Soluzioni Cap. 4

Charles K. Alexander, Matthew N.O. Sadiku, Circuiti elettrici, 4e

© 2014 McGraw-Hill Education (Italy) srl

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C apitolo 4

4 .

1

1 V

+

−

1 Ω i

8 Ω

5 Ω i o

3

8 ( 5 + 3 ) = 4 Ω , i =

1

1

+ 4

=

1

5 i o

=

1 i =

1

= 0.1A

2 10

Since the resistance remains the same we get i = 10/5 = 2A which leads to i o

= (1/2)i = (1/2)2 = 1A .

Ω

COSMOS: Complete Online Solutions Manual Organization System

4 .

2

6 ( 4 + 2 ) = 3 Ω , i

1

= i

2

=

1

2

A i o

=

1

2 i

1

=

1

4

, v o

= 2 i o

= 0.5V

1 A

If i s

= 1 µ A, then v o

= 0.5

µ V i

2

8 Ω

5 Ω i

1

6 Ω

4 Ω i o

2 Ω

C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl

Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku

© 2007 The McGraw-Hill Companies.


4 .

3

R

3R i o

V s

+

−

3R 3R

R

+ v

− o

1 V

(a)

(a) We transform the Y sub-circuit to the equivalent ∆ .

+

−

R 3 R =

3 R 2

=

3

R ,

3

R +

3

R =

3

R

4 R 4 4 4 2 v o

= v s independent of R

2 i o

= v o

/(R)

When v s

= 1V, v o

= 0.5V

, i o

= 0.5A

(b) When v s

= 10V, v o

(c) When v s

= 5V , i o

= 5A

= 10V and R = 10 Ω , v o

= 5V , i o

= 10/(10) = 500mA

(b)

3R 1.5R





4 .

4

If I o

= 1, the voltage across the 6 Ω resistor is 6V so that the current through the 3 Ω resistor is 2A.

2A

1A 3A

2 Ω

3A

2 Ω i

1

3 Ω 6 Ω 4 Ω I s 2 Ω v

1

+

−

4 Ω I s

(a) (b)

3 6 = 2 Ω , v o

= 3(4) = 12V, i

1

= v o =

4

Hence I s

= 3 + 3 = 6A

If I s

= 6A

I s

= 9A I

I o o

= 1

= 9/6 = 1.5A

3 A .





4 .

5

2 Ω v

1

V s

+

− 6 Ω

If v o

= 1V, V

1

=

1

3

+ 1 = 2 V

3 Ω

V s

= 2





2

3

If v s

=

10

3

v o

= 1

+ v

1

Then v s

= 15

=

10

3

v o

=

3

10 x 15 = 4.5V

v o

6 Ω 6 Ω





4 .

6

Due to linearity, from the first experiment,

V o

=

1

3

V s

Applying this to other experiments, we obtain:

Experiment V s

V o

3 1 V

4

0.333 V

-6 V -2V





4.7

If V o

= 1V, then the current through the 2-

Ω

and 4-

Ω

resistors is ½ = 0.5. The voltage across the 3-

Ω

resistor is ½ (4 + 2) = 3 V. The total current through the 1-

Ω

resistor is

0.5 +3/3 = 1.5 A. Hence the source voltage v s

= 1 1.5 3 4.5 V

If v s

= 4.5

⎯⎯→ 1 V

Then v s

= 4 ⎯⎯→

1

4.5

x =

=

888.9 mV

.





4 .

8

Let V o

V

1

= V

1

+ V

2

, where V

1

and V

, consider the circuit below.

2

are due to 9-V and 3-V sources respectively. To find

9 Ω

V

1

3 Ω

1 Ω

+

_

9 V

9 −

3

V

1 =

V

9

1 +

V

1

1 → V

1

=

To find V

2


V

1

9 Ω 3 Ω

+

_

3 V

V

9

2 +

V

3

2 =

3 −

1

V

2 → V

2

=

V o

= V

1

+ V

2

= 4.1538 V





4 .

9

Let v o v

1

= v

1

+ v

2

, where v

using the circuit below.

