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Sept 20134.10 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 v1as in

Fig. 4.45 which is equal to VTh.v1−25

5+

v1

20−3=0

v1=vTh=32V

فيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها أو 9 4444 260 برسالةاإللكتروني بالبريد

Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economyشعبان. حمادة محلولة info@ eng-hs.com 260 4444 9م ومسائل -eng بالموقعين مجاناشرح

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تكون أن يجببكثير أكبر حياتك

وجود مجرد .من

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Sept 20132) Calculate the short-circuit current isc as in

Fig. 4.46.v2−25

5+

v2

20−3+

v2

4=0

v2=16 V

isc=164

=4 A

3) Calculate the Thevenin resistance which is the

ratio of the open-circuit voltage to the short-

circuit current as in Fig. 4.47.

RTh=V Th

isc=32

4=8 Ω

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 short-circuit current at the terminals of interest, and the Norton resistance

is identical to the Thevenin resistance.

Using Source Transformations (independent sources):




تخافمنه ماداومت إذا سيقع

تفكيرك على



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Sept 2013

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.

Finding the Thevenin Equivalent of a Circuit with a Dependent Sourceفيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها أو 9 4444 260 برسالة

اإللكتروني بالبريد Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy

شعبان. حمادة محلولة info@ eng-hs.com 260 4444 9م ومسائل -eng بالموقعين مجاناشرحhs. com , eng-hs. net

من يكون قدعليك الواجب



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Sept 2013Example 4.10:

Find the Thevenin equivalent for the

circuit containing dependent sources

shown in Fig. 4.49.

Solution:

1. Calculating VTh (Open-Circuit):

Applying KVL at left loop:−5+2000 i+3 v=0

2000 i+3 v=5(1)

Applying KVL at right loop:−v−25∗20i=0

500 i+v=0(2)

Solving, i=0.01 A , v=−5V =V Th .

2. Calculating isc (Short-Circuit):

Applying KVL for the outer right loop (with no resistor, no voltage drop) isc=−20 i

v=0(no voltage drop)¿equation 1 ,i= 52000

=2.5∗10−3 A

isc=−0.05 A

3. Calculating RTh:

RTh=V Th

isc= −5

−0.05=100 Ω

The Deactivation Method (for independent Sources Only)

The technique for determining RTh that

we discussed and illustrated earlier is not




البعضأمورا يتمنىبالفعل، يمتلكونها هم

يدركون ال ولكنهم

+-

b

a

5v

100 Ω



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Sept 2013always the easiest method available. Two other

methods are generally simpler to use.

The first is useful if the network contains only independent sources. 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,

Rab=RTh=4+ 5 ×2025

=8 Ω

4.11 The Test Method (for Independent and Dependent Sources)

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.

Finding the Thevenin Equivalent Using a Test Source

Example 4.11:

Find the Thevenin resistance RTh for the circuit in Fig. 4.49, using the alternative

method described.




الناس من كثيرالخطى تسرع

Figure 4.52 A circuit used to illustrate a Thévenin equivalent



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Sept 2013Solution:

We first deactivate the independent voltage source from the circuit and then excite

the circuit from the terminals a, b with either a test voltage source or a test current

source. If we apply a test voltage source, we will know the voltage of the dependent

voltage source and hence the controlling current i. Therefore we opt for the rest

voltage source.

The externally applied test voltage source is denoted vT , and the current that it

delivers to the circuit is labeled iT. To find the Thevenin resistance, we simply solve

the circuit for the ratio of the voltage to the current at the test source; that is, RTh=vT / iT

iT=vT

25+20 i(node−voltage)i=

−3 vT

2mA (¿)iT=

vT

25–

60 vT

2000=

vT

100RTh=

vT

iT=100 Ω

In general, these computations are easier than those involved in computing the

short-circuit current. Moreover, in a network containing only resistors and

independent sources, you must use the alternative method because the ratio of the

Thevenin voltage to the short-circuit current is indeterminate. That is, it is the

ratio 0/0.

