‫‪October 2010‬‬
‫‪Chapter 3‬‬
‫‪Simple Resistive‬‬
‫‪Circuits‬‬
‫مذكرات شرح وتمارين محلولة‬
‫امتحانات سابقة للعديد من المواد أدناه‬
‫متاحة مجانا على الموقعين المذكورين أدناه‬
‫الخوف هو أحد األسلحة التي‬
‫سلطها هللا للمسلم على عدوه‬
‫فال تستخدمه ضد نفسك‪.‬‬
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October 2010
3.1 Resistors in Series:
 Series-connected circuit elements carry the same current.
𝑖𝑠 = 𝑖1 = −𝑖2 = 𝑖3 = 𝑖4 = −𝑖5 = −𝑖6 = 𝑖7
Thus we can redraw Fig. 3.1 as shown in Fig. 3.2.
 To find 𝑖𝑠 , we apply Kirchhoff’s voltage law around the loop:
−𝑣𝑠 + 𝑖𝑠 𝑅1 + 𝑖𝑠 𝑅2 + 𝑖𝑠 𝑅3 + 𝑖𝑠 𝑅4 + 𝑖𝑠 𝑅5 + 𝑖𝑠 𝑅6 + 𝑖𝑠 𝑅7 = 0
𝑣𝑠 = 𝑖𝑠 (𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 + 𝑅5 + 𝑅6 + 𝑅7 )
 The seven resistors can be replaced by a single resistor.
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 + 𝑅5 + 𝑅6 + 𝑅7
and 𝑣𝑠 = 𝑖𝑠 𝑅𝑒𝑞
Thus we can redraw Fig. 3.2 as shown in Fig. 3.3.
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October 2010
3.2 Resistors in Parallel:
 The circuit shown in Fig. 3.5 illustrates resistors connected in parallel. Don't
make the mistake of assuming that two elements are parallel connected
merely because they are lined up in parallel in a circuit diagram. The defining
characteristic of parallel-connected elements is that they have the same
voltage across their terminals. In Fig. 3.6, you can see that R1 and R3, are not
parallel connected because, between their respective terminals, another
resistor dissipates some of the voltage.
Resistors in parallel can be
reduced to a single equivalent resistor
using Kirchhoff's current law and
Ohm's law. In the circuit shown in Fig.
3.5, we let the currents i1, i2, i3, and i4
be the currents in the resistors R1
through
R4,
respectively.
From
Kirchhoff's current law,
𝑖𝑠 = 𝑖1 + 𝑖2 + 𝑖3 + 𝑖4
(3.7)
The voltage across each resistor must be the same. Hence, from Ohm's law,
𝑖1 𝑅1 = 𝑖2 𝑅2 = 𝑖3 𝑅3 = 𝑖4 𝑅4 = 𝑣𝑠
(3.8)
Therefore,
𝑣
𝑖1 = 𝑅𝑠
1
𝑣
𝑖2 = 𝑅𝑠
2
𝑣
𝑖3 = 𝑅𝑠
3
𝑣
𝑖4 = 𝑅𝑠
4
(3.9)
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October 2010
Substituting Eq. 3.9 into Eq. 3.7 yields
𝑖𝑠 = 𝑣𝑠 (
1
𝑅1
+
1
𝑅2
+
1
𝑅3
+
1
𝑅4
)
𝑖𝑠
1
1
1
1
1
=
=( +
+
+ )
𝑣𝑠 𝑅𝑒𝑞
𝑅1 𝑅2 𝑅3 𝑅4
𝑘
1
1
1
1
1
=∑ =
+
+⋯+
𝑅𝑒𝑞
𝑅𝑖 𝑅1 𝑅2
𝑅4
𝑖=1
 Note that the resistance of the equivalent resistor is always smaller than the
resistance of the smallest resistor in the parallel connection.
Sometimes, using conductance
𝑘
𝐺𝑒𝑞 = ∑ 𝐺𝑖 = 𝐺1 + 𝐺2 + ⋯ + 𝐺𝑘
(3.13)
𝑖=1
 Many times only two resistors are connected in parallel. Figure 3.8 illustrates
this special case.
