‫‪July 2013‬‬
‫‪Chapter 9‬‬
‫‪Sinusoidal Steady‬‬‫‪State Analysis‬‬
‫ٍزمشاخ ششح ٗتَاسٌِ ٍحي٘ىح‬
‫اٍتحاّاخ ساتقح ىيؼذٌذ ٍِ اىَ٘اد أدّآ‬
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‫فشئ ّٗجاح اَخشٌِ‪.‬‬
‫اىْ٘تاخ ٍجاٍّح ىيْفغ اىؼاً فٍشجى اىَسإَح تاإلتالؽ ػِ أي خطأ أٗ ٍالحظاخ تشإا ضشٗسٌح تشساىح ّصٍح ‪ 9 4444 260‬أٗ تاىثشٌذ اإلىنتشًّٗ‬
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July 2013
 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?
 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 + ∅
𝑖 0 = 10 A.
But
Therefore
10 = 20 cos ∅ and ∅ = 60𝑜 .
Thus the expression for 𝑖 𝑡 becomes:
𝑖 𝑡 = 20 cos 2000𝜋𝑡 + 60𝑜 .
d) The rms value of a sinusoidal current is 𝐼𝑚
is 20
2. Therefore the rms value
2, or 14.14 A.
‫إرا اتتسٌ اىَٖضًٗ فقذ‬
.‫اىَْتصش ىزج اىف٘ص‬
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July 2013
 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?
 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.
‫إرا أسدخ أُ تْتقٌ ٍَِ أساء إىٍل تطشٌقح‬
.ْٔ‫اػف ػ‬
،ٔ‫تجؼئ ٌتضاءه أٍاً ّفس‬
ُ
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July 2013
 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 .
 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.
‫أسشع طشٌق إلّٖاء اىَؼشمح‬
.‫أُ تستسيٌ ىيٖضٌَح‬
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July 2013
 Continued (Example 9.7):
1
= 0.1 S
10
1
6 − 𝑗8
𝑌2 =
=
= 0.06 − 𝑗0.08 S
6 + 𝑗8
100
1
𝑌3 =
= 𝑗0.2 S
−𝑗5
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:
1
𝑍 = = 5 − 36.87o
𝑌
The Voltage V is
𝑌1 =
V = ZI = 40 − 36.87o V
Hence
40 − 36.87o
I1 =
= 4 − 36.87o = 3.2 − 𝑗2.4 A
10
40 − 36.87o
I2 =
= 4 − 90𝑜
6 + 𝑗8
I3 =
40 − 36.87o
5−
90o
= 8 53.13o
= −𝑗4 A
= 4.8 + 𝑗6.4 A
We check the computations at this point by verifying that
I1 + I2 + I3 = I
Specifically,
3.2 − 𝑗2.4 − 𝑗4 + 4.8 + 𝑗6.4 = 8 + 𝑗0
The corresponding steady-state time-domain expressions are
𝑣 = 40 cos 200,000𝑡 − 36.87o V
𝑖1 = 4 cos 200,000𝑡 − 36.87o A
𝑖2 = 4 cos 200,000𝑡 − 90o A
𝑖3 = 8 cos 200,000𝑡 + 53.13o A
‫أسشع طشٌق إلّٖاء اىَؼشمح‬
.‫أُ تستسيٌ ىيٖضٌَح‬
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July 2013
 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)?
 Solution:
a)
𝑤 = 2000 rad s
−1
= − 20 Ω
𝑤𝐶
20 𝑗10
= 20 𝑗10 + 5 − 𝑗20 =
+ 5 − 𝑗20
20 + 𝑗10
𝑤𝐿 = 10 Ω
𝑍𝑥𝑦
= 4 + 𝑗8 + 5 − 𝑗20 = 9 − 𝑗12 Ω
b)
𝑤 = 8000 rad s
−1
𝑤𝐿 = 40 Ω
𝑤𝐶
=−5Ω
𝑍𝑥𝑦 = 5 − 𝑗5 + 20 𝑗40 = 5 − 𝑗5 +
20 𝑗40
20 + 𝑗40
= 5 − 𝑗5 + 16 + 𝑗8 = 21 + 𝑗3 Ω
c) 𝑍𝑥𝑦 =
20(𝑗𝑤𝐿 )
20+𝑗𝑤𝐿
+ 5−
𝑗 10 6
25𝑤
=
20𝑤 2 𝐿2
+5−
2
400+𝑤 2 𝐿
𝑗 10 6
25𝑤
+
𝑗 400𝑤𝐿
400+𝑤 2 𝐿2
The impedance will be purely resistive when the 𝑗 terms cancel, i.e.,
400𝑤𝐿
106
=
400 + 𝑤 2 𝐿2 25𝑤
Solving for 𝑤 yields 𝑤 = 4000 rad s
d) 𝑍𝑥𝑦
20𝑤 2 𝐿2
=
+ 5 = 10 + 5 = 15 Ω
400 + 𝑤 2 𝐿2
ٌ‫أسٍش اىحشب ٕ٘ سجو ٌحاٗه قتيل ٗى‬
.ٔ‫ ٍِٗ ثٌ ٌسأىل أال تقتي‬،‫ٌستطغ‬
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July 2013
 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?
