‫‪April 2015‬‬
‫‪Electric Circuits‬‬
‫‪Week 5‬‬
‫‪Techniques Of Circuit Analysis‬‬
‫)‪Part (2‬‬
‫(الشرح واألفكار الرئيسية)‬
‫نوت السيركيت‬
‫ضمان الجودة (‪ 03‬يوما ً)‬
‫تتكون النوت من عشرة أسابيع‪.‬‬
‫يحتوي كل نوت على شرح وحلول‬
‫(‪ )51‬سؤاالً من هوموركات‬
‫وامتحانات سابقة‪.‬‬
‫تم إعداد هذا النوت كي تكون مرجعا ً‬
‫أساسيا ً لتفوقك الباهر بالمادة‪.‬‬
‫إن لم تحز رضاك أعدها واستعد ثمنها‪.‬‬
‫نوتات األمثلة والتمارين اإلضافية‬
‫لماذا ال تقتني األحدث؟‬
‫يحتوي الموقع (‪ )eng-hs.net‬على‬
‫ملفات إضافية بكل منها حلول لعشرين‬
‫سؤاالً أخرى من هوموركات‬
‫وامتحانات سابقة‪.‬‬
‫بعض الناس تحزن إن لم تجد‬
‫مشاكل تتحدث عنها‪.‬‬
‫النوت يتم تنقيحها وتحديثها بشكل‬
‫أسبوعي‪ ،‬راجع (‪)eng-hs.net‬‬
‫للتأكد من شرائك اإلصدار األحدث‪.‬‬
‫)‪(1‬‬
April 2015
4.10 Thevenin and Norton Equivalents
Thevenin and Norton equivalents are circuit
simplifications techniques that focus on terminal
behavior. We can best describe a Thevenin
equivalent circuit by reference to Fig. 4.44, which
represents any circuit made up of sources
(both independent and dependent) and resistors.
The letters a and b denote the pair of terminals of interest.
Figure 4.44(b) shows the Thevenin equivalent. Thus, a Thevenin equivalent circuit
is an independent voltage source (VTh ) in series with a resistor (RTh,) which replaces
an interconnection of sources and resistors. This series combination of (VTh) and
(RTh) is equivalent to the original circuit in the sense that, if we connect the same
load across the terminals (a), (b) of each circuit, we get the same voltage and current
at the terminals of the load.
Finding a Thevenin Equivalent
1) Calculate the open-circuit voltage (
as
in Fig. 4.45 which is equal to (VTh).
.‫ اآلن‬،‫يمكن البدء في تحقيق حلمك بقرار‬
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April 2015
2) Calculate the short-circuit current
as in Fig. 4.46.
,
3) Calculate the Thevenin resistance which is the
ratio of the open-circuit voltage to the shortcircuit current as in Fig. 4.47.
The Norton Equivalent
A Norton equivalent circuit consists of an
independent current source in parallel with the Norton
equivalent resistance. We can derive it from a Thevenin
equivalent
circuit
simply
by
making
a
source
transformation. Thus the Norton current equals the shortcircuit current at the terminals of interest, and the Norton
resistance is identical to the Thevenin resistance.
Using Source Transformations (independent sources):
Sometimes we can make effective use of source
transformations to derive a Thevenin or Norton equivalent
circuit.
For example, we can derive the Thevenin and Norton
equivalents of the circuit shown in Fig 4.45 by making the
series of source transformations shown in Fig. 4.48.
This technique is most useful when the network
contains only independent sources.
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April 2015
4. More on Deriving a the venin equivalent
To calculate (RTh) for such a network, we first
deactivate all independent sources and then calculate the
resistance seen looking into the network at the terminal
pair. A voltage source is deactivated by replacing it
with a short circuit. A current source is deactivated by
replacing it with a short circuit. For example, consider
the circuit shown in Fig. 4.52. Deactivating the
independent sources simplifies the circuit to the one
shown in Fig. 4.53. The resistance seen looking into the
terminals (a), (b) is denoted (Rab), which consists of the
(4 Ω) resistor in series with the parallel combinations of
the (5 Ω) and (20 Ω) resistors. Thus,
If the circuit or network contains dependent and independent sources, an
alternative procedure for finding the Thevenin resistance (RTh) is as follows. We
first deactivate all independent sources, and we then apply either a test voltage source
or a test current source to the Thevenin terminals (a), (b). The Thevenin resistance
equals the ratio of the voltage across the test sources to the current delivered by the
test source.
(Thevenin Method) ‫ملخص بطرق الحل بطريقة‬
Method
Basic Method
Source Transformation
Deactivation
Test (alternative)
Indep. Only




Dep. Only

----
(4)
Indep + Dep.
------
‫ليس هناك خطوة واحدة عمالقة تحقق‬
.‫ إنما مجموعة خطوات صغيرة‬،‫اإلنجاز‬
April 2015
Solution:
The (10 mA) current source and the (10 kΩ)
resistor will have no effect on the behavior of
the circuit with respect to the terminals (a),
(b). This is because they are in parallel with
an ideal voltage source.
Which can be transformed to:
Which can be simplified to Norton equivalent:
.‫ هذا هو شعار كل فاشل‬،‫أنا حزين وأنتم أوغاد‬
(5)
April 2015
Solution:
a)
Open circuit:
Short circuit:
b)
‫عمل بدون تخطيط كإطالق‬
.‫الرصاص بدون تصويب‬
(6)
April 2015
Solution:
After making a source transformation the circuit becomes
Solving,
and
Using deactivation method to get (RTh):
Ω
The final Thevenin circuit is:
‫يعكف أغلب الناس على تغيير‬
.‫النتائج وليس تغيير األسباب‬
(7)
April 2015
Solution:
Open circuit:
Short circuit:
Applying mesh method to get (isc):
Solving,
.ً ‫دائما ً ما يسمع الجبان أصواتا‬
(8)
April 2015
Solution:
a)
Use source transformations to simplify the left side of the circuit, as follows:
1) Transfer (16 V in series with 4 kΩ) to (4 mA in parallel with 4 kΩ).
