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July 2013

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Chapter 9

Sinusoidal Steady-

State Analysis

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اىاد أدا اتحااخ ساتقح ىيؼذذ

ا ػيى أدااىقؼ اىزمس تاحح جاا

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July 2013




𝑖 𝑡 = 20 cos 2000𝜋𝑡 + 60𝑜 .

Solution:

a) From the statement of the problem, 𝑇 = 1 ms;

Hence 𝑓 = 1 𝑇 = 1000 Hz.

b) 𝜔 = 2𝜋𝑓 = 2000 𝜋 rad s .

c) We have 𝑖 𝑡 = 𝐼𝑚 cos 𝜔𝑡 + ∅ = 20 cos 2000𝜋𝑡 + ∅

But 𝑖 0 = 10 A.

Therefore 10 = 20 cos ∅ and ∅ = 60𝑜 .

Thus the expression for 𝑖 𝑡 becomes:

d) The rms value of a sinusoidal current is 𝐼𝑚 2 . Therefore the rms value

is 20 2 , or 14.14 A.

Example 9.1: Finding the Characteristics of a Sinusoidal Current

A sinusoidal current has a maximum amplitude of 20 A. The current passes through one

complete cycle in 1 ms. The magnitude of the current at zero time is 10 A.

a) What is the frequency of the current in hertz?

b) What is the frequency in radians per second?

c) Write the expression for 𝑖 𝑡 using the cosine function. Express ∅ in degrees.

d) What is the rms value of the current?

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Solution:

a) From the expression for 𝑣, 𝜔 = 120𝜋 rad s .

Because 𝜔 = 2 𝜋 𝑇 , 𝑇 = 2 𝜋 𝜔 =1

60 s = 16.667 ms.

b) The frequency is 1 𝑇 = 60 Hz.

c) From a , 𝜔 = 2𝜋 16.667 ;

Thus, at 𝑡 = 2.778 ms, 𝜔𝑡 ≈ 1.047 rad or 60𝑜 .

Therefore, 𝑣 2.778 ms = 300 cos 60𝑜 + 30𝑜 = 0 V.

d) 𝑉𝑟𝑚𝑠 = 300 2 = 212.13 V.

Example 9.2: Finding the Characteristics of a Sinusoidal Voltage

A sinusoidal voltage is given by the expression 𝑣 = 300 cos 120𝜋𝑡 + 30𝑜 .

a) What is the period of the voltage in milliseconds?

b) What is the frequency in hertz?

c) What is the magnitude of 𝑣 at 𝑡 = 2.778 ms?

d) What is the rms value of 𝑣?

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Solution:

a) The phasor transform of the current source is 8 0o ; the resistors transform directly

to the frequency domain as 10 and 6 Ω; the 40 𝜇H inductor has an impedance of

𝑗8 Ω at the given frequency of 200,000 rad s ; and at this frequency the 1 𝜇F

capacitor has an impedance of – 𝑗5 Ω. Figure 9.19 shows the frequency-domain

equivalent circuit and symbols representing the phasor transforms of the unknowns.

b) The circuit shown in Fig. 9.19 indicates that we can easily obtain the voltage across

the current source once we know the equivalent impedance of the three parallel

branches. Moreover, once we know 𝐕, we can calculate the three phasor currents

𝐈𝟏, 𝐈𝟐, and 𝐈𝟑. To find the equivalent impedance of the three branches, we first find

the equivalent admittance simply by adding the admittances of each branch.

Example 9.7: Combining Impedances in Series and in Parallel

The sinusoidal current source in the circuit shown in Fig. 9.18 produces the current

𝑖𝑠 = 8 cos 200,000𝑡 A.

a) Construct the frequency-domain equivalent circuit.

b) Find the steady-state expressions for 𝑣, 𝑖1, 𝑖2 and 𝑖3.

