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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 𝑣1as in

Fig. 4.45 which is equal to VTh.

𝑣1 − 25

5+

𝑣1

20 − 3 = 0

𝑣1 = 𝑣𝑇ℎ = 32 V

جب أ حىى دبحه أوبش

.بىثش جشد وجىد

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2) Calculate the short-circuit current 𝑖𝑠𝑐 as in

Fig. 4.46.

𝑣2 − 25

5+

𝑣2

20− 3 +

𝑣2

4= 0

𝑣2 = 16 V

𝑖𝑠𝑐 = 16

4= 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 =𝑉𝑇ℎ

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.

ب حخبف ه عمغ إرا

. ػ حفىشن فهداوج

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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.

لذ ىى اىاجب

.ػه احخبر امشاس ا

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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:

−5 + 2000𝑖 + 3𝑣 = 0

2000𝑖 + 3𝑣 = 5 (1)

Applying KVL at right loop:

−𝑣 − 25 ∗ 20𝑖 = 0

500𝑖 + 𝑣 = 0 (2)

Solving, 𝑖 = 0.01 A, 𝑣 = −5 V = 𝑉𝑇ℎ .

2. Calculating 𝑖𝑠𝑐 (Short-Circuit):

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

𝑖𝑠𝑐 = −20𝑖

𝑣 = 0 (no voltage drop)

From equation 1 , 𝑖 =5

2000= 2.5 ∗ 10−3 A

𝑖𝑠𝑐 = −0.05 A

3. Calculating 𝑅𝑇ℎ :


𝑖𝑠𝑐=

−5

−0.05= 100 Ω

هب خ ابؼض أىسا ه خىى

.ببفؼ، وىه ال ذسوى ره

+

-

b

a

𝟓 𝒗

𝟏𝟎𝟎𝛀

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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,

𝑅𝑎𝑏 = 𝑅𝑇ℎ = 4 +5 × 20

25= 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.

وثش ابط حغشع

. ف ىبهباخط

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 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+ 20𝑖 (node − voltage)

𝑖 =−3𝑣𝑇

2 mA (left loop)

𝑖𝑇 =𝑣𝑇

25 –

60𝑣𝑇

2000=

𝑣𝑇

100

𝑅𝑇ℎ =𝑣𝑇

𝑖𝑇= 100 Ω

ػمه اببط ال غخخذ اطك،

.إه مب ب حه ػه

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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) خص بطشق اذ بطشمت

Method Indep. Only Dep. Only Indep + Dep.

Basic Method ---

Source Transformation --- ---

Deactivation --- ---

Test (alternative)

إه اذبلت أ حؤ بمذسة

.بؼض اشبء ػ إذبق اري به

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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 find 𝑣𝑇ℎ :

The node voltage equations are:

𝑣1 − 72

5 +

𝑣1

20 +

𝑣1 − 𝑣𝑇ℎ

8 = 0

𝑣𝑇ℎ − 𝑣1

8+

𝑣𝑇ℎ − 72

12 = 0

Solving,

𝑣1 = 60 V and 𝑣𝑇ℎ = 64.8 V

.إرا اؼذج شهخه، فال ح اطؼب

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.

أب وب اشء از حطبه بصذق،

.عخذص ػه، إرا آج أه حغخذمه

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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 mA 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:

𝑣𝐴𝐵 =100,000

125,000 150 = 120 V

اػخمذ ف اخش عىف حذص ػه،

.اػخمذ ف اشش عىف بي ه

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Find the Thévenin equivalent circuit with respect


Solution:

Calculating the open circuit voltage, which is also vTh,

𝑣𝑇ℎ

8+ 4 + 3𝑖𝑥 +

𝑣𝑡ℎ − 24

2= 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 + 3𝑖𝑥 +𝑣𝑇

2= 4𝑖𝑥 +

𝑣𝑇

2 𝑖𝑥 =

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.