1
𝑅𝑒𝑞
1
1
=𝑅 +𝑅 =
1
2
𝑅2 +𝑅1
𝑅1 𝑅2
(3.14)
Or
𝑅 𝑅
𝑅𝑒𝑞 = 𝑅 1+𝑅2
1
2
(3.15)
‫من وجد هللا ماذا فقد؟ ومن فقد هللا ماذا وجد؟؟‬
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October 2010
 Example 3.1: Applying Series-Parallel Simplification
Find is , i1 , and i2 in the circuit shown shown in Fig. 3.9.
 Solution:
 The 3 Ω resistor is in series with the 6
Ω resistor. We therefore replace this
series combination as in Fig. (a).
 We
now
replace
the
parallel
combination of the 9 Ω and 18 Ω
with
a
single
resistance
of
(18 × 9)⁄(18 + 9) = 6 Ω. Fig. (b).
𝑖𝑠 =
𝑣
120
=
= 12 A
𝑅 (4 + 6)
 We added the voltage v1 to help clarify
the
subsequent
discussion.
Using
Ohm's law we compute the value of 𝑣𝑟 :
𝑣1 = (12)(6) = 72 V
 𝑣1 is the voltage drop from node x to
node y, so in Fig. (a)
𝑖1 =
𝑖2 =
𝑣1 72
=
=4 A
18 18
𝑣1 72
=
=8 A
9
9
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October 2010
 Another Solution for example 3.1:
Applyong KCL:
𝑖𝑠 − 𝑖1 − 𝑖2 = 0
(1)
Applying KVL for the left loop:
− 120 + 4 𝑖𝑠 + 18𝑖1 = 0
(2)
Applying KVL for the outer loop:
− 120 + 4 𝑖𝑠 + 9 𝑖2 = 0
(3)
Solving,
𝑖1 = 4 A,
𝑖2 = 8 A,
𝑖𝑠 = 12 A
،‫ أسرية‬،‫ دينية‬:‫لتكن أهدافك منوعة‬
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October 2010
 Assessment 3.1: Be able to recognize resistors connected in series and in parallel
For the circuit shown, find (a) the
voltage 𝑣, (b) the power delivered to the
circuit by the current source, and (c) the
power dissipated in the 10 Ω resistor.
 Solution:
a)
‖ ‫) بينما يتم التعبير عن مقاومتين على التوازى على شكل‬+( ‫يتم التعبير عن مقاومتين على التوالى على شكل‬
𝑅𝑒𝑞 = (((6 + 10) ‖ 64) + 7.2) ‖ 30
= ((16 ‖ 64) + 7.2) ‖ 30
=(
16 × 64
+ 7.2) ‖ 30
16 + 64
= (12.8 + 7.2) ‖ 30
= 20 ‖ 30 =
20 × 30
= 12 Ω
20 + 30
𝑣 = 𝑖𝑅 = 5 × 12 = 60 V
b) 𝑃 = 𝑣𝑖 = 60 × 5 = 300 W
c) Applying KVL for the left loop:
−60 + 𝑖𝐴 × 30 = 0 → 𝑖𝐴 = 2 𝐴
Applying KCL for node a
−5 + 𝑖𝐴 + 𝑖𝐵 = 0
𝑖𝐵 = 3 A
a
b
Applying KVL for the outer loop:
−60 + 7.2 𝑖𝐵 + (6 + 10) × 𝑖𝐶 = 0
𝑖𝐶 = 2.4 A
𝑃10Ω = 𝑖 2 𝑅 = (2.4)2 × 10 = 57.6 W
‫إن لم يكن لك هدف فإلى أي‬
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October 2010
3.3 The Voltage-Divider and Current-Divider Circuits:
Applying Kirchhoff's voltage law around the closed loop yields:
𝑣𝑠 = 𝑖𝑅1 + 𝑖𝑅2
or
𝑖=
𝑣𝑠
𝑅1 + 𝑅2
Now we can use Ohm's law to calculate 𝑣1 and 𝑣2 :
𝑣1 = 𝑖𝑅1 = 𝑣𝑠
𝑅1
𝑅1 + 𝑅2
𝑣2 = 𝑖𝑅2 = 𝑣𝑠
𝑅2
𝑅1 + 𝑅2
The Current-Divider Circuits:
 The current-divider circuit shown in Fig.