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July 2013




𝑌1 =1

10= 0.1 S

𝑌2 =1

6 + 𝑗8=

6 − 𝑗8

100= 0.06 − 𝑗0.08 S

𝑌3 =1

−𝑗5= 𝑗0.2 S

𝑍 =1

𝑌= 5 − 36.87o

V = ZI = 40 − 36.87o V

I1 =40 − 36.87o

10= 4 − 36.87o = 3.2 − 𝑗2.4 A

I2 =40 − 36.87o

6 + 𝑗8= 4 − 90𝑜 = −𝑗4 A

I3 =40 − 36.87o

5 − 90o = 8 53.13o = 4.8 + 𝑗6.4 A

I1 + I2 + I3 = I

3.2 − 𝑗2.4 − 𝑗4 + 4.8 + 𝑗6.4 = 8 + 𝑗0

𝑖1 = 4 cos 200,000𝑡 − 36.87o A

𝑖2 = 4 cos 200,000𝑡 − 90o A

𝑖3 = 8 cos 200,000𝑡 + 53.13o A

Continued (Example 9.7):

The admittance of the three branches is:

𝑌 = 𝑌1 + 𝑌2 + 𝑌3 = 0.16 + 𝑗0.12 = 0.2 36.87o

The impedance at the current source is:

The Voltage V is

Hence

We check the computations at this point by verifying that

Specifically,

The corresponding steady-state time-domain expressions are

𝑣 = 40 cos 200,000𝑡 − 36.87o V

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𝑤𝐿 = 10 Ω − 1

𝑤𝐶= − 20 Ω

𝑍𝑥𝑦 = 20 𝑗10 + 5 − 𝑗20 =20 𝑗10

20 + 𝑗10 + 5 − 𝑗20

= 4 + 𝑗8 + 5 − 𝑗20 = 9 − 𝑗12 Ω

𝑍𝑥𝑦 = 5 − 𝑗5 + 20 𝑗40 = 5 − 𝑗5 + 20 𝑗40

20 + 𝑗40

= 5 − 𝑗5 + 16 + 𝑗8 = 21 + 𝑗3 Ω

400𝑤𝐿

400 + 𝑤2𝐿2=

106

25𝑤

d) 𝑍𝑥𝑦 =20𝑤2𝐿2

400 + 𝑤2𝐿2+ 5 = 10 + 5 = 15 Ω

Solution:

a) 𝑤 = 2000 rad s

b) 𝑤 = 8000 rad s

𝑤𝐿 = 40 Ω − 1

𝑤𝐶= − 5 Ω

c) 𝑍𝑥𝑦 = 20(𝑗𝑤𝐿 )

20+𝑗𝑤𝐿 + 5 −

𝑗106

25𝑤 =

20𝑤2𝐿2

400+𝑤2𝐿2+ 5 −

𝑗106

25𝑤+

𝑗400𝑤𝐿

400+𝑤2𝐿2

The impedance will be purely resistive when the 𝑗 terms cancel, i.e.,

Solving for 𝑤 yields 𝑤 = 4000 rad s

Assessment 9.7:

A 20 Ω resistor is connected in parallel with a 5 mH inductor. This parallel combination is

connected in series with a 5 Ω resistor and a 25 𝜇F capacitor.

a) Calculate the impedance of this interconnection if the frequency is 2 krad s .

b) Repeat (a) for a frequency of 8 krad s .

c) At what finite frequency does the impedance of the interconnection become purely

resistive?

d) What is the impedance at the frequency found in (c)?

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July 2013




V = 150 0o , Is =V

Zxy=

150 0o

15= 10 0o A

I𝐿 =20

20 + 𝑗20 10 = 5 − 𝑗5 = 7.07 − 45o A

𝑖𝐿 = 7.07 cos 4000𝑡 − 45𝑜 A, 𝐼𝑚 = 7.07 A

Solution:

The frequency 4000 rad s was found to give 𝑍𝑥𝑦 = 15 Ω in Assessment

problem 9.7. Thus,

Using current division,

Assessment 9.8:

The interconnection described in Assessment Problem 9.7 is connected across the terminals

of a voltage source that is generating 𝑣 = 150 cos 4000𝑡 V. What is the maximum

amplitude of the current in the 5 mH inductor?

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July 2013




𝑌𝑝 =1

6 − 𝑗2+

1

4 + 𝑗12+

1

5+

1

𝑗10 = 0.375 − 𝑗0.125 S

𝑍𝑝 =1

𝑌𝑝=

1

0.375 − 𝑗0.125 = 2.4 + 𝑗0.8 Ω

𝑍𝑎𝑏 = − 𝑗12.8 + 2.4 + 𝑗0.8 + 13.6 = 16 − 𝑗12 Ω

𝑌𝑎𝑏 =1

𝑍𝑎𝑏=

1

16 − 𝑗12= 0.04 + 𝑗0.03 S = 40 + 𝑗30 mS = 50 36.87o mS

Solution:

First find the admittance of the parallel branches

Problem 9.26:

Find the admittance 𝑌𝑎𝑏 in the circuit seen in Fig. P9.26. Express 𝑌𝑎𝑏 in both polar and

rectangular form. Give the value of 𝑌𝑎𝑏 in millisiemens.