.فىش ف اؼشال اخ حىاجه طىدبحه، وعىف حذذد ه، دخب

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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 = 48 V

The final circuit of Thevenin is:

.لشس جبده، حؼب ووأه لذ ولغ ببفؼ

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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:

اؼمبت ف طشك حفىله

.حى ف ػمه اببط فمظ

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Solution:

a)

Open circuit:

𝑣2 − 9

20+

𝑣2

70− 1.8 = 0 ∴ 𝑣2 = 35 V

𝑣𝑇ℎ =60

70∗ 𝑣2 = 30 V

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 𝑜𝑅 =𝑣2

10

𝑅𝑇ℎ =30

1.5= 20 Ω

b)

𝑅𝑇ℎ = 20 + 10 ||60 =30 + 60

90= 20 Ω

بؼض ابط غخخؼى بب

.ه جب اهخب اخش به

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Solution:

After making a source transformation the circuit becomes

−500 + 8 𝑖1 − 𝑖2 + 12𝑖1 = 0

−300 + 30𝑖2 + 5.2𝑖2 + 8 𝑖2 − 𝑖1 = 0

Solving,

𝑖1 = 30 A and 𝑖2 = 12.5 A

𝑣𝑇ℎ = 12𝑖1 + 5.2𝑖2 = 425 V

Using deactivation method to get RTh:

𝑅𝑇ℎ = (8| 12 + 5.2 ||30 = 7.5 Ω

The final Thevenin circuit is:

.إ أفض وعت ه حه راث اجهىد ال

30 Ω

40 Ω 5.2 Ω 8 Ω

12 Ω

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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.

أج حغخذك أوثش ب أج

.ػه ا، ال أشه ف ره

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Continue 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).

𝑣𝑇ℎ = 48 V

𝑅𝑇ℎ = 6 kΩ

𝑣𝑚𝑒𝑎𝑛𝑠 =100

106 48 = 45.28 V

b)

%𝑒𝑟𝑟𝑜𝑟 = 45.28 − 48

48 × 100 = −5.67%

حزوش دائب أ از حبذذ

.ػه هى أضب بذذ ػه

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Solution:

Open circuit:

Applying node − voltage method to get VTh knowing that:

Short circuit:

Applying mesh method to get isc

−40 + 2000𝑖1 + 20,000 𝑖1 − 𝑖2 = 0

5000𝑖2 + 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

ال حغخ أدذ أ ذشه

. ف اذبة غبخه

𝑣1 − 40

2000+

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 = 280 V,

VTh = 𝑣3 = 280 V.

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Solution:

Open circuit:

Applying Ohm’s law for right loop:

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

4 × 10−5 𝑣2 = −160𝑖𝑏

Applying KVL for middle loop:

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

Applying KCL at top left node:

500 μA =1150𝑖𝑏

100+ 𝑖𝑏 𝑖𝑏 = 40 μA

Solving, 𝑣2 = −160 V, 𝑖𝑏 = 40 μA.

Short circuit:

𝑣2 = 0 𝑖𝑠𝑐 = −80𝑖𝑏

𝑖𝑏 = 500 × 10−6100

(100 + 1310)= 35.46 μA

𝑖𝑠𝑐 = −80 35.46 = −2837 mA

𝑅𝑇ℎ =−160

−2837 × 10−6= 56.4 kΩ

The final Thevenin circuit is:

.حغ اصىسة ػ أف وت

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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).

2)4 𝑘Ω || 6 kΩ = 4 ∗ 6

4 + 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= 1 mA

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 − 7.62

7.62 × 100 = −1.57%

اعخ خش ببالخخالف ؼه،

.إهب أفض وعت خطىش راحه

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Solution:

VTh = 0, since there are no independent sources,

we must apply the test method:

𝑣1 − 10𝑖∆

10+

𝑣1

2.5+

𝑣1 − 𝑣𝑇

12= 0

𝑣𝑇 − 𝑣1

12+

𝑣𝑇 − 10𝑖∆6

− 1 = 0

𝑖∆ =𝑣𝑇 − 𝑣1

12

Solving,

𝑣1 = 2 V, 𝑣𝑇 = 8 V, 𝑖∆ = 0.5 A.


