4.10 thevenin and norton equivalents - موقع المهندس حماده شعبان technique is...

Sept 2013

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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.

يجب أن تكون حياتك أكبر

.بكثير من مجرد وجود

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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 short-

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

is identical to the Thevenin resistance.

مااا تفاااه مناا سااي ع ا

.على تفكيرك في داومت


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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.

قااااد يكااااون ماااان الواجااااب

.عليك اتفا ال رار اآلن


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Finding the Thevenin Equivalent of a Circuit with a Dependent Source

Example 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:

Applying KVL at right loop:

Solving,

2. Calculating (Short-Circuit):

Applying KVL for the outer right loop (with no resistor, no voltage drop)

3. Calculating :

ها نيتمنى البعض أمورا هم يمتلكو

.بالفعل، ولكنهم ال يدركون لك

+

-

b

a


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The Deactivation Method (for independent Sources Only)

The technique for determining RTh that

we discussed and illustrated earlier is not

always 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,

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.

كثير من الناس تسرع

.في مكانها طىالف

Figure 4.52 A circuit used to illustrate

a Thévenin equivalent


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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.

Solution:

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 , and the current that it

delivers to the circuit is labeled . 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,

ع لك الباطن ال يستفدم المنطق،

. ن ي بل ما تملي علي


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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) ملفص بطرق الحل بطري ة

Method Indep. Only Dep. Only Indep + Dep.

Basic Method ---

Source Transformation --- ---

Deactivation --- ---

Test (alternative)

نااا مااان الحماقاااة أن تااا من ب ااادرة

.بك بعض األشياء على لحاق األ ى


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Assessment Problem 4.16:

Solution:

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

Using node voltage analysis to find :

The node voltage equations are:

Solving,

and

. ا انعدمت شهيتك، فال تلم الطعام

Find the Thévenin equivalent circuit with respect

to the terminals a, b for the circuit shown.


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Find the Norton equivalent circuit with respect


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.

أيااا كااان الااايء الاا ي تطلباا بصاادق،

.ستحصل علي ، ا آمنت أنك تستح


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A voltmeter with an internal resistance of 100 kΩ

is used to measure the voltage 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

source 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:

اعت د في الفير سوه تحصل علي ،

.اعت د في الار سوه ينال منك


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Find the Thévenin equivalent circuit with respect


Solution:

Calculating the open circuit voltage, which is also vTh,

Solving,

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:

Solving,

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

.فكر في العراقيل التي تواج طموحاتك، وسوه تحدث لك، حتما


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Solution:

Using deactivating method, we can calculate RTh

as follows:

Using voltage divider:

The final circuit of Thevenin is:

.قرر نجاحك، تعامل وكأن قد وقع بالفعل


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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:

الع باااة فاااي طرياااق تفوقاااك

.تكمن في ع لك الباطن ف ط


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Solution:

a)

Open circuit:

Short circuit:

b)

بعض الناس يستمتعون باأللم

.ألن يجلب اهتمام اآلخرين بهم


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Solution:

After making a source transformation the circuit becomes

Solving,

and

Using deactivation method to get RTh:

Ω

The final Thevenin circuit is:

. ن أفضل وسيلة هي تلك ات المجهود األقل


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Solution:

a)

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

1. Converting 30 V along with series 10 k

resistor to 3 mA current source parallel to

10 k resistor, then combining 10 k || 40 k

to 8 k .

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.

أنت تستحق أكثر مما أنت

.علي اآلن، ال أشك في لك


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Continue soluation (Problem 4.71):

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

with 12 k , 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).

Ω

b)

ت كر دائما أن ال ي تبحث

.عن هو أيضا يبحث عنك


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Solution:

Open circuit:

Short circuit:

Applying mesh method to get isc

Solving,

ألي أحد أن يحرمك ال تسمح

.في الحياة من غايتك


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Solution:

Open circuit:

Applying Ohm’s law for right loop:

Applying KVL for middle loop:

Applying KCL at top left node:

Solving, .

Short circuit:

The final Thevenin circuit is:

.تغني الصورة عن ألف كلمة


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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).

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:

Applying KVL for left loop:

b)

اسمح لآلخرين باالختاله معك،

. نها أفضل وسيلة لتطوير اتك


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Solution:

VTh = 0, since there are no independent sources,

we must apply the test method:

Solving,

The equivalent Thevenin circuit is:

Note: is zero since there are no independent sources.

ن أهم معنى للصفح عن اآلخرين

.هو زاحة األلم عن نفسك

1A


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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:

نك تستحق أفضل بكثير من

.تلك الحياة التي تعياها اآلن


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Solution:

a)

Since maximum power will be delivered to the 9 Ω resistor

when

b)

يجببدائما بب اتواع وجبب ارصببوئا بب ااام ،اكنامس عدمائما ا،اس كوواهنب اا احيثاالاع وج ارصوئه اتبدم


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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.

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,

كثيرا ما تكون بعض المااكل

.مصباحا لبداية طريق أفضل

Fig. 4.59


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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

Solving this equation yields

.

