Chapter9 Assignments

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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku

Copyright ©2004 The McGraw-Hill Companies Inc.

1

Chapter 9, Problem 32 (26).

Two elements are connected in series as shown in Fig. 9.40.

If i = 12 cos (2t - 30°) A, find the element values.

Figure 9.40

Chapter 9, Solution 32 (26).

V = 180∠10°, I = 12∠-30°, w = 2

Ω+=°∠=°∠°∠

== 642.9 j49.11041530-12

01180

I

VZ

One element is a resistor with R = 11.49Ω.The other element is an inductor with wL = 9.642 or L = 4.821 H.





2


If i s = 5 cos (10t + 40°) A in the circuit in Fig. 9.53, find io.

Figure 9.53


°∠= → °+= 405)40t10cos(5i ss I

j-)1.0)(10( j

1

C j

1F1.0 ==

ω →

2 j)2.0)(10( jL jH2.0 ==ω →

Let 6.1 j8.02 j4

8 j2 j||41 +=+

==Z , j32 −=Z

)405(6.0 j8.3

j1.60.8s

21

1

o °∠++

=+

= IZZ

ZI

°∠=°∠

°∠°∠= 46.94325.2

97.8847.3

)405)(43.63789.1(oI

Thus, =)t(io 2.325 cos(10t + 94.46°) A





3


Find Z in the network of Fig. 9.62, given that Vo = 4∠0° V

Figure 9.62


-j0.5

8 j

4

8 j

o

1 ===V

I

j8 j4-

)8 j((-j0.5)

j4-

)8 j(1

2 +=+

=+

=ZZZI

I

5.0 j8

j8

-j0.521 +=++=+=ZZ

III

)8 j(12 j20- 1 ++= ZII

)8 j(2 j-

2 j

812 j20- ++

+= ZZ

−=2

1 j

2

3 j26-4- Z °∠=

°∠°∠

=−

= 279.6864.1618.43-5811.1

25.26131.26

2

1 j

2

3

j26-4-Z

Z = 2.798 – j16.403 Ω

-j4 Ω

I I1

8 Ω

12 Ω Z

+

−

I2

-j20 V

+

Vo

−





4


For the circuit in Fig. 9.73, calculate ZT and Vab.

) j12(145

170

5 j60

)10 j40)(5 j20()10 j40(||)5 j20(T −=

++−

=+−=Z

=TZ 14.069 – j1.172 Ω = 14.118∠-4.76°

°∠=°∠

°∠== 76.9425.4

76.4-118.14

9060

TZ

VI

III j12

2 j8

5 j60

10 j401 +

+=

++

=

III j12

j4

5 j60

5 j202 +

−

=+

−

=

21ab 10 j20- IIV += IIV j12

40 j10

j12

)40 j(160-ab +

++

++

=

IIV145

j)(150)-12(

j12

150-ab

+=

+= )76.9725.4)(24.175457.12(ab °∠°∠=V

=abV 52.94∠273° V

+Vab

10 Ω20 ΩI2I1

I





5


(a) Calculate the phase shift of the circuit in Fig. 9.82.

(b) State whether the phase shift is leading or lagging (output with respect to input).

(c) Determine the magnitude of the output when the input is 120 V.

Figure 9.82


Consider the circuit as shown.

21 j390 j30

)60 j30)(30 j()60 j30(||30 j1 +=

++

=+=Z

°∠=+=+

+=+= 21.80028.9896.8 j535.1

31 j43

)21 j43)(10 j()40(||10 j 12 ZZ

Let °∠= 01iV .

896.8 j535.21

)01)(21.80028.9(

20 i

2

2

2 +°∠°∠

=+

= VZ

ZV

°∠= 77.573875.02V

20 Ω 30 Ω

10 Ω 60 Ω+

Vi

−

+

Vo

−

40 Ω

30 Ω

V2

Z2

V1

Z1





6

°∠°∠°∠

=+

+=

+=

03.2685.47

)77.573875.0)(87.81213.21(

21 j43

21 j3

40 22

1

1

1 VVZ

ZV

°∠= 61.1131718.01V

111o ) j2(5

2

2 j1

2 j

60 j30

60 jVVVV +=

+=

+=

)6.1131718.0)(56.268944.0(o °∠°∠=V

°∠= 2.1401536.0oV

Therefore, the phase shift is 140.2° The phase shift is leading.

If V120i =V , then

°∠=°∠= 2.14043.18)2.1401536.0)(120(oV V

and the magnitude is 18.43 V.





7


The ac bridge shown in Fig. 9.84 is known as a Maxwell bridge and is used for

accurate measurement of inductance and resistance of a coil in terms of a standard

capacitance C s. Show that when the bridge is balanced,

L x = R2 R3 C s and 23

1

x

R R R

R=

Find L x and R x for R1 = 40 k Ω, R2 = 1.6 k Ω, R3 = 4 k Ω, and C s = 0.45 µ F.

Figure 9.84


Lets

11 C j

1||R

ω=Z , 22 R =Z , 33 R =Z , and xxx L jR ω+=Z .

1CR j

R

C j

1R

C j

R

s1

1

s

1

s

1

1 +ω=

ω+

ω=Z

Since 2

1

3

x ZZ

ZZ = ,

)CR j1(R

R R

R

1CR jR R L jR s1

1

32

1

s1

32xx ω+=+ω

=ω+

Equating the real and imaginary components,





8

1

32

x R

R R R =

)CR (R

R R L

s11

32

xω=ω implies that

s32x CR R L =

Given that Ω= k 40R 1 , Ω= k 6.1R 2 , Ω= k 4R 3 , and F45.0Cs µ=

=Ω=Ω== k 16.0k 40

)4)(6.1(

R

R R R

1

32

x 160 Ω

=== )45.0)(4)(6.1(CR R L s32x 2.88 H





9


An industrial coil is modeled as a series combination of an inductance L and

resistance R, as shown in Fig. 9.90. Since an ac voltmeter measures only the

magnitude of a sinusoid, the following measurements are taken at 60 Hz when

the circuit operates in the steady state:

1V 145V, V 50V, V 110V s o= = =

Use these measurements to determine the values of L and R.

Figure 9.90


Let °∠= 0145sV , L377 jL)60)(2( jL jX =π=ω=

jXR 80

0145

jXR 80

s

++°∠

=++

=V

I

jXR 80

)145)(80(801 ++

== IV

jXR 80

)145)(80(50

++=

(1)

jXR 80

)0145)( jXR () jXR (o ++

°∠+=+= IV





10

jXR 80

)145)( jXR (110

+++

=

(2)

From (1) and (2),

jXR

80

110

50

+=

=+5

11)80( jXR

30976XR 22 =+

(3)

From (1),

23250

)145)(80(

jXR 80 ==++ 53824XR R 1606400 22 =+++

47424XR R 160 22 =++

(4)

Subtracting (3) from (4),

= → = R 16448R 160 102.8 Ω

From (3),

204081056830976X2 =−=

= → == LL37786.142X 0.3789 H

Chapter9 Assignments

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