage division,

16 672 11

672 11 + = 42.18 V

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6 x 8 12 12 x 6

4 x 8 x 2 x

Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω

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Combining resistors in parallel,

, 368 7 38

280 28

43

7 x 36 7 36 Ω

=

30

3 x 27 3

+

567 26

7 2 x 18 867 5 7 2 18

7 2 x 18

868 5 18

7 2 x 868 5

Rcn

) 14 5964 0 ( ) 368 7 977 3 ( 829 1 4

= 5 829 + 11 346 14 5964 = 12.21 Ω

i = 20/(Req) = 1.64 A

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Three lightbulbs are connected in series to a 100-V battery as shown in Fig 2.123 Find

the current I through the bulbs

TOTAL POWER P = 30 + 40 + 50 + 120 W = VI

OR I = P/(V) = 120/(100) = 1.2 A

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R1 = 80Ω, cost = $0.60 (standard size)

R2 = 90Ω, cost = $0.90 (standard size)

R3 = 100 Ω, cost = $0.75 (nonstandard size)

The system should be designed for minimum cost such that I = 1.2 A ± 5 percent

i2 = 70/47.37 = 1.4777 or i = 1.2156 (which is within our range), cost = $1.65

Note that cases (b) and (c) satisfy the current range criteria and (b) is the cheaper

of the two, hence the correct choice is:

R1 and R3

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100 x 10 x R

I I

I

3

3 m

m m

In = I - Im = 4.998 A

p = I2R ( 4 998 )2( 0 04 ) 0 9992

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The potentiometer (adjustable resistor) Rx in Fig 2.126 is to be designed to adjust current

Ix from 1 A to 10 A Calculate the values of R and Rx to achieve this

When Rx = 0, ix = 10 A R = = 11Ω

10110

When Rx is maximum, ix = 1A + = = 110 Ω

1

110 R

full-Chapter 2, Solution 65

= Ω

50 R

A 20-kΩ/V voltmeter reads 10 V full scale,

(a) What series resistance is required to make the meter read 50 V full scale?

(b) What power will the series resistor dissipate when the meter reads full scale?

20 kΩ/V = sensitivity =

fsI 1

The intended resistance Rm = = 10 ( 20 k Ω / V ) = 200 k Ω

I

V

fs fs

V 50 R

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(c) Obtain the voltage vo in the circuit of Fig 2.127

(d) Determine the voltage v’o measured when a voltmeter with 6-kΩ internal

resistance is connected as shown in Fig 2.127

(e) The finite resistance of the meter introduces an error into the measurement

Calculate the percent error as

% 100 ' ×

−

o

o o

v

v v

5

i'

+ +

=

= Ω

28.57%

(f) k 36 k Ω = 3 6 k Ω By current division,

mA 042 1 ) mA 2 ( 5 6 3 1

5

i'

+ +

=

V 75 3 ) mA 042 1 )(

k 6 3 (

% 100 x v

v v

0

'

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(f) Find the current i in the circuit of Fig 2.128(a)

(g) An ammeter with an internal resistance of 1 Ω is inserted in the network to

measure i' as shown in Fig 2.128 (b) What is i"

?

(h) Calculate the percent error introduced by the meter as

% 100 ' ×

−

i

i i

(F) 40 = 24 60 Ω

+ 24 16

4

0.1 A

+ +

=

24 1 16

4

(H) % error = − x 100 % =

1 0

09756 0 1 0

2.44%

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m 2

R R R R

R R V

+ +

= where Rm = 100 kΩ without the voltmeter,

S S 2 1

2

R R R

R V

+ +

=

(a) When R2 = 1 kΩ, = k Ω

101

100 R

Rm 2

+

) 40 ( 30 101

R2 m

+ 30 ( 40 ) 091

9

091 9

9.30 V (with)

+ 30 ( 40 ) 10

10

10 V (without)

(c) When R2 = 100 kΩ, R2 Rm = k 50 Ω

= +

= ( 40 )

30 50

50

+ 30 ( 40 ) 100

100

30.77 V (without)

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Find Vo in the two-way power divider circuit in Fig 2.132

in o

in

Z V

Z

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An ammeter model consists of an ideal ammeter in series with a 20-Ω resistor It is

connected with a current source and an unknown resistor Rx as shown in Fig 2.133 The ammeter reading is noted When a potentiometer R is added and adjusted until the

ammeter reading drops to one half its previous reading, then R = 65 Ω What is the value

of Rx?