(Thevenin Method) بطريقة الحل ملخصبطرق

MethodIndep. OnlyDep. OnlyIndep + Dep.

Basic Method---

Source Transformation------




ال الباطن عقلكإنه المنطق، يستخدم

. عليه تمليه ما يقبل



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Sept 2013

Deactivation------

Test (alternative)

Assessment Problem 4.16:

Solution:

To find RTh, replace the 72 V source with a short circuit:

RTh=((5 20)+8)∨¿12=6 Ω

Using node voltage analysis to findvTh:




أن الحماقة من إنهبعض بقدرة تؤمنإلحاق على األشياء

Find the Thévenin equivalent circuit with respect to the terminals a, b for the circuit shown.



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Sept 2013The node voltage equations are:

v1−725

+v1

20+

v1−vTh

8=0

vTh−v1

8+

vTh−7212

=0

Solving,v1=60 V and vTh=64.8 V




فال انعدمتشهيتك، إذا. الطعام تلم



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Sept 2013


Find the Norton equivalent circuit with respect

to the terminals a, b for the circuit shown.

Solution:

We perform a source transformation, turning the parallel combination of the 15 A

source and 8 Ω resistor into a series combination of a 120 V source and an 8 Ω

resistor. Next, combine the 2 Ω, 8 Ω and 10 Ω resistors in series to give an

equivalent 20 Ω resistance. Then transform the series combination of the 120 V

source and the 20 Ω equivalent resistance into a parallel combination of a 6 A source

and a 20 Ω resistor.

Finally, combine the 20 V and 12 Ω parallel resistors to give RN = 20||12 = 7.5 Ω.

Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a

7.5 Ω resistor.




الذي الشيء كان أياستحصل بصدق، تطلبه

أنك آمنت إذا عليه،



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Sept 2013


A voltmeter with an internal resistance of 100 kΩ

is used to measure the voltage vAB in the circuit

shown. What is the voltmeter reading?

Solution:

Using source transformations, convert the series

combination of the -36 V source and 12 kΩ

resistor into a parallel combination of a -3 mA

source and 12 kΩ resistor.

Combine the two parallel current sources and the two parallel resistors to give a

−3+18=15 mAsource in parallel with a 12 k ||60 k =

10 kΩ resistor. Transform the 15 mA source in

parallel with the 10 kΩ resistor into a 150 V

source in series with a 10 kΩ resistor, and combine

this

10 kΩ resistor in series with the 15 kΩ resistor.

The Thevenin equivalent is thus a 150 V source in series with a 25 kΩ resistor

Using voltage division:

vAB=100,000125,000

(150 )=120 V




سوف الخير في اعتقدفي اعتقد عليه، تحصل

. منك ينال سوف الشر



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Sept 2013


Find the Thévenin equivalent circuit with respect

to the terminals a, b for the circuit shown.

Solution:

Calculating the open circuit voltage, which is also vTh,vTh

8+4+3 ix+

v th−242

=0

ix=vTh

8

Solving, vTh=8 V

Using the test source method to calculate RTh, replace

the voltage source with a short circuit, the current

source with an open circuit:

Applying KCL equation at the middle node:

iT=ix+3i x+vT

2=4 i x+

vT

2ix=

vT

8

Solving,iT=4 ( vT /8 )+vT /2=vT RTh=vT / iT=1 Ω

The Thevenin equivalent is an 8 V source in series with a 1 Ω resistor.