1

and v

2

are due to 6-A and 20-V sources respectively. We find

6 A

2 Ω v

+

1

_

1 Ω

2 Ω

4 Ω

2//2 = 1 Ω , v

1

= 1 x

4

We find v

2


2 Ω v

2

=

1

+ v

2

_

1 Ω

2 Ω

+

_

4 Ω

18 V v o

= v

1

+ v

2

= 4 + 3 = 7 V





4 .

10

Let v ab

= v ab1

+ v ab2

where v respectively. ab1

and v ab2

are due to the 4-V and the 2-A sources

10 Ω

3v

+ − ab1

10 Ω

3v

+ − ab2

4V

+

−

+ v ab1

2 A

(a)

−

(b)

For v ab1

, consider Fig. (a). Applying KVL gives,

- v ab1

– 3 v ab1

+ 10x0 + 4 = 0, which leads to v ab1

= 1 V

For v ab2

, consider Fig. (b). Applying KVL gives,

v ab2

– 3v ab2

+ 10x2 = 0, which leads to v ab2

= 5

+ v ab2

− v ab

= 1 + 5 = 6 V





4 .

11

Let v o find v

1

= v

1

+ v

2

, where v

1

and v

2

are due to the 6-A and 80-V sources respectively. To


6 A v a

40 Ω

I

1

+

10

V

1

Ω

_ v b

4 i

20

1

Ω

At node a,

6 = v a

40

+ v a

−

10 v b → v a

− 4 v b

(1)

At node b,

–I

1

– 4I

1

+ (v b

– 0)/20 = 0 or v b

= 100I

1

But i

1

= v a

− v b

10

which leads to 100(v a

–v b

)10 = v b

or v b

= 0.9091v

a

(2)

Substituting (2) into (1),

5v a

– 3.636v

a

= 240 or v a

= 175.95 and v b

= 159.96

However, v

1

= v a

– v b

= 15.99 V.

To find v

2





COSMOS: Complete Online Solutions Manual Organization System i o

+

10 v

2

Ω

_ f v c e

40 Ω 4 i o

But i o

0

=

−

50 v c

(0

+

−

50 v c

4 i o

)

+

( 30 −

20 v c

)

= 0 i

−

2 v

2

5

50

= v c

0

=

−

−

50

10 i

2

(30 v c

=

+

20

= v

0 10 1

2 V c

+

)

50

= 0

=

5

→ v c

= − 10 V

20 Ω

–

+

30 V v o

= v

1

+ v

2

=15.99 + 2 = 17.99 V and i o

= v o

/10= 1.799 A .




4 .

12

Let v o

For v o1

= v o1

+ v o2

+ v o3

, where v o1


, v o2

, and v o3

are due to the 2-A, 12-V, and 19-V sources respectively.

2A

2A

6 Ω 3 Ω

5

+ v

Ω o1

−

4

12

Ω

Ω i o

5

+ v

5

Ω o1

Ω

−

6||3 = 2 ohms, 4||12 = 3 ohms. Hence, i o

= 2/2 = 1, v o1

= 5io = 5 V

For v o2


6 Ω 5 Ω

+ v o2

−

12V

+

−

3 Ω

4

12

Ω

Ω

12V

+

−

6 Ω

+ v

1

3||8 = 24/11, v

1

= [(24/11)/(6 + 24/11)]12 = 16/5 v o2

= (5/8)v

1

= (5/8)(16/5) = 2 V

For v o3

, consider the circuit shown below.

5 Ω

+ v o3

−

4 Ω

+

5 Ω o3

−

4 Ω

6 Ω

3 Ω

12 Ω

−

19V 2 Ω

+ v

12 Ω

+ v

2

7||12 = (84/19) ohms, v

2

= [(84/19)/(4 + 84/19)]19 = 9.975 v = (-5/7)v2 = -7.125 v o

= 5 + 2 – 7.125 = -125 mV

5

+ v

3

Ω o2

Ω

−

+

−

3

19V

Ω



4 .

13

Let v o

= v v

2

+ v

3

, where v

1

, v

2

, and v

3

are due to the independent sources. To find v

1


2 A 10 Ω v

1

= 5 x

10 x =

8 Ω

5 Ω v

+

1

_

To find v

2


10 Ω v

2

= 5 x

8 x =

8

4 A

Ω

5 Ω

+ v

2

_

To find v

3






8 Ω

12 V

+ –

10 Ω v

3

= − 12



 +

5

+





= − 2.6087

v o

= v v

2

+ v

3

=

5 Ω

8.6956 V = 8.696V

.