3.15 consists of two resistors connected in
parallel across a current source.
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‫تعلم قول (ال) عندما‬
.‫ينبغي أن تقول ذلك‬
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October 2010
 The current divider is designed to divide the current 𝑖𝑠 between 𝑅1 and 𝑅2 . We
find the relationship between the current 𝑖𝑠 and the current in each resistor by
directly applying Ohm's law and Kirchhoff’s current law.
The voltage across the parallel resistors is
𝑣 = 𝑖1 𝑅1 = 𝑖2 𝑅2 = 𝑅𝑒𝑞 𝑖𝑠 =
𝑖1 =
𝑅2
𝑖
𝑅1 + 𝑅2 𝑠
𝑖2 =
𝑅1
𝑖
𝑅1 + 𝑅2 𝑠
𝑅1 𝑅2
𝑖
𝑅1 + 𝑅2 𝑠
‫إذا رأيت فأر يسخر من هر‬
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October 2010
 Example 3.2: Analyzing the Voltage-Divider Circuit
The resistors used in the voltage-divider circuit shown in Fig. 3.14 have a
tolerance of •± 10%. Find the maximum and minimum value of 𝑣𝑜 .
 Solution:
𝑣𝑜 = 𝑣𝑠
𝑅2
𝑅1 + 𝑅2
(R1) ‫( وتصغير المقام قدر اإلمكان‬R2) ‫للحصول على أكبر قيمة 𝑜𝑣 يجب تكبير البسط قدر اإلمكان‬
𝑣𝑜 (max) = 100 ×
110
= 82.02 V
(110 + 22.5)
‫( وتكبير المقام قدر اإلمكان‬R2) ‫للحصول على أصغر قيمة للـ 𝑜𝑣 يجب تصغير البسط قدر اإلمكان‬
(R1)
𝑣𝑜 (min) = 100 ×
90
= 76.60 V
(90 + 27.5)
‫ ركز‬،‫اعرف أولوياتك‬
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October 2010
 Example 3.3: Analyzing a Current-Divider Circuit
Find the power dissipated in the 6 Ω resistor shown in Fig. 3.16.
 Solution:
First, we must find the current in the
resistor by simplifying the circuit with
series-parallel reductions. Thus, the
circuit shown in Fig. 3.16 reduces to
the one shown in Fig. 3.17. We find the
current i0 by using the formula for
current division:
𝑖0 =
16
(10) = 8 A
16 + 4
Note that i0 is the current in the 1.6 Ω
resistor in Fig. 3.16. We now can
further divide i0 between the 6 Ω and 4
Ω resistors. The current in the 6 Ω
resistor is:
𝑖6 =
4
(8) = 3.2 A
6+4
and the power dissipated in the 6 Ω
resistor is:
𝑝 = (3.2)2 (6) = 61.44 W
‫حدد مفهوم السعادة بالنسبة‬
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October 2010
 Assessment problem 3.2: Know how to design simple voltage-divider and current-divider circuits
a) Find the no-load value of 𝑣𝑜 in the circuit
shown.
b) Find 𝑣𝑜 when 𝑅𝐿 is 150 kΩ is.
c) How much power is dissipated in the 25 kΩ
resistor if the load terminals are accidentally
short-circuited?
d) What is the maximum power dissipated in
the 75 kΩ resistor?
Solution:
a) 𝑣𝑜 = 𝑣2 = 𝑣𝑠
= 200 ×
b) 𝑅𝑒𝑞 =
𝑅2
𝑅1 +𝑅2
75
= 150 V
25 + 75
75×150
75+150
𝑣2 = 𝑣𝑠
= 50 kΩ
50
= 133.33 V
(25 + 50)
c) Load terminals are short circuited means only 25 kΩ is dissipating power
𝑖=
200
= 8 mA
25 × 103
𝑃 = 𝑖 2 𝑅 = (8 × 10−3 )2 × (25 × 103 ) = 1.6 W
d) To get 𝑃𝑚𝑎𝑥𝑖𝑚𝑢𝑛 in the 75 kΩ resistor, 𝑅𝑒𝑞 should be maximum, 𝑅𝐿 = 0
𝑖=
200
= 2 mA
(25 + 75) × 103
𝑃75 = 𝑖 2 𝑅 = (2 × 10−3 )2 × (75 × 103 ) = 0.3 W
‫توقع السعادة أو التعاسة فستحصل حتما‬
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October 2010
 Assessment 3.3: Know how to design simple voltage-divider and current-divider circuits
a) Find the value of R that will cause 4 A of
current to flow through the 80 Ω resistor
in the circuit shown.