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𝑍 = 4 + 𝑗 50 0.24 − 𝑗1000

50 2.5 = 5.66 45o Ω

I𝑜 =0.1 − 90o × 10−3

5.66 45o = 17.67 − 135o mA

𝑖𝑜 𝑡 = 17.67 sin 50𝑡 − 135o mA

Solution:

Problem 9.31:

Find the steady-state expression for 𝑖𝑜 𝑡 in the circuit in Fig. P9.31 if

𝑣𝑠 = 100 sin 50𝑡 mV.

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V𝑔 = 40 − 15o V; I𝑔 = 40 − 68.13o mA

𝑍 =V𝑔

I𝑔= 1000 53.13o Ω = 600 + 𝑗800 Ω

𝑍 = 600 + 𝑗 3.2𝑤 −0.4 × 106

𝑤

∴ 3.2𝑤 −0.4 × 106

𝑤= 800

∴ 𝑤2 − 250𝑤 − 125,000 = 0

𝑤 > 0 ∴ 𝑤 = 500 rad s

Solution:

𝑖𝑜 = 40 sin 𝑤𝑡 + 21.87o mA

𝑣𝑔 = 40 cos 𝑤𝑡 − 15o V

Problem 9.28:

The circuit shown in Fig. P9.28 is operating in the sinusoidal steady state. Find the value

of 𝑤 if

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𝑍𝐿 = 𝑗 2000 60 × 10−3 = 𝑗120 Ω

𝑍𝐶 =− 𝑗

2000 12.5 × 10−6 = − 𝑗40 Ω

I = 120 − 𝑗40

120 − 𝑗40 + 40 + 𝑗120 0.5 = 0.25 − 𝑗0.25 A

V𝑜 = 𝑗120 I = 30 + 𝑗30 = 42.43 45𝑜

𝑣𝑜 = 42.43 cos 2000𝑡 + 45o V

Solution:

Construct the phasor domain equivalent circuit:

Using current division:

Problem 9.33:

Find the steady-state expression for 𝑣𝑜 in the circuit of Fig. P9.33 if

𝑖𝑔 = 500 cos 2000𝑡 mA.

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1

𝑗𝑤𝐶=

106

1 50,000 = − 𝑗20 Ω

𝑗𝑤𝐿 = 𝑗50,000 1.2 10−3 = 𝑗60 Ω

V𝑔 = 40 0𝑜 V

𝑍𝑒 = 30 𝑗60

30 + 𝑗60= 24 + 𝑗12 Ω

𝑍𝑇 = 24 + 𝑗12 − 𝑗20 = 24 − 𝑗8 Ω

I𝑔 =40 0𝑜

24 − 𝑗8= 1.5 + 𝑗0.6 A

Vo = 𝑍𝑔I𝑔 = 24 + 𝑗12 1.5 + 𝑗0.5 = 30 + 𝑗30 = 42.43 45𝑜 V

𝑣𝑜 = 42.43 cos 50,000𝑡 + 45o V

Solution:

Problem 9.29:

The circuit in Fig. P9.29 is operating in the sinusoidal steady state. Find the steady-state

expression for 𝑣𝑜 𝑡 if 𝑣𝑔 = 40 cos 50,000𝑡 V.

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a) 𝑍𝑔 = 500 − 𝑗106

𝑤+

103 𝑗0.5 𝑤

103 + 𝑗0.5 𝑤= 500 − 𝑗

106

𝑤+

0.5 × 103𝑗 𝑤 103 − 𝑗0.5 𝑤

106 + 0.25𝑤2

= 500 − 𝑗106

𝑤+

250𝑤2

106 + 0.25𝑤2+ 𝑗

5 × 105𝑤

106 + 0.25𝑤2

∴ 106

𝑤=

0.5 × 106𝑤

106 + 0.25𝑤2

106 + 0.25𝑤2 = 0.5𝑤2 ⟹ 𝑤2 = 4 × 106 ⟹ 𝑤 = 2000 rad s

b) When 𝑤 = 2000 rad s

𝑍𝑔 = 500 +250 2000 2

106 + 0.25 2000 2= 1000 Ω

∴ I𝑔 =20 0o

1000= 20 0o mA

V𝑜 = V𝑔 − I𝑔Z1

𝑍1 = 500 − 𝑗106

1 × 2000= 500 − 𝑗500 Ω

V𝑜 = 20 0o − 0.02 0o 500 − 𝑗500 = 10 + 𝑗10 = 14.14 45o

𝑣𝑜 = 14.14 cos 2000𝑡 + 45o V

Solution:

Problem 9.38:

a) The frequency of the source voltage in the circuit in Fig. P9.38 is adjusted until 𝑖𝑔 is in

phase with 𝑣𝑔 . What is the value of 𝑤 in radians per second?

b) If 𝑣𝑔 = 20 cos 𝑤𝑡 V (where 𝑤 is the frequency found in [a]), what is the steady-state

expression for 𝑣𝑜?

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a) 𝑌1 =1

5000= 0.2 × 10−3 S

𝑌2 =1

1200 + 𝑗0.2 𝑤=

1200

1.44 × 106 + 0.04𝑤2− 𝑗

0.2𝑤

1.44 × 106 + 0.04𝑤2

𝑌3 = 𝑗𝑤50 × 10−9

𝑌𝑇 = 𝑌1 + 𝑌2 + 𝑌3

𝑤50 × 10−9 =0.2𝑤

1.44 × 106 + 0.04𝑤2

or 0.04𝑤2 + 1.44 × 106 =0.2 × 109

50= 4 × 106

∴ 0.04𝑤2 = 2.56 × 106 ∴ 𝑤 = 8000 rad s = 8 krad s

b) 𝑌𝑇 = 0.2 × 10−3 +1200

1.44 × 106 + 0.04 64 × 106= 0.5 × 10−3 S

∴ 𝑍𝑇 = 2000 Ω

V𝑜 = 2.5 × 10−3 0o 2000 = 5 0o

𝑣𝑜 = 5 cos 8000𝑡 V

Solution:

For 𝑖𝑔 and 𝑣𝑜 to be in phase the 𝑗 component of 𝑌𝑇 must be zero; thus,

Problem 9.42:

The frequency of the sinusoidal current source in the circuit in Fig. P9.42 is adjusted until

𝑣𝑜 is in phase with 𝑖𝑔 .

a) What is the value of 𝑤 in radians per second?

b) If 𝑖𝑔 = 2.5 cos 𝑤𝑡 mA (where 𝑤 is the frequency found in [a]), what is the steady-

state expression for 𝑣𝑜?

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𝑍𝑜 = 12,000 − 𝑗109

20,000 3.125 = 12,000 − 𝑗16,000 Ω

𝑍𝑇 = 6000 + 𝑗40,000 + 12,000 − 𝑗16,000 = 18,000 + 𝑗24,000 Ω

= 30,000 53.13o Ω

V𝑜 = V𝑔

𝑍𝑜

𝑍𝑇=

75 0o 20,000 − 53.13o

30,000 53.13o = 50 − 106.26o V

𝑣𝑜 = 50 cos 20,000𝑡 − 106.26o V

Solution:

Problem 9.66:

Use the concept of voltage division to find the steady-state expression for 𝑣𝑜 𝑡 in the

circuit in Fig.P9.66 if 𝑣𝑔 = 75 cos 20,000𝑡 V.

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1

𝑗𝑤𝐶= − 𝑗

109

12,500 (800)= − 𝑗100 Ω

𝑗𝑤𝐿 = 𝑗 12,500 0.04 = 𝑗500 Ω

Let 𝑍1 = 50 − 𝑗100 Ω; 𝑍2 = 250 + 𝑗500 Ω

I𝑔 = 125 0o mA

I𝑜 =−I𝑔𝑍2

𝑍1 + 𝑍2=

−125 0o 250 + 𝑗500

300 + 𝑗400

= − 137.5 − 𝑗25 mA = 139.75 − 169.7o mA

𝑖𝑜 = 139.75 cos 12,500𝑡 − 169.7𝑜 mA

Solution:

Problem 9.65:

Use the concept of current division to find the steady-state expression for 𝑖𝑜 in the circuit

in Fig. P9.65 if 𝑖𝑔 = 125 cos 12,500𝑡 mA.

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sinusoidal steady- state analysis · 2013-08-20 · sinusoidal steady-state analysis حح سات...

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