طموحاتك، تواجه التي العراقيل في فكر. حتما لك، وسوفتحدث



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Sept 2013

Solution:

Using deactivating method, we can calculate RTh

as follows:

RTh=8+(40 )(10)

50 =16 Ω

Using voltage divider:

vTh=40

(10+40 )∗60=48V

The final circuit of Thevenin is:




وكأنه تعامل نجاحك، قرر. بالفعل وقع قد



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Sept 2013

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:




فيطريق العقبةتكمنفي تفوقك

الباطن عقلك



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Sept 2013

Solution:

a)

Open circuit: v2−9

20+

v2

70−1.8=0∴v2=35 V

vTh=6070

∗v2=30V

Short circuit: v2−9

20+

v2

10−1.8=0∴v2=15 V

ia=9−15

20=−0.3 A

isc=1.8−0.3=1.5 A(¿= v2

10 )RTh=

301.5

=20 Ω

b)

RTh=(20+10 )∨¿60=30+6090

=20 Ω




بعضالناس ألنه باأللم يستمتعون



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Sept 2013

Solution:

After making a source transformation the circuit becomes−500+8 (i1−i2 )+12 i1=0

−300+30 i2+5.2i2+8 (i2−i1 )=0

Solving,

i1=30 A and i2=12.5 A

vT h=12 i1+5.2 i2=425 V

Using deactivation method to get RTh:RT h=¿

The final Thevenin circuit is:




ذات تلك هي وسيلة أفضل إن. األقل المجهود

30 Ω

40 Ω 5.2 Ω8 Ω

12Ω



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Sept 2013

Solution:

a)

Using source transformation method, we can get vTh through the following steps:

1. Converting 30 V along with series 10 kΩ

resistor to 3 mA current source parallel to

10 kΩ resistor, then combining 10 kΩ || 40 kΩ

to 8 kΩ.

2. Converting 3 mA current source along with

parallel 8 kΩ resistor to 24 V voltage source in

series along with 8 kΩ resistor.

3. Converting 24 V voltage source in series

along with 12 kΩ resistor to 2 mA current

source in parallel along with 12 kΩ resistor,

and adding current sources together.




أكثر تستحق أنتاآلن، عليه أنت مما

. ذلك في أشك ال



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Sept 2013Continue soluation (Problem 4.71):

4. Combining (10 mA with 12 kΩ) to (120 V

with 12 kΩ ¿, adding (12 kΩ with 3 kΩ) to

15 kΩ, transferring (120 V with 15 kΩ) to

(8 mA with 15 kΩ), combining resulting

(15 kΩ || 10 kΩ) to 6 kΩ, finally transferring

(8 mA and 6 kΩ) to (48 V and 6 kΩ).

vT h=48V

RT h=6 k Ω

vmeans=100106

(48 )=45.28 V

b)

% error=( 45.28−4848 )×100=−5.67 %




الذي أن دائما تذكرأيضا هو عنه تبحث

. يبحثعنك



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Sept 2013

Solution:

Open circuit:Applyingnode−voltage method ¿get V Th knowing that :

Short circuit:

Applying mesh method to get isc

−40+2000i1+20,000 (i1−i2 )=0 5000 i2+50,000 (i2−isc )+20,000 (i2−i1)=0

50,000 (isc−i2)+10,000(isc−30ii ∆)=0 i∆= i1−i2

Solving,

i1=13.6 mA , i2=12.96 mA ,

isc=14 mA , i∆=640 μA .

RTh=280

0.014=20 kΩ




أن أحد ألي تسمح الغايتك من يحرمك

. الحياة في

v1−402000

+v1

20,000+

v1−v2

5000=0

v2−v1

5000+

v2

50,000+

v2−v3

10,000+30

v1

20,000=0

v3−v2

10,000+

v3

40,000−30

v1

20,000=0

Solving , v1=24 V , v2=−10 V , v3=280V ,

V Th=v3=280V .



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Sept 2013

Solution:

Open circuit:

Applying Ohm’s law for right loop:

v2=−iR=−(80 ib ) (50×103 )=−40 × 105ib

4 ×10−5 v2=−160 ib

Applying KVL for middle loop:

4 ×10−5 v2+1310 ib=1310 ib−160 ib=1150 ib=100i100

Applying KCL at top left node:

500 μA=1150 ib

100+ibib=40 μA

Solving, v2=−160 V ,ib=40 μA.