b) How much power will the resistor R from
part (a) need to dissipate?
c) How much power will the current source
generate for the value of R from part (a)?
 Solution:
a) 𝑖1 =
4=
𝑅2
𝑅1 +𝑅2
𝑖
𝑅
× 20
𝑅 + 120
4𝑅 + 480 = 20𝑅
𝑅 = 30 Ω
b) When 𝑖1 in the 80 Ω resistor is 4 A, 𝑖2 in the R resistor is 16 A
𝑃𝑅 = 𝑖 2 𝑅 = (16)2 × 30 = 7,680 W
c) 𝑅𝑒𝑞 = 60 + (40 + 80) ‖ 𝑅
𝑅𝑒𝑞 = 60 + 120 ‖ 30
= 60 +
120 × 30
= 60 + 24 = 84 Ω
120 + 30
𝑃20𝐴 = 𝑖 2 𝑅 = (20)2 × 84 = 33,600 W
‫اعلم أن الفشل كلمة قبيحة سيئة ابتدعها األشخاص‬
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October 2010
3.4 Voltage Division and Current Division:
We can generalize the results from analyzing the
voltage-divider and current-divider circuit. The
generalizations will yield two additional and
very useful circuit analysis techniques known as
voltage division and current division. Consider
the circuit shown in Fig. 3.18.
The box on the left contains a single voltage to the right of the box are n
resistors connected in series. We are interested in finding the voltage drop 𝑣𝑗 across
an arbitrary resistor 𝑅𝑗 in terms of the voltage 𝑣. We start by using Ohm's law to
calculate 𝑖, the current through all of the resistors in series, in terms of the current 𝑣
and the n resistors:
𝑣
𝑣
𝑖=
=
𝑅1 + 𝑅2 + ⋯ + 𝑅𝑛 𝑅𝑒𝑞
We apply Ohm's law a second time to calculate the voltage drop 𝑣𝑗 across the
resistor 𝑅𝑗 , using the current 𝑖 calculated:
Voltage-division equation 
𝑣𝑗 = 𝑖𝑅𝑗 =
𝑅𝑗
𝑅𝑒𝑞
𝑣
(3.30)
Now consider the circuit shown in
Fig. 3.19. The box on the left contains a
single current source to the right of the
box are n resistors connected in parallel.
We are interested in finding the current
𝑖𝑗 through an arbitrary resistor 𝑅𝑗 in
terms of the current 𝑖. We start by using Ohm's law to calculate 𝑣, the voltage drop
across each of the resistors in parallel, in terms of the current 𝑖 and the n resistors:
𝑣 = 𝑖(𝑅1 ‖𝑅2 ‖ … ‖𝑅𝑛 ) = 𝑖𝑅𝑒𝑞
Current-division equation 
𝑣
𝑖𝑖 = 𝑅 =
𝑗
𝑅𝑒𝑞
𝑅𝑗
𝑖
Note that the constant of proportionality in the current division equation is the
inverse of the constant of proportionality in the voltage division equation!
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 Example 3.4: Using Voltage Division and Current Division to Solve a Circuit
Use current division to find the current 𝑖𝑜 and use voltage division to find the voltage 𝑣𝑜
for the circuit in Fig. 3.20.