1 𝐴= 8 Ω

The equivalent Thevenin circuit is:

Note: 𝑉𝑇ℎ is zero since there are no independent sources.

إ أه ؼ صفخ ػ اخش

.هى إصادت ا ػ فغه

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:

𝑣 − 12

12,000 +

𝑣 − 10

20,000+

𝑣

12,500= 0

Solving: 𝑣 = 7.03125 V

𝑣10 KΩ =10,000

12,500(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 = 956

957− 1 100 = −0.1%

إه حغخذك أفض بىثش

.حه اذبة اخ حؼشهب ا

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𝑝 =(30)2

6= 150 W

Solution:

a)

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

when 𝑅0 = 0

b)

يجب دائما أن تتوقع وجود فرصة ما، كن مستعدا دائما ، ستكون هناك

فرصة حيث ال تتوقع وجودها أبدا

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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,

𝑝 = 𝑖2𝑅𝐿 = 𝑉𝑇ℎ

𝑅𝑇ℎ + 𝑅𝐿

2

𝑅𝐿 وثشا ب حىى بؼض اشبو

.صببدب بذات طشك أفض

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:

𝑑𝑝

𝑑𝑅𝐿= 𝑉𝑇ℎ

2 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 =𝑉𝑇ℎ

2 𝑅𝐿

2𝑅𝐿 2

=𝑉𝑇ℎ

2

4𝑅𝐿=

𝑉𝑇ℎ2

4𝑅𝑇ℎ

.

أبط وثشو حش دشوخه

.ػمببث ه وضؼىهب بأفغه

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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

𝑉𝑇ℎ

150+

𝑉𝑇ℎ − 360

30= 0

𝑉𝑇ℎ = 300 V

Using Deactivation method:

𝑅𝑇ℎ = 150 (30)

180= 25 Ω = 𝑅𝐿

b) 𝑝𝑚𝑎𝑥 =𝑉𝑇ℎ

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

𝑖 𝑠 =360 − 150

30=

210

30= 7 A

𝑝𝑠 = −𝑖𝑠 360 = −2520 W

percentage =900

2520× 100 = 35.71 %

.إرا أسدث لخ اخىف شء، أصذه بفؼه

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Sept 2013




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 𝑣, and write two

nodal equations with 𝑉𝑇ℎ as the right node voltage.)

Solution:

Using the node voltage method

𝑣

60+

𝑣 − (𝑣𝑇ℎ + 160𝑖∆)

20− 4 = 0

𝑣𝑇ℎ

40+

𝑣𝑇ℎ

80+

𝑣𝑇ℎ + 160𝑖∆ − 𝑣

20= 0

Solving with:

𝑖∆ =𝑣𝑇ℎ

40

𝑣 = 172.5 V 𝑣𝑇ℎ = 30 V

Using test source method to calculate the test

current and thus 𝑅𝑇ℎ . 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 + 160𝑖∆80

Solving with:

𝑖∆ =𝑉𝑇ℎ

40 ⟹ 𝑖𝑇 =

𝑣𝑇

10


𝑖𝑇= 10 Ω

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

.عخىى بجذب طبب اػخمذث ره

𝑖Δ

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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:

𝑣𝑇ℎ − (100 + 𝑣∅)

4+

𝑣𝑇ℎ − 𝑣1

4 = 0

𝑣1 − 100

4+

𝑣1 − 20

4+

𝑣1 − 𝑣𝑇ℎ

4 = 0

𝑣∅ − 𝑣1 + 20 = 0

Solving,

𝑣𝑇ℎ = 120 V, 𝑣1 = 80 V, 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∅ + 4𝑖2 + 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.

