February 5, 2006

CHAPTER 10

P.P.10.1 2,010)t2sin(10 =ω°∠⎯→⎯

4jLjH2 =ω⎯→⎯

5.2j-Cj

1F2.0 =

ω⎯→⎯

Hence, the circuit in the frequency domain is as shown below.

At node 1, j2.5-2

10 211 VVV −+=

21 4j)4j5(100 VV −+= (1)

At node 2, 4

3j2.5-4j

2x212 VVVVV −+

−= where 1x VV =

)3(5.2)(4jj2.5- 21212 VVVVV −+−=

21 )5.1j5.2()4j5.7(-0 VV +++= (2) Put (1) and (2) in matrix form.

⎥⎦⎤

⎢⎣

⎡=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡++

+0

1005.1j5.2j4)(7.5-

j4-4j5

2

1

VV

where °∠=−=+−++=Δ 29.05-74.255.12j5.22j4))-j4)(-(7.5()15.j5.2)(4j5(

⎥⎦

⎤⎢⎣

⎡−

⎥⎦

⎤⎢⎣

⎡++

+

=⎥⎦

⎤⎢⎣

⎡0

1005.12j5.22

4j5j47.5j45.1j5.2

2

1

VV

°∠=°∠°∠

=−+

= 01.6032.11)100(29.05-74.25

96.30915.2)100(

5.12j5.225.1j5.2

1V

°∠=°∠

°∠=

−+

= 12.5702.33)100(29.05-74.2507.285.8

)100(5.12j5.22

4j5.72V

-j2.5 Ω

+

Vx

−

4 Ω

j4 Ω10∠0° A + − 3Vx2 Ω

V1 V2

In the time domain, =)t(v1 11.32 sin(2t + 60.01°) V =)t(v2 33.02 sin(2t + 57.12°) V P.P.10.2 The only non-reference node is a supernode.

2j-4j4

15 2211 VVVV++=

−

2211 24jj-15 VVVV ++=− 21 )4j2()j1(15 VV ++−= (1) The supernode gives the constraint of (2) °∠+= 602021 VV Substituting (2) into (1) gives

2)3j3()6020)(j1(15 V++°∠−=

°∠=°∠°∠

=+

°∠−−= 7.165376.3

45243.472.210327.14

3j3)6020)(j1(15

2V

)32.17j10()8327.0j272.3-(602021 +++=°∠+= VV 154.18j728.61 +=V

Therefore, =1V 19.36∠69.67° V, =2V 3.376∠165.7° V P.P.10.3 Consider the circuit below.

For mesh 1, 04j)4j2j8( 21 =−+− II (1) 21 4j)2j8( II =+

6 Ω

j4 Ω8 Ω

-j2 Ω

I1 I2+ − 10∠30° V

I3

2∠0° A

For mesh 2, 0301064j)4j6( 312 =°∠+−−+ III For mesh 3, 2-3 =I Thus, the equation for mesh 2 becomes °∠−=−+ 301012-4j)4j6( 12 II (2)

From (1), 112 )2j5.0(4j

2j8III −=

+= (3)

Substituting (3) into (2), °∠−=−−+ 301012-4j)2j5.0()4j6( 11 II )5j66.20(-)14j11( 1 +=− I

14j11

)5j66.20(-1 −

+=I

Hence, °∠°∠

=−+

==51.84-8.1713.6256.21

14j115j66.20

- 1o II

=oI 1.194∠65.44° A P.P.10.4 Meshes 2 and 3 form a supermesh as shown in the circuit below.

For mesh 1, 05)4j()4j15(50 321 =−−−−+− III

5054j)4j15( 321 =−+− III (1) For the supermesh, 0)4j5()6j5()4j8j( 132 =−−−+− III (2) Also, 223 += II (3)

10 Ω

50∠0° V + −

I1

-j4 Ω

5 Ω

j8 Ω

-j6 Ω I3

I2

Eliminating from (1) and (2) 3I60)4j5-()4j15( 21 =++− II (4)

12j10-)2j5()4j5-( 21 +=−++ II (5) From (4) and (5),

⎥⎦

⎤⎢⎣

⎡+

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+

+−j1210-

60j2-5j45-j45-4j15

2

1

II

°∠=−=+

+−=Δ 9.78-86.5810j58

j2-5j45-j45-4j15

°∠=−=+

+=Δ 3.84-67.29820j298

j2-5j1210-j45-60

1

Thus, =ΔΔ

== 11o II 5.074∠5.94° A

P.P.10.5 Let , where and are due to the voltage source and current source respectively. For consider the circuit in Fig. (a).