 Solution:
𝑅𝑒𝑞 = (36 + 44) ‖ 10 ‖ (40 + 10 + 30) ‖ 24
= 80 ‖ 10 ‖ 80 ‖ 24
1
1
1
1
1
1
=
+
+
+
=
𝑅𝑒𝑞 80 10 80 24 6
𝑅𝑒𝑞 = 6 𝛺
𝑖𝑗 =
𝑅𝑒𝑞
𝑖
𝑅𝑗
𝑖𝑜 =
6
×8=2A
24
We can use Ohm's law to find the voltage drop across the 24 Ω resistor:
𝑣 = (24)(2) = 48 V
This is also the voltage drop across the branch containing the 40 Ω, 10 Ω,
and 30 Ω resistors in series. We can then use voltage division to determine
the voltage drop vo across the 30 Ω resistor:
𝑣𝑜 =
30
(48 V) = 18 V
80
‫اعتبر نفسك شخصا تريد الحيااة أن يشاق‬
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October 2010
 Assessment problem 3.4:
a) Use voltage division to determine the
voltage 𝑣𝑜 across the 40 Ω resistor in the
circuit shown.
b) Use 𝑣𝑜 from part (a) to determine the
current through the 40 Ω resistor, and
use this current and current division to
calculate the current in the 30 Ω resistor.
c) How much power is absorbed by the 50
Ω resistor?
 Solution:
a) We can get equivalent resistance of 20 Ω, 30 Ω, 50 Ω and 10 Ω as follows:
𝑅𝑒𝑞 ′ = 20 ‖ 30 ‖ (50 + 10) = 20 ‖ 30 ‖ 60
1
1
1
1
=
+
+
→ 𝑅𝑒𝑞 ′ = 10 Ω
′
20 30 60
𝑅𝑒𝑞
𝑅𝑒𝑞 = 40 + 10 + 70 = 120 Ω
𝑣𝑗 =
𝑅𝑗
𝑣→
𝑅𝑒𝑞
𝑣𝑜 =
40
𝑣 = 20 V
120
b) 𝑣𝑜 = 40 𝑖𝑜
𝑖𝑜 = 0.5 A
We can get equivalent resistance of 20 Ω, 50 Ω and 10 Ω
20 × 60
= 15 Ω
20 + 60
15
=
× 0.5 = 0.167 A
15 + 30
𝑅𝑒𝑞 ″ = 20‖(50 + 10) =
𝑖2 =
𝑅1
𝑖 → 𝑖30𝛺
𝑅2 + 𝑅2
c) To get current in 50Ω and 10Ω we can get equivalent resistance of 20Ω and 30Ω
𝑅𝑒𝑞 ‴ = 20 ‖ 30 =
20 × 30
= 12 Ω
20 + 30
12
× 0.5 = 0.0833 A
12 + 60
= 𝑖 2 𝑅 = 0.347 W
𝑖50𝛺 =
𝑃50𝛺
‫ هي‬،‫أن يكون لديك أهداف قابلة للتحقيق‬
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October 2010
 Question 3.24:
a) Find the voltage 𝑣𝑥 in the circuit in Fig. P3.24.
b) Replace the 30 V source with a general voltage source equal to 𝑉𝑠 . Assume 𝑉𝑠 is
positive at the upper terminal. Find 𝑣𝑥 as a function of 𝑉𝑠 .
 Solution:
a) Using voltage divider concept knowing that the volt different in either
loop is 30V:
𝑣1 = 𝑣𝑠
𝑅1
𝑅1 + 𝑅2
5
= 25 V
5+1
60
= 30 ×
= 24 V
60 + 15
𝑣1kΩ = 30 ×
𝑣15kΩ
𝑣𝑥 = 25 − 24
=1 V
b) Repeating above calculation, but with 30 V replaced by 𝑣𝑠 :
5
5𝑣𝑠
=
5+1
6
60
4𝑣𝑠
𝑣15𝑘𝛺 = 𝑣𝑠 ×
=
60 + 15
5
5𝑣𝑠 4𝑣𝑠 𝑣𝑠
𝑣𝑥 =
−
=
6
5
30
𝑣1𝑘𝛺 = 𝑣𝑠 ×
(same as a)
‫كونك ممتنا وسعيدا بما‬
‫تحققه طبقا إلمكاناتك يعد‬
.‫وسيلة جيدة لبناء الثقة‬
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October 2010
 Question 3.25:
Find 𝑣1 and 𝑣2 in the circuit in Fig. P3.25.