Short circuit: v2=0 isc=−80 ib

ib=500×10−6 100(100+1310)

=35.46 μAisc=−80 (35.46 )=−2837 mA

RTh=−160

−2837× 10−6 =56.4 kΩ

The final Thevenin circuit is:




عن الصورة تغني. ألفكلمة



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Sept 2013

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 series with 4 kΩ).

2¿4 k Ω∨¿6k Ω=4∗64+6

=2.4 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:

ib=10−7.5

2.5=1mA

let R0=7.5 /0.8=9.375 kΩ

R0=( Rmeter ) (10 )Rmeter+10

=9.375

Rmeter=(9.375 ) (10 )

0.625=150 kΩ

Applying KVL for left loop:

−10+(2.5∗103 ) ib+(10∗103 )∗0.8ib=0

ib=0.952 mA

b)

ve= iR=(0.8∗0.9524∗10−3 )∗(10∗103 )=7.62 V

% error=( 7.5−7.627.62 )×100=−1.57 %




لآلخرين اسمحإنها باالختالفمعك،

لتطوير وسيلة أفضل



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Sept 2013

Solution:

VTh = 0, since there are no independent sources,

we must apply the test method:

v1−10 i∆

10+

v1

2.5+

v1−vT

12=0

vT−v1

12+

vT−10 i∆

6−1=0

i∆=vT−v1

12

Solving,v1=2V ,vT=8 V , i∆=0.5 A .

RTh=vT

1 A=8 Ω

The equivalent Thevenin circuit is:

Note: V Th is zero since there are no independent sources.




للصفح معنى أهم إنإزاحة هو اآلخرين عن

. نفسك عن األلم

1A



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Sept 2013

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:

v−1212,000

+ v−1020,000

+ v12,500

=0

Solving: v=7.03125 V

v10 KΩ=10,00012,500

(7.01325)=5.625 V

vTh=v−10=−4.375 V

RTh=R0=( (12,000 20 ,000 )+2500 )∨¿10,000=5 kΩ

b¿ pmax=(−437.5 ×10−6)2(500)=957 μW

c ¿v4.7 KΩ=4700

4700+5000(−4.375)=−2.12 V

p4.7 KΩ=(−2.12)2

4700=956.12 μW

% error=( 956957

−1) (100 )=−0.1 %




أفضل تستحق إنكالحياة تلك من بكثير

. اآلن تعيشها التي



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Sept 2013




Solution :

a)

Since 0≤ R0≤ ∞ maximum power will be delivered to the 9 Ω resistor when R0=0

b)

p=(30)2

6=150 W

تتوقع أن دائما يجبكن ما، فرصة وجودستكون ، دائما مستعدا

ال حيث فرصة هناك



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Sept 20134.12 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.

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.

Thus,

p=i2 RL=( V Th

RTh+RL)

2

RL




بعض تكون ما كثيرامصباحا المشاكل

. أفضل طريق لبداية

Fig. 4.59



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Sept 2013Next, 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:

dpd RL

=V Th2 [ ( RTh+ RL )2−RL∗2 ( RTh+RL)

( RTh+RL )4 ]The derivative is zero and p is maximized when

( RTh+RL )2=2 RL(RTh+RL)

Solving this equation yields

RL=RTh

pmax=V Th

2 RL

(2 RL)2=

V Th2

4 RL=

V Th2

4 RTh.

Example 4.12: Calculating the Condition for Maximum Power Transfer




تشل أناسكثيرونعقباتهم حركتهم

. بأنفسهم وضعوها



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Sept 2013a) For the circuit shown, find the value of RL that

results in maximum power being transferred to RL.

b) Calculate the maximum power that can be

delivered to RL.

c) When RL is adjusted for maximum power transfer, what percentage of the power

delivered by the 360 V source reaches RL?