"o

'oo III += '

oI "oI

'oI

For mesh 1, 04j)2j8( 21 =−+ II (1) 12 )2j5.0( II −= For mesh 2, 030104j)4j6( 12 =°∠−−+ II (2) Substituting (1) into (2), °∠=−−+ 30104j)2j5.0)(4j6( 11 II

556.0j08.014j11

30101

'o +=

−°∠

== II

8 Ω

-j2 Ω 6 Ω

I1 10∠30° V + − j4 Ω I2

Io'

(a)

For consider the circuit in Fig. (b). "oI

Let , Ω−= 2j81Z Ω+=+

== 769.2j846.14j6

24j4j||62Z

53.0j4164.077.0j846.9

)769.2j846.1)(2()2(21

2"o +=

++

=+

=ZZ

ZI

Therefore, 086.1j4961.0"

o'oo +=+= III

=oI 1.1939∠65.45° A P.P.10.6 Let , where is due to the voltage source and is due to the current source. For , we remove the current source.

"o

'oo vvv += '

ov "ov

'ov

5,030)t5sin(30 =ω°∠⎯→⎯

j-)2.0)(5(j

1Cj

1F2.0 ==

ω⎯→⎯

5j)1)(5(jLjH1 ==ω⎯→⎯ The circuit in the frequency domain is shown in Fig. (a).

6 Ω

j4 Ω8 Ω

-j2 Ω

2∠0° A

Io"

(b)

+

Vo'

−

+ − 30∠0° V

8 Ω

-j Ω j5 Ω

(a)

Note that -j1.25j5||j- = By voltage division,

°∠=−

= 81.12-631.4)30(25.1j8

j1.25-'oV

Thus, )12.81t5sin(631.4v'o °−=

For , we remove the voltage source. "

ov 10,02)t10cos(2 =ω°∠⎯→⎯

5.0j-)2.0)(10(j

1Cj

1F2.0 ==

ω⎯→⎯

10j)1)(10(jLjH1 ==ω⎯→⎯ The corresponding circuit in the frequency domain is shown in Fig (b).

Let , 5.0j-1 =Z 9.3j878.410j8

80j10j||82 +=

+==Z

By current division,

)2(21

2

ZZZ

I+

=

j3.44.878j3.9)(4.877j-

-j0.5))(2(-j0.5)(21

2"o +

+=

+==

ZZZ

IV

°∠=°∠°∠

= 86.24-051.188.3494.5

51.36-245.6"oV

Thus, )24.86t10cos(051.1v"o °−=

Therefore, "

o'oo vvv +=

=ov 4.631 sin(5t – 81.12°) + 1.051 cos(10t – 86.24°) V

+

Vo"

−

-j0.5 Ωj10 Ω8 Ω 2∠0°

I

(b)

P.P.10.7 If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a).

(a)

VS+ −

4 Ω j Ω-j3 Ω 2 Ω

-j2 Ω

1 Ω

Io

j5 Ω

16j12)3j4)(4j(sss +=−== ZIV We transform the voltage source to a current source as shown in Fig. (b).

Let 2j6j23j4 −=++−=Z . Then, 3j5.12j616j12s

s +=−+

==ZV

I .

Io

6 1 Ω ΩIS j5 Ω

-j2 -j2 Ω Ω

(b)

Note that )j1(3

103j6

)5j)(2j6(5j|| +=

+−

=Z .

By current division,

)3j5.1()2j1()j1(

310

)j1(3

10

o +−++

+=I

°∠°∠

=++−

=1.17602.1356.11672.44

4j1340j20

oI

=oI 3.288∠99.46° A

P.P.10.8 When the voltage source is set equal to zero,

)2j6(||)4j-(10th ++=Z

j2-6j2)-j4)(6(

10th

++=Z

2.3j4.210th −+=Z =thZ 12.4 – j3.2 Ω

By voltage division,

2j6)2030)(4j-(

)2030(4j2j6

4j-th −

°∠=°∠

−+=V

°∠°∠°∠

=43.18-324.6

)2030)(90-4(thV

=thV 18.97∠-51.57° V P.P.10.9 To find , consider the circuit in Fig. (a). thV

At node 1, 4j8

52j4

0 211

+−

+=−− VVV

))(5.0j1(50)j2(- 211 VVV −−+=+

12 )5.0j3()5.0j1(50 VV +−−= (1)

At node 2, 04j8

2.05 21o =

+−

++VV

V , where . 21o VVV −=

Hence, the equation for node 2 becomes

V1V2

4 – j2

8 + j4

a

b

5∠0°

0.2Vo

(a)

+ − Vo

4 – j2

8 + j4

a

b

0.2Vo

(b)

+ − 1∠0°

+ −Vo

Is VS

04j8

)(2.05 2121 =

+−

+−+VV

VV

5.0j350

21 +−= VV (2)

Substituting (2) into (1),

5.0j35.0j3

)50()5.0j3()5.0j1(50 22 −+

++−−= VV

)12j35(3750

)j2(50-0 2 +++−= V

°∠=++

= 9.7235.7j2

22.16j702.2-2V

== 2th VV 7.35∠72.9° V To find , we remove the independent source and insert a 1-V voltage source between terminals a-b, as shown in Fig. (b).

thZ

At node a, 2j44j8

-0.2 sos −+++=

VVI

But, and –1s =V so 2j44j8VV

−+++

=4j8

So, 2j12

8.0j6.22j122j12

4j8(0.2)s ++

=+

+++

=1I

and °∠°∠

=++

===10.1772.246.9166.12

8.0j6.22j121

ss

sth II

VZ

=thZ 4.473∠–7.64° Ω

P.P.10.10 To find , consider the circuit in Fig. (a). NZ

j13

)3j9)(2j4()3j9(||)2j4(N −

−+=−+=Z

=NZ 3.176 + j0.706 Ω To find , short-circuit terminals a-b as shown in Fig. (b). Notice that meshes 1 and 2 form a supermesh.