 Solution:
 Getting equivalent resistance for 30 Ω and 60 Ω:
𝑅𝑒𝑞 =
30 × 60
= 20 Ω
30 + 60
 Using current divider, we can get current in 30 Ω :
𝑖30𝛺 =
(50 + 25)
× 25 = 15 A
(30 + 20) + (50 + 25)
𝑣2 = 𝑖𝑅
= 15 × 20 = 300 V
 Applying KVL for the left loop :
𝑣1 − 12 × 25 − 15 × (30 + 20) = 0
𝑣1 = 1,050 V
،‫إن الذي ينتصر على غيره قوي‬
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October 2010
 Question 3.26:
Find 𝑣𝑜 in the circuit in Fig. P3.26.
 Solution:
No current in 2 kΩ and 4 kΩ , so we can neglect them decreasing the
resistances to (15 + 16) kΩ parallel to (12 + 13) kΩ
Using current divider, we can get currents in 15 kΩ and 12 kΩ :
𝑖15𝑘𝛺 =
(12 + 3)
× 18 = 6.75 mA
(15 + 10) + (12 + 3)
𝑖12𝑘𝛺 =
(15 + 10)
× 18 = 11.25 mA
(15 + 10) + (12 + 3)
Applying KVL for the lower loop:
(15 × 103 ) × (6.75 × 10−3 ) + 𝑣𝑜 − (12 × 103 ) × 11.25 × 10−3 = 0
𝑣𝑜 = 33.75 V
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October 2010
 Question 3.27:
Find 𝑖𝑜 and 𝑖𝑔 in the circuit in Fig. P3.27.
 Solution:
 Getting 𝑅𝑒𝑞 for 27 Ω and 54 Ω which are in parallel:
27 × 54
𝑅𝑒𝑞 =
= 18 Ω
27 + 54
18 Ω
 Getting 𝑅𝑒𝑞 ′ for (2+18) Ω with (10+15+35) Ω which are in parallel :
20 × 60
𝑅𝑒𝑞 ′ =
= 15 Ω
20 + 60
 Applying ohm’s law for the circuit :
𝑣
675
𝑖𝑔 = =
= 15 A
𝑅 30 + 15
15 Ω
 Using current divider :
60
𝑖2𝛺 =
× 15 = 11.25 A
20 + 60
 Using current divider :
54
𝑖𝑜 =
× 11.25 = 7.5 A
27 + 54
.‫ وال تكتفي بالكالم واألماني‬،‫قم بخطوات صغيرة في كل عمل‬
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October 2010
 Question 3.28:
For the circuit in Fig. P3.28, calculate (a) 𝑖𝑜 and (b) the power dissipated in
the 15 Ω resistor.
 Solution:
 Getting equivalent resistance for 40 Ω and 10 Ω which are in parallel:
𝑅40 ‖ 10 =
40 × 10
=8Ω
40 + 10
 Getting equivalent resistance for 15 Ω and (2 Ω+8 Ω) which are in parallel :
𝑅15 ‖ 10 =
15 × 10
=6Ω
15 + 10
 Getting equivalent resistance for 30 Ω with (4 Ω+6 Ω) which are in parallel :
30 × 10
= 7.5 Ω
30 + 10
𝑣 120
= =
= 16 A
𝑅 7.5
𝑅30 ‖ 10 =
𝑖120 V
 Using current divider:
30
× 16 = 12 A
30 + 10
15
=
× 12 = 7.2 A
15 + 10
10
=
× 7.2 = 1.44 A
40 + 10
𝑖4𝛺 =
𝑖2𝛺
𝑖40𝛺
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October 2010
3.7 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits:
The resistors 𝑅𝑎 , 𝑅𝑏 , and 𝑅𝑐 in the
circuit shown in Fig. 3.28 are referred to
as a delta (∆) interconnection because
the interconnection looks like the Greek
letter ∆. It also is referred to as a pi
interconnection because the ∆ can be shaped into a π without disturbing the
electrical equivalence of the two configurations.