Solution:

a) Applying the node-voltage equationV Th

150+

V Th−36030

=0

V Th=300V

Using Deactivation method:

RTh=(150 )(30)

180 =25 Ω=RLb¿ pmax=V Th

2

4 RTh=

(300 )2

4∗25=900 W

c) When RL equals 25 Ω, the voltage vab is

vab=vS∗R

Req=300∗25

(25+25 )=150 V

is=360−150

30=210

30=7 A

ps=−is (360 )=−2520 W

percentage= 9002520

×100=35.71 %

Assessment Problem 4.20

Find the Thévenin equivalent circuit with respect to

the terminals a, b for the circuit shown. (Hint: Define




الخوفمنشيء، قتل أردت إذا

iΔ



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Sept 2013the voltage at the leftmost node as v, and write two

nodal equations with V Th as the right node voltage.)

Solution:

Using the node voltage method

v60

+v−(vTh+160 i∆)

20−4=0

vTh

40+

vTh

80+

vTh+160 i∆−v20

=0

Solving with:

i∆=vTh

40

v=172.5 V vTh=30 V

Using test source method to calculate the test

current and thusRTh. Replace the current source

with a short circuit and apply the test source.

Applying KCL equation at the rightmost node:

iT=vT

80+

vT

40+

vT+160 i∆

80

Solving with:

i∆=V Th

40⟹iT=

vT

10

RTh=vT

iT=10 Ω

Thus, the Thevenin equivalent is a 30 V source in series with a 10 Ω resistor.




طالما ناجحا ستكون. ذلك اعتقدت



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Sept 2013


a) Find the value or R that enables the

circuit shown to deliver maximum power

to the terminals a, b.

b) Find the maximum power delivered to R.

Solution:

Applying node-voltage to get VTh:vTh−(100+v∅)

4+

vTh−v1

4=0

v1−1004

+v1−20

4+

v1−vTh

4=0 v∅−v1+20=0

Solving,vTh=120 V ,v1=80V , v∅=60 V .

Creating a short circuit between nodes a and b and use

the mesh current method to get isc

−100+4 (i1−i2 )+v∅+20=0

−v∅+4 i2+4 ( i2−isc )+4 (i2−i1 )=0

−20−v∅+4 (isc−i2 )=0

v∅=4 (i1−isc)

Solving, i1=45 A , i2=30 A , isc=40 A ,v∅=20 V .

Continue soluation (Assessment Problem 4.21):




يكن لم إذا فقط الرجل نحصيعمر إننا. يحصيه آخر شيء لديه



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Sept 2013Thus,

RTh=vTh

isc=120

40=3Ω

a)

For maximum power transfer, R=R th=3Ω

b)

The Thevenin voltage, vTh=120 V , is divided equally between the Thevenin

resistance and the load resistance, so

vload=1202

=60 V

pmax=v load

2

R load= 602

3=1200 W

¿ pmax=V Th

2

4 Rload=

(120)2

4 (3)=1,200 W =1.2 KW




العالم أن المؤكد منالتي إسهاماتك ينتظرمتى أجلها، خلقتمن



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Sept 2013


Assume that the circuit in Assessment problem 4.21

is delivering maximum power to the load resistor R.

a) How much power is the 100 V source

delivering to the network?

b) Repeat (a) for the dependent voltage source.

c) What percentage of the total power generated

by these two sources is delivered to the load

resistor R?

Solution:

According to Assessment 4.21, R=3 Ω

Using mesh current method:−100+4 (i1−i2 )+v∅+20=0

−v∅+4 i2+4 ( i2−i3 )+4 (i2−i1 )=0

−20−v∅+4 (i3−i2 )+3 i3=0

v∅=4 (i1−i3)

Solving,i1=30 A , i2=20 A ,i3=20 A , v∅=40 V .

a) P100 v=−(100 ) (i1 )=−3000 W (delivers)

b) Pdependent=−(v∅ ) (i2 )=−800 W (delivers)

c ¿P3 Ω=i2 R=(20 )2 (3 )=1200 W

% delivered= 12003000+800

=31.6%




ال قليلة، بكميات لكن تشاء، ما كل



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Sept 2013

Solution:

We need to find the Thevenin equivalent with respect to R0.