Ammeter model

By the current division principle, the current through the ammeter will be

one-half its previous value when

R = 20 + Rx

65 = 20 + Rx Rx = 45 Ω

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The circuit in Fig 2.134 is to control the speed of a motor such that the motor draws currents 5 A, 3 A, and 1 A when the switch is at high, medium, and low positions,

respectively The motor can be modeled as a load resistance of 20 mΩ Determine the

series dropping resistances R1, R2, and R3

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Find Zab in the four-way power divider circuit in Fig 2.135 Assume each element is 1Ω

1

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Repeat Prob 2.75 for the eight-way divider shown in Fig 2.136

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Suppose your circuit laboratory has the following standard commercially available

resistors in large quantities:

i.e., one 300Ω resistor in series with 1.8Ω resistor and

a parallel combination of two 20Ω resistors

i.e., A series combination of a 20Ω resistor, 300Ω resistor, 24kΩ resistor, and a parallel combination of two 56kΩ resistors.

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-V S

R

+

V 0 -(1- α)R

2

1 V R ) 1 ( R

R ) 1 (

− +

1 V

V

S0

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An electric pencil sharpener rated 240 mW, 6 V is connected to a 9-V battery as shown in

Fig 2.138 Calculate the value of the series-dropping resistor Rx needed to power the sharpener

Rs

9 V

– +

IRx = Vx = 9 - 6 = 3 V

Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 Ω

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A loudspeaker is connected to an amplifier as shown in Fig 2.139 If a 10-Ω

loudspeaker draws the maximum power of 12 W from the amplifier, determine the

maximum power a 4-Ω loudspeaker will draw

+ V

-R 2 CASE 2

2

R

R p

p = = = ( 12 ) =

4

10 p R

R

2 1

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Req = 1 + 2 (1)

1 2

2 S

0

R R 5

R 5 V

2

2 2

R 5

R 5 R 5

+

= or R2 = 3.333 kΩ From (1), 40 = R1 + 2 R1 = 38 kΩ

Thus R1 = 38 kΩ, R2 = 3.333 kΩ

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The pin diagram of a resistance array is shown in Fig 2.141 Find the equivalent

resistance between the following:

(a) 1 and 2 (b) 1 and 3 (c) 1 and 4

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Two delicate devices are rated as shown in Fig 2.142 Find the values of the resistors R1

and R2 needed to power the devices using a 24-V battery

The voltage across the fuse should be negligible when compared with 24

V (this can be checked later when we check to see if the fuse rating is exceeded in the final circuit) We can calculate the current through the devices

V 9

mW 45 V

Let R3 represent the resistance of the first device, we can solve for its value from

knowing the voltage across it and the current through it

R3 = 9/0.005 = 1,800 Ω

This is an interesting problem in that it essentially has two unknowns, R1 and R2 but only one condition that need to be met and that the voltage across R3 must equal 9 volts Since the circuit is powered by a battery we could choose the value of R2 which draws the least current, R2 = ∞ Thus we can calculate the value of R1 that give 9 volts across R3

9 = (24/(R1 + 1800))1800 or R1 = (24/9)1800 – 1800 = 3,000Ω

This value of R1 means that we only have a total of 25 mA flowing out of the battery through the fuse which means it will not open and produces a voltage drop across it of 0.05V This is indeed negligible when compared with the 24-volt source

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