The resistors 𝑅1 , 𝑅2 , and 𝑅3 in the
circuit shown in Fig. 3.30 are referred to
as a wye (Y) interconnection because the
interconnection can be shaped to look like
the letter Y. The Y configuration also is
referred to as a tee (T) interconnection because the Y structure can be shaped into
a T structure without disturbing the electrical equivalence of the two structures.
Figure 3.31 illustrates the ∆-to-Y (or
π -to-T) equivalent circuit transformation.
Note that we cannot transform the ∆
interconnection into the Y interconnection
simply by changing the shape of the
interconnections. Saying the ∆-connected circuit is equivalent to the Y-connected
circuit means that the ∆ configuration can be replaced with a Y configuration to
make the terminal behavior of the two configurations identical.
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October 2010
The Y-connected resistors in terms of the ∆-connected resistors:
𝑅1 =
𝑅𝑏 𝑅𝑐
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐,
𝑅2 =
𝑅𝑐 𝑅𝑎
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐,
𝑅3 =
𝑅𝑎 𝑅𝑏
𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐,
The three ∆-connected resistors as functions of the three Y connected resistors are:
𝑅𝑎 =
𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1
𝑅1
𝑅𝑏 =
𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1
𝑅2
𝑅𝑐 =
𝑅1 𝑅2 + 𝑅2 𝑅3 + 𝑅3 𝑅1
𝑅3
.‫اإلرادة نصف الطريق‬
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October 2010
 Example 3.7: Applying a Delta-to-Wye Transform
Find the current and power supplied
by the 40 V source in the circuit
shown in Fig. 3.32.
 Solution:
 We find equivalent resistance after replacing
either the upper ∆ (100, 125, 25 Ω) or the lower
∆ (40, 25, 37.5 Ω) with its equivalent Y. We
choose to replace the upper ∆.
100 × 125
= 50 Ω
250
125 × 25
𝑅2 =
= 12.5 Ω
250
100 × 25
𝑅3 =
= 10 Ω
250
𝑅1 =
 we can calculate the resistance across the terminals of the 40 V source by
series-parallel simplifications:
(50)(50)
𝑅𝑒𝑞 = 55 +
= 80 Ω
100
 The final step is to note that the circuit reduces to
an 80 Ω resistor across a 40 V source, as shown in
Fig. 3.35.
𝑣 40
𝑖= =
= 0.5 A
𝑅 80
𝑝 = 𝑣𝑖 = 40 × 0.5 = 20 W
.‫الخطوة األولى يعول عليها كثيرا‬
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October 2010
 Assessment problem 3.8: Know when and how to use delta-to-wye equivalent circuits
Use a Y-to-∆ transformation to find
the voltage 𝑣 in the circuit shown.
 Solution:
 Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three ∆-connected
resistors 𝑅𝑎 , 𝑅𝑏 and 𝑅𝑐 .
(5)(10) + (5)(20) + (10)(20)
= 17.5 Ω
20
(5)(10) + (5)(20) + (10)(20)
𝑅𝑏 =
= 35 Ω
10
(5)(10) + (5)(20) + (10)(20)
𝑅𝑐 =
= 70 Ω
5
 We can get the 𝑅𝑒𝑞 for the new shape :
𝑅𝑎 =
𝑅𝑒𝑞 = ((28 ‖ 70) + (17.5 ‖ 105))‖ 35
28 × 70 17.5 × 105
=(
+
) ‖ 35
28 + 70 17.5 + 105
= (20 + 15) ‖ 35
35 × 35
=
= 17.5 Ω
35 + 35
𝑣 = 𝑖𝑅 = 2 × 17.5 = 35 V
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October 2010
 Question 3.50:
Find the power dissipated in the 18 Ω
resistor in the circuit in Fig. P3.50.