. إرا ى ذه شء آخش ذصه فمظإب ذص ػش اشج

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Continue soluation (Assessment Problem 4.21):

Thus,

𝑅𝑇ℎ =𝑣𝑇ℎ

𝑖𝑠𝑐=

120

40= 3 Ω

a)

For maximum power transfer, 𝑅 = 𝑅𝑡ℎ = 3 Ω

b)

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

resistance and the load resistance, so

𝑣𝑙𝑜𝑎𝑑 =120

2= 60 V

𝑝𝑚𝑎𝑥 = 𝑣𝑙𝑜𝑎𝑑

2

𝑅𝑙𝑜𝑎𝑑=

602

3= 1200 W

or 𝑝𝑚𝑎𝑥 =𝑉𝑇ℎ

2

4𝑅𝑙𝑜𝑎𝑑=

(120)2

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

اؤوذ أ اؼب خظش إعهببحه

اخ خمج أجهب، خ حبذأ؟

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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, 𝑅 = 3 Ω

Using mesh current method:

−100 + 4 𝑖1 − 𝑖2 + 𝑣∅ + 20 = 0

−𝑣∅ + 4𝑖2 + 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) 𝑃100𝑣 = − 100 𝑖1 = −3000 W (delivers)

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

c) 𝑃3Ω = 𝑖2𝑅 = 20 2 3 = 1200 W

% 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 =1200

3000 + 800= 31.6 %

ب حشبء، ى بىبث لت، ال حذفش لبشن بأعبه .و

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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 − 30𝑖𝑥

50 𝑖𝑥 = 2 A

𝑉𝑇ℎ = 20𝑖𝑥 + 30𝑖𝑥 = 50𝑖𝑥 = 100 V

Using the test-source method to find the

Thevenin resistance gives

𝑖𝑇 =𝑣𝑇

30+

𝑣𝑇 − 30(−𝑣𝑇 30)

20

𝑖𝑇𝑣𝑇

=1

30+

1

10=

4

30=

2

15


𝑖𝑇=

15

2= 7.5 Ω

Solving for 𝑅0:

𝑝 = 𝑖2𝑅0

𝑝 = 100

7.5 + 𝑅0

2

𝑅0 = 250

𝑅0 = 22.5 Ω OR 𝑅0 = 2.5 Ω .اطشك اخطأ أفض وثشاا خالي االطشك

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Solution:

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

Open circuit voltage:

إذا استطعنا أن نتخيل نتيجة التغيير ونؤمن بأننا جديرون بهذه النتائج فاألرجح أننا سنبلغ غايتنا

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440 − 220 = 5𝑖𝑎 − 2𝑖𝑏 − 3𝑖𝑐

0 = −2𝑖𝑎 + 10𝑖𝑏 − 1𝑖𝑐

𝑖𝑐 = 0.5𝑣∆ 𝑣∆ = 2(𝑖𝑎 − 𝑖𝑏)

∴ 𝑣𝑡ℎ = 7𝑖𝑏 = 184.8 V

440 − 220 = 5𝑖𝑎 − 2𝑖𝑆𝐶 − 3𝑖𝑐

0 = −2𝑖𝑎 + 10𝑖𝑠𝑐 − 1𝑖𝑐

𝑖𝑐 = 0.5𝑣∆

𝑣∆ = 2(𝑖𝑎 − 𝑖𝑠𝑐)

𝑅𝑡ℎ = 𝑣𝑡ℎ 𝑖𝑠𝑐 = 184.8 60 = 3.08 Ω

Continued Solution (problem 4.85):

Solving, 𝑖𝑏 = 26.4 A

Short circuit current:

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

فإذا ضاع ،الوقت هو الحياة الوقت ضاعت الحياة

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𝑝𝑚𝑎𝑥 =(92.4)2

3.08= 2772 W

440 − 220 = 5𝑖𝑎 − 2𝑖𝑏 − 35𝑖𝑐


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

to the load is:

From this circuit,

With 𝑅0 equal to 3.08 the original circuit becomes:

يكتب للمرء ما نوى

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Sept 2013




𝑖𝑐 = 0.5𝑣∆ 𝑣∆ = 2(𝑖𝑎 − 𝑖𝑏)

−92.4 = −2𝑖𝑎 + 3𝑖𝑏 − 1𝑖𝑐

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) = 15,711.52 W

% 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 =2772

38,896= 7.13 %


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

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

supplied is 38,896 W.