NI

For the supermesh, 0)3j9()3j1(820- 321 =−−−++ III (1) Also, 4j21 += II (2) For mesh 3, 0)3j1(8)j13( 213 =−−−− III (3) Solving for , we obtain 2I

°∠°∠

=−−

==43.18-487.911.51-65.79

3j962j50

2N II

=NI 8.396∠-32.68° A Using the Norton equivalent, we can find as in Fig. (c). oI

j2 Ω 4 Ω

-j3 Ω 8 Ω 1 Ω

ZN

a

b(a)

j2 Ω 4 Ω

-j3 Ω8 Ω 1 Ω a

b(b)

INI2+ − 20∠0° I1

I3

-j4 Ω

IN 10 – j5 ΩZN

(c)

Io

By current division,

)68.32-396.8(294.4j176.13706.0j176.3

5j10 NN

No °∠

−+

=−+

= IZ

ZI

°∠

°∠°∠=

18.05-858.13)32.68-396.8)(53.12254.3(

oI

=oI 1.971∠-2.10° A P.P.10.11

Ω=××

=ω

⎯→⎯ k-j20)1010)(105(j

1Cj

1nF10 9-3

1

Ω=××

=ω

⎯→⎯ k-j10)1020)(105(j

1Cj1

nF20 9-32

Consider the circuit in the frequency domain as shown below.

As a voltage follower, o2 VV =

At node 1, 2020j-10

2 o1o11 VVVVV −+

−=

−

o1 )j1()j3(4 VV +−+= (1)

At node 2, 10j-

020

oo1 −=

− VVV

(2) o1 )2j1( VV += Substituting (2) into (1) gives

or o6j4 V= °∠= 90-32

oV

+−

+ −

10 kΩ 20 kΩ

-j20 kΩ

-j10 kΩ

Io Vo

V2

V1

2∠0° V

Hence, V)90t5000cos(667.0)t(vo °−= =)t(vo 0.667 sin(5000t) V

Now, k20j-

1oo

VVI

−=

But from (2) 34-

2j- o1o ==− VVV

A66.66j-k20j-34-

o μ==I

Hence, A)90t5000cos(67.66)t(io μ°−= =)t(io 66.67 sin(5000t) μA

P.P.10.12 Let RCj1

RCj

1||RZ

ω+=

ω=

ZR

RVV

o

s

+=

The loop gain is

RCj2RCj1

RCj1RR

RZR

RVV

G/1o

sω+ω+

=

ω++

=+

==

where 10)101)(1010)(1000(RC 6-3 =××=ω

°∠°∠

=++

=69.782.1029.8405.10

10j210j1G/1

G = 1.0147∠–5.6°

P.P.10.13 The schematic is shown below.

Since . Setup/Analysis/AC Sweep as Linear for 1 point starting and ending at a frequency of 447.465 Hz. When the schematic is saved and run, the output file includes

Hz465.477fs/rad3000f2 =⎯→⎯=π=ω

Frequency IM(V_PRINT1) IP(V_PRINT1) 4.775E+02 5.440E-04 -5.512E+01

Frequency VM($N_0005) VP($N_0005) 4.775E+02 2.683E-01 -1.546E+02

From the output file, we obtain 0.2682∠-154.6° V and =oV =oI 0.544∠-55.12° mA Therefore, =)t(vo 0.2682 cos(3000t – 154.6°) V =)t(io 0.544 cos(3000t – 55.12°) mA

P.P.10.14 The schematic is shown below.

We select ω = 1 rad/s and f = 0.15915 Hz. We use this to obtain the values of capacitances, where cX1C ω= , and inductances, where ω= LXL . Note that IAC does not allow for an AC PHASE component; thus, we have used VAC in conjunction with G to create an AC current source with a magnitude and a phase. To obtain the desired output use Setup/Analysis/AC Sweep as Linear for 1 point starting and ending at a frequency of 0.15915 Hz. When the schematic is saved and run, the output file includes

Frequency IM(V_PRINT1) IP(V_PRINT1) 1.592E-01 2.584E+00 1.580E+02

Frequency VM($N_0004) VP($N_0004) 1.592E-01 9.842E+00 4.478E+01

From the output file, we obtain =xV 9.842∠44.78° V and =xI 2.584∠158° A

P.P.10.15 ( =×⎟⎠⎞

⎜⎝⎛

××

+=⎟⎠

⎞⎜⎝

⎛+= 9-

3

6

1

2eq 1010

10101010

1CRR

1C ) 10 μF

P.P.10.16 If Ω=== k5.2RRR 21 and nF1CCC 21 ===

=××π

=π

=)101)(105.2)(2(

1RC21

f 9-3o 63.66 kHz