 Solution:
 We can change the above ∆ to Υ as shown :
30 × 10
= 3.33 Ω
10 + 30 + 50
10 × 50
𝑅2 =
= 5.56 Ω
10 + 30 + 50
30 × 50
𝑅3 =
= 16.67 Ω
10 + 30 + 50
𝑅1 =
R1
R2
R3
6Ω
18 Ω
 Now we can get 𝑅𝑒𝑞 for all resistances:
𝑅𝑒𝑞 = (3 + 3.33) + (𝑅2 + 6 ‖ 𝑅3 + 18)
= 6.33 + (11.56 ‖ 34.67)
= 6.33 +
11.56 × 34.67
11.56 + 34.67
= 6.33 + 8.67 = 15 Ω
𝑣 300
=
= 20 A
𝑅
15
11.56
=
× 20 = 5
(11.56 + 34.67)
𝑖𝑜 =
𝑖18Ω
𝑃18Ω = 𝑖 2 𝑅 = (5)2 × 18
A
= 450 W
‫اقرأ قصص ومواقف‬
.‫عن اإلبداع والمبدعين‬
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October 2010
 Question 3.54:
Find the equivalent resistance 𝑅𝑎𝑏 in
the circuit in Fig. P3.54.
 Solution:
Changing upper and lower ∆-to-Y equivalent circuits:
(25)(10)
𝑅1U =
=5 Ω
50
(10)(15)
𝑅2U =
=3 Ω
50
(25)(15)
𝑅3U =
= 7.5 Ω
50
(125)(25)
𝑅1L =
= 12.5 Ω
250
(25)(100)
𝑅2L =
= 10 Ω
250
(125)(100)
𝑅3L =
= 50 Ω
250
R2U
R1U
R3U
R1L
R3L
R2L
The resulting circuit is shown below:
We can get the equivalent resistance for the whole circuit:
𝑅𝑒𝑞 = 15 + 5 + ((3 + 17 + 30 + 10) ‖ (7.5 + 12.5)) + 50 + 14
= 20 + (60 ‖ 20) + 64
60 × 20
= 20 +
+ 64
60 + 20
= 20 + 15 + 64
= 99 Ω
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October 2010
 Question 3.59:
Use a Y-to-∆ transformation to find (a) 𝑖𝑜 ; (b) 𝑖1 , (c) 𝑖2 ; and (d) the power
delivered by the ideal current source in the circuit in Fig. P3.59.
a
 Solution:
We can Υ-to-∆ transformation to reach the following shape :
20 × 50 + 20 × 100 + 50 × 100
= 80 Ω
100
20 × 50 + 20 × 100 + 50 × 100
𝑅𝑏 =
= 160 Ω
50
20 × 50 + 20 × 100 + 50 × 100
𝑅𝑐 =
= 400 Ω
20
𝑅𝑎 =
We can get the equivalent of each two parallel resistance :
320 × 80
= 64 Ω
320 + 80
240 × 160
=
= 96 Ω
240 + 160
400 × 600
=
= 240 Ω
400 + 600
𝑅320 ‖ 80 =
𝑅240 ‖ 160
𝑅400 ‖ 600
.‫ تفاءل‬،‫في كل صباح تشرق الشمس في كل مكان‬
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October 2010
 Continued Solution(Question 3.59):
𝑖=
96
× 1 = 0.24 A
(96 + 64 + 240)
𝑖𝑜 =
400
× 0.24 = 0.096 A
(600 + 400)
𝑖1 =
80
×𝑖
(80 + 320)
= 0.048 A
Applying KCL at node (a) in original circuit:
−𝑖1 + 𝑖𝑜 + 𝑖50Ω = 0
−0.048 + 0.096 − 𝑖50Ω = 0
𝑖50Ω = 0.048 Ω
Applying KVL for the right loop:
− 100 𝑖2 + 50 𝑖50Ω + 600 𝑖𝑜 = 0
−100 𝑖2 + 50 × 0.048 + 600 × 0.096 = 0
𝑖2 = 0.6 A
We can get the equivalent resistance for the whole circuit:
𝑅𝑒𝑞 =
96 × 304
= 72.96 Ω
(96 + 304)
𝑃 = 𝑖 2 𝑅 = (1)2 × 72.96 = 72.96 W
.‫متى توفرت اإلرادة سهلت الطريقة‬
‫ أو بالبريد االلكتروني‬SMS ‫ديناران هدية عنـد التنبيه على كـل خطـأ بمذكرات الموقع برسالة‬
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C++, Java, MATLAB, Data Structures, Algorithms, Discrete Math, Digital Logic, Concepts :‫مواد كمبيوتر‬
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