يمكنك أن تخدع اآلخرين أحيانا، لكن ماذا ستقول لنفسك؟

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Solution:

a)

Finding the Thevenin equivalent:

Open circuit voltage:


−240 + 3 𝑖1 − 𝑖2 + 20 𝑖1 − 𝑖3 + 2𝑖1 = 0

2𝑖2 + 4 𝑖2 − 𝑖3 + 3 𝑖2 − 𝑖1 = 0

10𝑖𝛽 + 1𝑖3 + 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.

𝑉𝑇ℎ = 20 𝑖1 − 𝑖3 = −24 V

ى أ حخخز غخشبس ه

.ف وبفت اجبالث، إهب اىخب

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Continued soluation (Problem 4.87):

Short-circuit current:


−240 + 3 𝑖1 − 𝑖2 + 2𝑖1 = 0

2𝑖2 + 4 𝑖2 − 𝑖3 + 3 𝑖2 − 𝑖1 = 0

10𝑖𝛽 + 1𝑖3 + 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


𝑖𝑠𝑐=

−24

−4 = 8 Ω

𝑅𝐿 = 𝑅𝑇ℎ = 6 Ω

b)

𝑝𝑚𝑎𝑥 =(12)2

6= 24 W

ىه أ حضبػف دبحه شاث

.وشاث، حىجذ طشق ػذذة

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Solution:

a) Finding the Thevenin equivalent:

(Open Circuit Voltage):


−100 + 4 𝑖1 − 𝑖2 + 80 𝑖1 − 𝑖3 + 16𝑖1 = 0

124𝑖Δ + 8 𝑖2 − 𝑖3 + 4 𝑖2 − 𝑖1 = 0

50 + 12𝑖3 + 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 = −80𝑖Δ = 48 V

(Short Circuit Current):


𝑖Δ = 0 124𝑖Δ = 0

−100 + 4 𝑖1 − 𝑖2 + 16𝑖1 = 0

8 𝑖2 − 𝑖3 + 4 𝑖2 − 𝑖1 = 0

50 + 12𝑖3 + 8 𝑖3 − 𝑖2 = 0

.البذ أ حخؼ شئب جذذا و ى

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Continue soluation (Problem 4.89):

Solving,

𝑖1 = 5 A,

𝑖2 = 0 A,

𝑖3 = −2.5 A.

𝑖𝑠𝑐 = 𝑖1 − 𝑖3 = 7.5 A 𝑅𝑇ℎ =48

7.5= 6.4 Ω

For maximum power transfer,

𝑅0 = 𝑅𝑇ℎ = 6.4 Ω

b)

𝑝𝑚𝑎𝑥 = 24 2

6.4= 90 W

c)

𝑖10Ω =48

6.4 + 10= 2.927 A

𝑝10Ω = 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.

إ ال أعخغ أبذا، دخ ػذب

.مىي ابط إ أجخ أبذا

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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.

𝑣1 − 120

6+

𝑣1

3+

𝑣1

2 + 4= 0 𝑣1 = 30 V

𝑖1′ =

120 − 30

6= 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 = −12 V 𝑣4 = −24 V

إرا ظج أه هضج

.فمذ هضج

Figure 4.63 The circuit shown in Fig. 4.62

with the current source deactivated.

𝑣3 𝑣4

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𝑖1′′ =

−𝑣3

6=

12

6= 2 A 𝑖2

′′ =𝑣3

3=

−12

3= −4 A

𝑖3′′ =

𝑣3 − 𝑣4

2=

−12 + 24

2= 6 A 𝑖4

′′ =𝑣4

4=

−24

4= −6 A

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

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.

