July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
Chapter 9
Sinusoidal Steady-
State Analysis
زمشاخ ششح تاس حيىح
اىاد أدا اتحااخ ساتقح ىيؼذذ
ا ػيى أدااىقؼ اىزمس تاحح جاا
: جتغ ػيى اىناسو ػقاتا
.فشي جاح اخش
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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?
إرا اتتس اىض فقذ
.اىتصش ىزج اىفص
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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?
إرا أسدخ أ تتق أساء إىل تطشقح
.تجؼي تضاءه أا فس، اػف ػ
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.
أسشع طشق إلاء اىؼشمح
.أ تستسي ىيضح
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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
أسشع طشق إلاء اىؼشمح
.أ تستسي ىيضح
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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)?
أسش اىحشب سجو حاه قتيل ى
.ستطغ، ث سأىل أال تقتي
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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?
تثذ األس أفضو مثشا إرا ظشا
.إىا اىطشف اخش
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.
تثذ األس أفضو مثشا إرا ظشا
.إىا اىطشف اخش
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.
تن أػيى دسجاخ االتاص ف
.مسش قاح ػذك د قتاه
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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
ستا تضطش ىخض اىؼشمح
.أمثش شج ىنسثا
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.
مو إسا ؼتقذ أ اىزي تيل اىحققح، أ
.اىحق ا ؼتقذ، زا سثة ضاػاخ اىؼاى
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.
ال ن ىؼاذج أ تذ إال
.إرا اتصش اىطشفا
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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?
ى نل االتصاس إرا ما
.تقذسك قثه اىضح
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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?
صف اىصش أ تثقى حتفظا
.تأػصاتل، ال تخثش تزىل ػذك
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.
July 2013
تاىثشذ اإلىنتش أ 9 4444 260 تشساىح صح تشاا ضشسح الحظاخأأي خطأ تاإلتالؽ ػ اىساحاىتاخ جاح ىيفغ اىؼا فشجى
Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy
ا ششح سائو حيىح [email protected] 260 4444 9 حادج شؼثا. eng-hs.com, eng-hs.net تاىقؼجاا
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.