1) Transfer (200 V in series with 25 Ω) to (8 A in parallel with 25 Ω).

2) 25 Ω∨¿100 Ω=20 Ω.

3) Transfer (8A in parallel with 20 Ω) to (160 V in series with 20 Ω).

4) 20 Ω+10 Ω=30 Ω.

ix=160−30i x

50ix=2 A

V Th=20 ix+30i x=50 ix=100 V

Using the test-source method to find the

Thevenin resistance gives

iT=vT

30+vT−30¿¿

iT

vT= 1

30+ 1

10= 4

30= 2

15

RTh=vT

iT=15

2=7.5 Ω

Solving forR0:

p=i2 R0

p=( 1007.5+R0 )

2

R0=250

R0=22.5 Ω∨R0=2.5 Ω




من كثيرا أفضل الخطأ الطريق. الالطريق خالل



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Solution :

Find the Th´evenin equivalent with respect to the terminals of R0

Open circuit voltage:

نتخيل أن استطعنا إذاونؤمن التغيير نتيجةبهذه جديرون بأننا

أننا فاألرجح النتائج



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Continued Solution (problem 4.85) :(440−220 )=5ia−2ib−3ic

0=−2 ia+10 ib−1 ic

ic=0.5 v ∆ v∆=2(ia−ib)

Solving, ib=26.4 A

∴ v th=7 ib=184.8 V

Short circuit current:

(440−220 )=5ia−2iSC−3ic

0=−2 ia+10 isc−1 ic

ic=0.5 v ∆

v∆=2(ia−isc)

Solving, isc=60 A ,ia=80 A ,ic=20 A .

Rth=v th/ isc=184.8/60=3.08Ω

الحياة، هو الوقتالوقت ضاع فإذاالحياة ضاعت



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Sept 2013




Continued Solution (problem 4.85) :

Therefore, the Th´evenin equivalent circuit configured for maximum power to the load is:

From this circuit,

pmax=(92.4)2

3.08=2772 W

With R0 equal to 3.08 the original circuit becomes:

440−220=5ia−2ib−35ic

ما للمرء يكتبنوى



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Sept 2013





ic=0.5 v ∆ v∆=2(ia−ib)

−92.4=−2 ia+3 ib−1ic

Solving, ia=88.4 A ,ib=43.2 A ,ic=45.2 A .

v∆=2 (88.4−43.2 )=90.4 V

p440 V=− (440 ) (88.4 )=−38,896 W

p440 V= (220 ) (88.4−45.2 )=9504 W

pdep . source=(440−92.4 ) (0.5 (90.4))=15,711.52W

Therefore, only the 440 V source supplies power to the circuit, and the power

supplied is 38,896 W.

% delivered= 277238,896

=7.13%

تخدع أن يمكنكلكن أحيانا، اآلخرين

لنفسك؟ ستقول ماذا



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Sept 2013

Solution:

a)

Finding the Thevenin equivalent:

Open circuit voltage:

Using mesh current method:

−240+3 (i1−i2 )+20 (i1−i3 )+2i1=0

2 i2+4 (i2−i3 )+3 (i2−i1 )=0

10 iβ+1 i3+20 ( i3−i1)+4 (i3−i2 )=0

The dependent source constraint equation is:iβ=i2−i1

Solving,i1=99.6 A , i2=78 A ,i3=100.8 A , iβ=−21.6 A .