 Solution:
The frequency 4000 rad s was found to give 𝑍𝑥𝑦 = 15 Ω in Assessment
problem 9.7. Thus,
o
V = 150 0 ,
V
150 0o
Is =
=
= 10 0o A
Zxy
15
Using current division,
I𝐿 =
20
10 = 5 − 𝑗5 = 7.07 − 45o A
20 + 𝑗20
𝑖𝐿 = 7.07 cos 4000𝑡 − 45𝑜 A,
𝐼𝑚 = 7.07 A
‫تثذٗ األٍ٘س أفضو مثٍشا إرا ّظشّا‬
.‫إىٍٖا ٍِ اىطشف اَخش‬
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July 2013
 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.
 Solution:
First find the admittance of the parallel branches
𝑌𝑝 =
1
1
1
1
+
+ +
= 0.375 − 𝑗0.125 S
6 − 𝑗2 4 + 𝑗12 5 𝑗10
𝑍𝑝 =
1
1
=
𝑌𝑝 0.375 − 𝑗0.125
= 2.4 + 𝑗0.8 Ω
𝑍𝑎𝑏 = − 𝑗12.8 + 2.4 + 𝑗0.8 + 13.6 = 16 − 𝑗12 Ω
𝑌𝑎𝑏 =
1
1
=
= 0.04 + 𝑗0.03 S = 40 + 𝑗30 mS = 50 36.87o mS
𝑍𝑎𝑏 16 − 𝑗12
‫تثذٗ األٍ٘س أفضو مثٍشا إرا ّظشّا‬
.‫إىٍٖا ٍِ اىطشف اَخش‬
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July 2013
 Problem 9.31:
Find the steady-state expression for 𝑖𝑜 𝑡
𝑣𝑠 = 100 sin 50𝑡 mV.
in the circuit in Fig. P9.31 if
 Solution:
𝑍 = 4 + 𝑗 50 0.24 − 𝑗
I𝑜 =
0.1 − 90o × 10−3
5.66
45o
1000
= 5.66 45o Ω
50 2.5
= 17.67 − 135o mA
𝑖𝑜 𝑡 = 17.67 sin 50𝑡 − 135o
mA
ً‫تنَِ أػيى دسجاخ االٍتٍاص ف‬
.‫مسش ٍقاٍٗح ػذٗك دُٗ قتاه‬
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July 2013
 Problem 9.28:
The circuit shown in Fig. P9.28 is operating in the sinusoidal steady state. Find the value
of 𝑤 if
𝑖𝑜 = 40 sin 𝑤𝑡 + 21.87o mA
𝑣𝑔 = 40 cos 𝑤𝑡 − 15o
V
 Solution:
V𝑔 = 40 − 15o V;
I𝑔 = 40 − 68.13o mA
V𝑔
𝑍=
= 1000 53.13o Ω = 600 + 𝑗800 Ω
I𝑔
0.4 × 106
𝑍 = 600 + 𝑗 3.2𝑤 −
𝑤
0.4 × 106
∴ 3.2𝑤 −
= 800
𝑤
∴ 𝑤 2 − 250𝑤 − 125,000 = 0
𝑤>0
∴ 𝑤 = 500 rad s
‫ستَا تضطش ىخ٘ض اىَؼشمح‬
.‫أمثش ٍِ ٍشج ىنسثٖا‬
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July 2013
 Problem 9.33:
Find the steady-state expression
𝑖𝑔 = 500 cos 2000𝑡 mA.
for
𝑣𝑜
in
the
circuit
of
Fig.
P9.33
if
 Solution:
𝑍𝐿 = 𝑗 2000 60 × 10−3 =
𝑍𝐶 =
𝑗120 Ω
−𝑗
= − 𝑗40 Ω
2000 12.5 × 10−6
Construct the phasor domain equivalent circuit:
Using current division:
I=
120 − 𝑗40
0.5 = 0.25 − 𝑗0.25 A
120 − 𝑗40 + 40 + 𝑗120
V𝑜 = 𝑗120 I = 30 + 𝑗30 = 42.43 45𝑜
𝑣𝑜 = 42.43 cos 2000𝑡 + 45o
V
ُ‫ ٗأ‬،‫مو إّساُ ٌؼتقذ أّٔ ٕ٘ اىزي ٌَتيل اىحقٍقح‬
.ٌ‫ ٕٗزا ٕ٘ سثة ّضاػاخ اىؼاى‬،‫اىحق ٕ٘ ٍا ٌؼتقذ‬
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July 2013
 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.