أفض وعت بش

.ببىػذ أ ال حؼذ

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Sept 2013




Example 4.13:

Use the principle of superposition to find 𝑣0 in the

circuit shown in Fig. 4.66

Solution:

𝑣0′ = 𝑣𝑠

𝑅

𝑅𝑒𝑞

= 1020

25= 8 V

Applying the node-voltage equations yield:

𝑣0

′′

20+

𝑣0′′

5− 0.4𝑣∆

′′ = 0

Or 5𝑣0′′ − 8𝑣∆

′′ = 0

0.4𝑣∆′′ +

𝑣𝑏 − 2𝑖∆′′

10− 5 = 0

Or 4𝑣∆′′ + 𝑣𝑏 − 2𝑖∆

′′ = 50

Solving with:

𝑣𝑏 = 2𝑖∆′′ + 𝑣∆

′′

5𝑣∆′′ = 50 Or 𝑣∆

′′ = 10 V

5𝑣0′′ = 80 Or 𝑣0

′′ = 16 V

The value of 𝑣0 is the sum of 𝑣0′ and 𝑣0

′′ , or 24 V

.إرا ابخغ اهضو فمذ اخصش زة افىص

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Solution:

a) 110 V source acting alone:

𝑅𝑒 = 10||14 =35

6 Ω

𝑖′ =110

5 + 35 6 =

132

13 A

𝑣 ′ = 35

6

132

13 =

770

13 V = 59.231 V

4 A source acting alone:

5||10 = 10 3 Ω

10 3 + 2 = 16 3 Ω

16 3 12 = 48 13 Ω

Our circuit reduces to:

𝑣𝑎′′ = 4 48 13 = (192 13) V

𝑣 ′′ =−𝑣𝑎

′′

(16 3) 10 3 = −9.31 V

𝑣 = 𝑣 ′ + 𝑣 ′′ = 50 V

b) 𝑝 =𝑣2

10= 250 W

ا أالا ف أ ىث شخص آخش .اذغذ هى أ حبخغ عب

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Sept 2013




Solution:

240 V source acting alone:

𝑅𝑒𝑞 = ( 4 + 1 ||20) + 5 + 7

= (5| 20 + 12 = 4 + 12 = 16 Ω

𝑣𝑜1 = 5||20

16 240 = 60 V

84 V source acting alone:

𝑅𝑒𝑞 = ( 5 + 7 ||20) + 4 + 1

= (12||20) + 5 = 12.5 Ω

𝑣𝑜2 = 12||20

12.5 −84 = −50.4 V

16 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 = 𝑣𝑜1 + 𝑣𝑜2 + 𝑣𝑜3 = 28 V

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Sept 2013




𝑖0′′′ = −5

2

8 = −1.25 mA ∴ 𝑖0 = 3.5 − 1.25 = 2.25 mA

Solution:

a) By hypothesis 𝑖0′ + 𝑖0

" = 3 mA

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Sept 2013




𝑣𝑏

6+

𝑣𝑏 − 8

2+ 5 − 10 = 0

∴ 𝑣𝑏 = 13.5 V

𝑖0 =𝑣𝑏

6= 2.25 mA


b)

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

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Sept 2013




Solution:

Voltage source acting alone:

Using node voltage method:

𝑣𝑜1

20,000+

𝑣𝑜1 − 25

4000− 2.2

𝑣𝑜1 − 25

4000 = 0

𝑣𝑜1 = 30 V

Current source acting alone:

Using node voltage method:

𝑣𝑜2

20,000+ 0.005 +

𝑣𝑜2

4000− 2.2

𝑣𝑜2

4000 = 0

𝑣𝑜2 = 20 V

𝑣𝑜 = 𝑣𝑜1 + 𝑣𝑜2 = 30 + 20 = 50 V

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Sept 2013




−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

Solution:

The mesh equations are:

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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

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

𝑣1 = 18.4 𝑖𝑐 − 𝑖𝑒 = 113.90 V

𝑣2 = 38.4 𝑖𝑑 − 𝑖𝑒 = 120.19 V


Place these equations in standard form:

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:

𝑣3 = 11.6𝑖𝑒 = 233.62 V

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