أناس كثيرون تال حركتهم

.ع بات هم وضعوها بأنفسهم


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Example 4.12: Calculating the Condition for Maximum Power Transfer

a) 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 equation

Using Deactivation method:

c) When RL equals , the voltage is

. ا أردت قتل الفوه من شيء، أنصحك بفعل


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Assessment Problem 4.20

Find the Thévenin equivalent circuit with respect to

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

the voltage at the leftmost node as , and write two

nodal equations with as the right node voltage.)

Solution:

Using the node voltage method

Solving with:

Using test source method to calculate the test

current and thus . Replace the current source

with a short circuit and apply the test source.

Applying KCL equation at the rightmost node:

Solving with:

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

.ستكون ناجحا طالما اعت دت لك


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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:

Solving,

Creating a short circuit between nodes a and b and use

the mesh current method to get

Solving,

. ا لم يكن لدي شيء آخر يحصي ف ط ننا نحصي عمر الرجل


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Continue soluation (Assessment Problem 4.21):

Thus,

a)

For maximum power transfer,

b)

The Thevenin voltage, , is divided equally between the Thevenin

resistance and the load resistance, so

من الم كد أن العالم ينتظر سهاماتك

التي خل ت من أجلها، متى تبدأ؟


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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,

Using mesh current method:

Solving,

a)

b)

.ما تااء، لكن بكميات قليلة، ال تحفر قبرك بأسنانك ل ك


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Solution:

We need to find the Thevenin equivalent with respect to .

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

2) .

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

4) .

Using the test-source method to find the

Thevenin resistance gives

Solving for :

.أفضل كثيرا من خالل الالطريقالطريق الفطأ


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Solution:

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

Open circuit voltage:

إذمامسبببببب نعن اتوان ييبببببب ان يجبببببب ااامل غيي ارنؤمنابأنن اصبدي روابذب ااا ملن اجا ألرصحاتنن اسنبلغاغ ي ن


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Continued Solution (problem 4.85):

Solving,

Short circuit current:

Solving,

إذماض عاا،ملوجتاهواملحي ة ملوجتاض عتاملحي ة


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Therefore, the Th´evenin equivalent circuit configured for maximum power

to the load is:

From this circuit,

With equal to 3.08 the original circuit becomes:

ايك دالل ءام انوى


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Solving,

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

supplied is 38,896 W.

ي كنكاتواعيدعامآلخ يناتحي ن ،ا س قولالنفسك؟لكنام ذما


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Solution:

a)

Finding the Thevenin equivalent:

Open circuit voltage:


The dependent source constraint equation is:

Solving,

يمكن أن تتف مستاارين لك

.في كافة المجاالت، نها الكتب


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Continued soluation (Problem 4.87):

Short-circuit current:


The dependent source constraint equation is:

Solving,

b)

تضاعف حياتك مرات يمكنك أن

.توجد طرق عديدةومرات،


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Solution:

a) Finding the Thevenin equivalent:

(Open Circuit Voltage):


Solving,

(Short Circuit Current):


.البد أن تتعلم شيئا جديدا كل يوم


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Continue soluation (Problem 4.89):

Solving,

For maximum power transfer,

b)

c)

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.

ننااي ال أستساالم أباادا، حتااى عناادما

.ي ول لي الناس نني لن أنجح أبدا


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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 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.

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.

ا ظننت أنك هزمت

.ف د هزمت

Figure 4.63 The circuit shown in Fig. 4.62

with the current source deactivated.


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The currents and in Fig. 4.62 are:

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

that the dependent sources are never deactivated.

وسيلة للبر أفضل

.بالوعد أن ال تعد


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Example 4.13:

Use the principle of superposition to find in the

circuit shown in Fig. 4.66

Solution:

Applying the node-voltage equations yield:

Solving with:

The value of is the sum of and

, or 24 V

. ا ابتسم المهزوم ف د المنتصر ل ة الفوز


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Solution:

a) 110 V source acting alone:

4 A source acting alone:

Our circuit reduces to:

.الحسد هو أن تبتلع سما أمال في أن يموت شفص آخر


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Solution:

240 V source acting alone:

84 V source acting alone:

16 A current source acting alone:

Solving,

Therefore,

.حاول أن تستمتع بمااكلك


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Solution:

a) By hypothesis

تن اتع فاك اشيءاإالانفسي


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b)

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

ايضحكاكثي مامنايضحكاتخي م


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Solution:

Voltage source acting alone:

Using node voltage method:

Current source acting alone:

Using node voltage method:

باألمس حلم األسد أن يجد فريسة اليوم، فما ا أنت حلمت؟


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Solution:

The mesh equations are:

ينببببب بي املحك ببببب اع بببببد ااا مي هذ ا يار وفاملك د


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Place these equations in standard form:

Solving,

Find the requested voltages:

مببب اتضبببي ا كببب امببب ا ئمماالاي س الك ا ك


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4.10 thevenin and norton equivalents - موقع المهندس حماده شعبان technique is...

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