 Solution:
1
106
=
= − 𝑗20 Ω
𝑗𝑤𝐶
1 50,000
𝑗𝑤𝐿 = 𝑗50,000 1.2 10−3 = 𝑗60 Ω
V𝑔 = 40 0𝑜 V
𝑍𝑒 =
30 𝑗60
= 24 + 𝑗12 Ω
30 + 𝑗60
𝑍𝑇 = 24 + 𝑗12 − 𝑗20 = 24 − 𝑗8 Ω
40 0𝑜
I𝑔 =
= 1.5 + 𝑗0.6 A
24 − 𝑗8
Vo = 𝑍𝑔 I𝑔 = 24 + 𝑗12 1.5 + 𝑗0.5 = 30 + 𝑗30 = 42.43 45𝑜 V
𝑣𝑜 = 42.43 cos 50,000𝑡 + 45o V
‫ال ٌَنِ ىَؼإذج أُ تذًٗ إال‬
.ُ‫إرا اّتصش اىطشفا‬
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July 2013
 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 ?
 Solution:
106 103 𝑗0.5 𝑤
106 0.5 × 103 𝑗 𝑤 103 − 𝑗0.5 𝑤
𝑍𝑔 = 500 − 𝑗
+ 3
= 500 − 𝑗
+
𝑤
10 + 𝑗0.5 𝑤
𝑤
106 + 0.25𝑤 2
a)
106
250𝑤 2
5 × 105 𝑤
= 500 − 𝑗
+ 6
+𝑗 6
𝑤
10 + 0.25𝑤 2
10 + 0.25𝑤 2
106
0.5 × 106 𝑤
∴
= 6
𝑤
10 + 0.25𝑤 2
106 + 0.25𝑤 2 = 0.5𝑤 2 ⟹
𝑤 2 = 4 × 106
⟹
𝑤 = 2000 rad s
b) When 𝑤 = 2000 rad s
250 2000 2
𝑍𝑔 = 500 + 6
10 + 0.25 2000
∴
2
= 1000 Ω
20 0o
I𝑔 =
= 20 0o mA
1000
V𝑜 = V𝑔 − I𝑔 Z1
106
𝑍1 = 500 − 𝑗
= 500 − 𝑗500 Ω
1 × 2000
V𝑜 = 20 0o − 0.02 0o 500 − 𝑗500 = 10 + 𝑗10 = 14.14 45o
𝑣𝑜 = 14.14 cos 2000𝑡 + 45o
V
ُ‫ىِ ٌَنْل االّتصاس إرا ما‬
.‫تَقذٗسك قث٘ه اىٖضٌَح‬
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July 2013
 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 steadystate expression for 𝑣𝑜 ?
 Solution:
1
= 0.2 × 10−3 S
5000
1
1200
0.2𝑤
𝑌2 =
=
−
𝑗
1200 + 𝑗0.2 𝑤 1.44 × 106 + 0.04𝑤 2
1.44 × 106 + 0.04𝑤 2
a) 𝑌1 =
𝑌3 = 𝑗𝑤50 × 10−9
𝑌𝑇 = 𝑌1 + 𝑌2 + 𝑌3
For 𝑖𝑔 and 𝑣𝑜 to be in phase the 𝑗 component of 𝑌𝑇 must be zero; thus,
𝑤50 × 10−9 =
b)
0.2𝑤
1.44 × 106 + 0.04𝑤 2
or
0.2 × 109
0.04𝑤 + 1.44 × 10 =
= 4 × 106
50
∴
0.04𝑤 2 = 2.56 × 106
2
𝑌𝑇 = 0.2 × 10−3 +
6
∴ 𝑤 = 8000 rad s = 8 krad s
1200
= 0.5 × 10−3 S
6
6
1.44 × 10 + 0.04 64 × 10
∴ 𝑍𝑇 = 2000 Ω
V𝑜 = 2.5 × 10−3 0o 2000 = 5 0o
𝑣𝑜 = 5 cos 8000𝑡 V
‫ّصف اىْصش أُ تثقى ٍحتفظا‬
.‫ ال تخثش تزىل ػذٗك‬،‫تأػصاتل‬
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July 2013
 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.
 Solution:
109
𝑍𝑜 = 12,000 − 𝑗
= 12,000 − 𝑗16,000 Ω
20,000 3.125
𝑍𝑇 = 6000 + 𝑗40,000 + 12,000 − 𝑗16,000 = 18,000 + 𝑗24,000 Ω
= 30,000 53.13o Ω
𝑍𝑜
75 0o 20,000 − 53.13o
o
V𝑜 = V𝑔
=
=
50
−
106.26
V
𝑍𝑇
30,000 53.13o
𝑣𝑜 = 50 cos 20,000𝑡 − 106.26o
V
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July 2013
 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.
 Solution:
1
109
=−𝑗
= − 𝑗100 Ω
𝑗𝑤𝐶
12,500 (800)
𝑗𝑤𝐿 = 𝑗 12,500 0.04 = 𝑗500 Ω
Let 𝑍1 = 50 − 𝑗100 Ω;
𝑍2 = 250 + 𝑗500 Ω
I𝑔 = 125 0o mA
−I𝑔 𝑍2
−125 0o 250 + 𝑗500
I𝑜 =
=
𝑍1 + 𝑍2
300 + 𝑗400
= − 137.5 − 𝑗25 mA = 139.75 − 169.7o mA
𝑖𝑜 = 139.75 cos 12,500𝑡 − 169.7𝑜
mA
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