3) Transfer (4 mA in parallel with 2.4 kΩ) to (9.6 V in series with 2.4 kΩ)
4) Add (9.6 V to 0.4 V) and (2.4 kΩ to 0.1 kΩ) to get the following shape:
b)
Applying (KVL) for the left loop:
‫من ال يخاف الموت يموت‬
.‫مرة واحدة فقط‬
(9)
April 2015
Solution:
(VTh = 0), since there are no independent sources:
1A
Solving,
The equivalent Thevenin circuit is:
Note: (
) is zero since there are no independent sources.
‫ فقد‬،‫احترس عند اختيارك‬
.‫تحصل على ما تختار‬
(11)
April 2015
Solution:
a) Open Circuit Voltage:
1) Convert (3 mA in parallel with 4 kΩ)
to (12 V in series with 4 kΩ)
2) 4 kΩ + 8 kΩ = 12 kΩ
Applying node voltage method:
Solving:
‫إنك تستحق أفضل بكثير من تلك‬
.‫الحياة التي تعيشها اآلن‬
(11)
April 2015
4.12 Maximum Power Transfer
Maximum power transfer can best be described
with the aid of the circuit shown in Fig. 4.58. We
assume a resistive network containing independent and
dependent sources and a designated pair of terminals,
(a), (b) to which a load, (RL) is to be connected.
The problem is to determine the value of (RL) that
permits maximum power delivery to (RL).
The first step in this process is to recognize that a resistive
network can always be replaced by its Thevenin equivalent.
Therefore, we redraw the circuit shown in Fig. 4.58 as the one shown in Fig. 4.59. Fig. 4.59
Replacing the original network by its Thevenin equivalent greatly simplifies the task of
finding (RL).
Derivation of (RL) requires expressing the power dissipated in (RL) as a function of the
three circuit parameters (VTh), (RTh), and (RL).
Next, we recognize that for a given circuit, (VTh) and (RTh) will be fixed.
Therefore the power dissipated is a function of the single variable RL.
To find the value of (RL) that maximizes the power, we use the elementary calculus.
We begin by writing an equation for the derivative of p with respect to (RL):
The derivative is zero and (p) is maximized when
.
،)ً‫األفضل أن تكون (أصال‬
.‫وليس (نسخة) من غيرك‬
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April 2015
 Solution:
Find the Th´evenin equivalent with respect to the terminals of (
).
Open circuit voltage:
Therefore, the Th´evenin equivalent circuit
configured for maximum power to the load is:
With (
) equal to (3.08 ) the original circuit
becomes:
Solving,
Short circuit current:
‫العقبات هي تلك األشياء المخيفة التي‬
.‫تراها عندما ترفع نظرك عن هدفك‬
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April 2015
 Continued Solution (problem 4.85):
Solving,
Solving,
Therefore, only the (440 V) source supplies power to the circuit, and the power
supplied is (
W).
(14)
‫ فكر ماذا لو‬،‫إذا أردت تقييم ذاتك‬
‫عرف عنك اآلخرون ما تخفيه‬
.‫عنهم‬
April 2015
4.13 Superposition
Whenever we have more than one independent source, we can study the effect
of one-by-one source and add these effects together to result in the same result of the
overall system.
We begin by finding the branch currents
resulting from the (120 V) voltage source.
Figure 4.63 The circuit shown in Fig. 4.62
with the current source deactivated.
To find the component of the branch currents
resulting from the current source, we deactivate
the ideal voltage source and solve the circuit
shown in Fig. 4.64.
Figure 4.64
The currents (
) and ( ) in Fig. 4.62 are:
(15)
‫سواء كنت تظن أنك قادر على اجتياز‬
.‫ فأنت على حق‬،‫الصعاب أم ال‬
April 2015
Solution:
a)
(110 V) source acting alone:
(4 A) source acting alone:
Our circuit reduces to:
.‫ تأكد أنك في الطريق الغير صحيح‬،‫في اليوم الذي ال تواجه فيه أية مشاكل‬
(16)
April 2015
Solution:
(240 V) source acting alone:
(84 V) source acting alone:
16 A) current source acting alone:
Solving,
Therefore,
،‫ َّو ِطن نفسك عليه‬،‫قَدِّر أسوأ احتمال‬
.‫يهن عليك ما هو أدنى من ذلك‬
(17)
April 2015
Solution:
Voltage source acting alone:
Using node voltage method:
Current source acting alone:
Using node voltage method:
.‫ فموصول بها فرج قريب‬،‫كل المصائب إذا عظمت‬
(18)
‫‪April 2015‬‬
‫‪ Solution:‬‬
‫‪The mesh equations are:‬‬
‫ال تستتتت يع أن تمنتتتع يتتتور الهتتتم أن ت لتتت فتتتو‬
‫رأسك‪ ،‬ولكنك تست يع أن تمنعها أن تعشش فيها‪.‬‬
‫)‪(19‬‬
April 2015
 Continued Solution (problem 4.99):
Place these equations in standard form:
Solving,
Find the requested voltages:
‫ال تسوء األمور أبدا لدرجة‬
.‫أنها ال يمكن أن تكون أسوأ‬
(21)