V Th=20 (i1−i3 )=−24 V




تتخذ أن يمكنلكفي مستشارينإنها المجاالت، كافة



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Sept 2013

Continued soluation (Problem 4.87):

Short-circuit current:

Using mesh current method:−240+3 (i1−i2 )+2 i1=0

2 i2+4 (i2−i3 )+3 (i2−i1 )=0

10 iβ+1 i3+4 (i3−i2 )=0

The dependent source constraint equation is:iβ=i2−i1

Solving,i1=92 A , i2=73.33 A , i3=96 A , iβ=−18.67 A .

isc=i1−i3=−4 A

RTh=V Th

isc=−24

−4=8 Ω

RL=RTh=6 Ω

b)

pmax=(12)2

6=24 W




تضاعف أن يمكنكومرات، مرات حياتك

. عديدة طرق توجد



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Sept 2013

Solution:

a) Finding the Thevenin equivalent:

(Open Circuit Voltage):

Using mesh current method:−100+4 (i1−i2 )+80 (i1−i3 )+16 i1=0

124 iΔ+8 (i2−i3 )+4 (i2−i1 )=0

50+12 i3+80 (i3−i1 )+8 (i3−i2 )=0 iΔ=i3−i1

Solving,i1=4.7 A ,i2=10.5 A , i3=4.1 A ,iΔ=−0.6 A . vTh=vab=−80 iΔ=48V

(Short Circuit Current):

Using mesh current method:

iΔ=0124 iΔ=0

−100+4 (i1−i2 )+16 i1=0

8 (i2−i3 )+4 (i2−i1 )=0 50+12 i3+8 (i3−i2 )=0




شيئا تتعلم أن البد. يوم كل جديدا



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Sept 2013

Continue soluation (Problem 4.89):

Solving,

i1=5 A ,

i2=0 A ,

i3=−2.5 A .

isc=i1−i3=7.5 A RTh=487.5

=6.4 Ω

For maximum power transfer,

R0=RTh=6.4 Ω

b)

pmax=(24 )2

6.4=90 W

c)

i10Ω= 486.4+10

=2.927 A

p10 Ω=10 (2.927)2=85.7 W

Thus, using a 10 resistor selected from Appendix H will cause 85.7 W of power to

be delivered to the load, compared to the maximum power of 90 W that will be

delivered if a 6.4 resistor is used.

4.13 Superposition




أبدا، أستسلم إننياللي يقول عندما حتى

أنجح لن الناسإنني



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Sept 2013Whenever 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 demonstrate the superposition principle by

using it to find the branch currents in the circuit

shown in Fig. 4.62. We begin by finding the

branch currents resulting from the 120 V voltage

source.

v1−1206

+v1

3+

v1

2+4=0 v1=30V i1

' =120−306

=15 A i2' =30

3=10 A i3

' =i4' =30

6=5 A

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.

v3

3+

v3

6+

v3−v4

2=0

v4−v3

2+

v4

4+12=0

v3=−12V v4=−24 V

i1' '=

−v3

6=12

6=2 A i2

' '=v3

3=−12

3=−4 Ai3

' '=v3−v 4

2=−12+24

2=6 A i4

' '=v 4

4=−24

4=−6 A

The currents i1 , i2 ,i3 and i4 in Fig. 4.62 are:




أنك ظننت إذا هزمتفقد

Figure 4.63 The circuit shown in Fig. 4.62 with the current source deactivated.

v3 v4



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Sept 2013 i1=i1

' + i1' '=15+2=17 A

i2=i2' + i2

' '=10−4=6 A

i3=i3' +i3

' '=5+6=11 A

i4=i4' +i4

' '=5−6=−1 A

For circuits containing both independent and dependent sources, you must recognize

that the dependent sources are never deactivated.

Example 4.13:

Use the principle of superposition to find v0 in the

circuit shown in Fig. 4.66




أفضلللبر وسيلة

ال أن بالوعد



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Sept 2013Solution:

v0' =v s

RReq

¿10 2025

=8 V

Applying the node-voltage equations yield: v0

' '

20+

v0' '

5−0.4 v∆

' '=0

¿5 v0' '−8v ∆

' '=0

0.4 v∆' '+

vb−2i∆' '

10−5=0

¿4 v∆' '+vb−2 i∆

' '=50

Solving with:

vb=2 i∆' '+v∆

' '

5 v∆' '=50∨v∆

' '=10V

5 v0' '=80∨v0

' '=16 V

The value of v0 is the sum of v0' andv0

' ', or 24 V

Solution:




فقد المهزوم ابتسم إذاالفوز لذة .المنتصر



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Sept 2013a) 110 V source acting alone:

Re=10∨¿14=356

Ω

i'= 1105+35/6

=13213

A

v'=( 356 )( 132

13 )=77013

V =59.231V

4 A source acting alone: 5∨¿10=10/3 Ω

10/3+2=16/3Ω

16 /3 12=48/13Ω

Our circuit reduces to:

va' '=4 (48 /13 )=¿¿

v' '=−v a

' '

¿¿¿

v=v '+v' '=50 V

b¿ p= v2

10=250 W

Solution:

240 V source acting alone:Req=( (4+1 )∨¿20)+5+7




أن في أمال سما تبتلع أن هو الحسد. يموتشخصآخر



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Sept 2013 ¿¿

vo1=5∨¿20

16(240 )=60 V

84 V source acting alone:Req=( (5+7 )∨¿20)+4+1

¿(12∨¿20)+5=12.5 Ω

vo 2=12∨¿20

12.5(−84 )=−50.4 V16 A current source acting alone:

(v1−v2)5

+v1

7−16=0

(v2−v1)5

+v2

20+

(v2−v3)4

=0

(v3−v2 )4

+v3

1+16=0

Solving, v2=18.4 V=vo3

Therefore, vo=vo 1+vo 2+vo3=28V




تستمتع أن حاولبمشاكلك.



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Solution :

a) By hypothesis i0' +i0

=3 m¿

i0' ' '=−5 (2 )

(8 )=−1.25 mA ∴i0=3.5−1.25=2.25 mA

شيء أعرفكل أنا

نفسي إال



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Sept 2013





b)

With all three sources in the circuit write a single node voltage equation.

vb

6+

vb−82

+5−10=0

∴ vb=13.5 V

i0=vb

6=2.25 mA

من كثيرا يضحك أخيرا يضحك



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Sept 2013

Solution:

Voltage source acting alone:

Using node voltage method:vo 1

20,000+

v o1−254000

−2.2( vo 1−254000 )=0

vo 1=30V

Current source acting alone:

Using node voltage method:vo 2

20,000+0.005+

vo 2

4000−2.2( vo 2

4000 )=0

vo 2=20V

vo=vo1+vo2=30+20=50 V




فريسة يجد أن األسد باألمسحلم



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Sept 2013




Solution :

The mesh equations are:−125+0.15 ia+18.4 (ia−ic )+0.25 (ia− ib )=0

−125+0.25 (ib−ia )+38.4 (ib−id )+0.15ib=0

0.15 ic+18.4 (ic−ie)+0.25 (ic−id )+18.4 (ic−ia )=0

الحكمة ينابيعفي مياهها تتدفق

الكتب رفوف



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Sept 2013





0.15 id+38.4 (id−ib )+0.25 (id−ic)+38.4 (id−ie )=0

11.6 ie+38.4 (ie−id )+18.4 (ie−ic )=0

Place these equations in standard form:ia (18.8 )+ ib (−0.25 )+ ic (−18.4 )+id (0 )+ ie (0 )=125

ia (−0.25 )+ib (38.8 )+ ic (0 )+id (−38.4 )+ ie (0 )=125

ia (−18.4 )+ib (0 )+ic (37.2 )+id (−0.25 )+ie (−18.4 )=0

ia (0 )+ib (−38.4 )+ic (−0.25 )+ id (77.2 )+ie (−38.4 )=0

ia (0 )+ib ( 0 )+ic (−18.4 )+id (−38.4 )+ie (68.4 )=0

Solving, ia=32.77 A , ib=26.46 A , ic=26.33 A ,

id=23.27 A ,ie=20.14 A .

Find the requested voltages:v1=18.4 (ic−ie)=113.90V

v2=38.4 (id−ie)=120.19V

v3=11.6 ie=233.62V

فكري أضيق مايتسع ال دام ما

فكر لكل



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