+ v

3

_




4 .

14

Let v o

For v o1

= v o1

+ v o2

+ v o3

, where v o1


, v o2

, and v o3

, are due to the 20-V, 1-A, and 2-A sources respectively.

4 Ω

6 Ω

2 Ω

+

−

20V v

+ o1

3 Ω

6||(4 + 2) = 3 ohms, v o1

= (½)20 = 10 V

For v o2


4 Ω

6 Ω

2 Ω

+

4V

− +

4 Ω

1A v o2

3 Ω

3||6 = 2 ohms, v o2

= [2/(4 + 2 + 2)]4 = 1 V

For v o3


4 Ω

6 Ω

2A

2 Ω v

+ o3

3 Ω

3

3

6

Ω

Ω

Ω

2

2A

Ω v

+ o2

3 Ω

6||(4 + 2) = 3, v o3

= (-1)3 = –3 v o

= 10 + 1 – 3 = 8 V

− v o3

+


4 .

15

Let i = i

1

+ i

2

+ i

3

, where i

1

, i

2

, and i

3

are due to the 20-V, 2-A, and 16-V sources. For i

1

, consider the circuit below. i o

20V

+

−

1 Ω i

1 4 Ω

2 Ω 3 Ω

4||(3 + 1) = 2 ohms, Then i o

= [20/(2 + 2)] = 5 A, i

1

= i o

/2 = 2.5 A

For i

3


2 Ω v o

+

’

− i

3

3

1

Ω

Ω 4 Ω

−

+

16V

2 Ω

2||(1 + 3) = 4/3, v o

’ = [(4/3)/((4/3) + 4)](-16) = -4 i

3

= v o

’/4 = -1

For i

2


1 Ω 2A

(4/3) Ω

1 Ω

3 i

2

Ω

4 Ω

3 i

2

Ω


2A


2||4 = 4/3, 3 + 4/3 = 13/3

Using the current division principle. i

2

= [1/(1 + 13/2)]2 = 3/8 = 0.375 i = 2.5 + 0.375 - 1 = 1.875 A p = i 2 R = (1.875) 2 3 = 10.55 watts





4 .

16

Let i o

= i o1

+ i o2

+ i o3

, where i o1

, i o2

, and i the 12-V, 4-A, and 2-A sources. For i o1 o3

are due to

, consider the circuit below. i o1

4 Ω 3 Ω 2 Ω

12V

+

−

10 Ω 5 Ω

10||(3 + 2 + 5) = 5 ohms, i o1

= 12/(5 + 4) = (12/9) A

For i o2


4A

4 Ω i o2

10 Ω

3 i

1

Ω 2 Ω

5 Ω

2 + 5 + 4||10 = 7 + 40/14 = 69/7 i

1

= [3/(3 + 69/7)]4 = 84/90, i o2

=[-10/(4 + 10)]i

1

= -6/9

For i o3

, consider the circuit below. i o3

3 Ω 2 Ω i

2

4 Ω 10 Ω 5 Ω

3 + 2 + 4||10 = 5 + 20/7 = 55/7 i

2

= [5/(5 + 55/7)]2 = 7/9, i o3

= [-10/(10 + 4)]i

2

= -5/9 i o

= (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA

2 A




90V

4 .

17

Let v x

= v x1

+ v x2 sources. For v x1

+ v x3

, where v x1

,v x2

, and v

, consider the circuit below. x3

are due to the 90-V, 6-A, and 40-V

−

+

30 Ω

60 Ω

10 Ω x1

+

30

−

Ω

20 Ω

3 A 20 Ω i o

10 Ω v x1

+

12

−

Ω

20||30 = 12 ohms, 60||30 = 20 ohms

By using current division, i o

= [20/(22 + 20)]3 = 60/42, v x1

= 10i o

= 600/42 = 14.286 V

For v x2


10 Ω i o

’

10 Ω i o

’

+ − + v x2

−

30 Ω 60 Ω 6A 30 Ω 20 Ω 20 Ω 6A

12 Ω i o

’ = [12/(12 + 30)]6 = 72/42, v x2

= –10i o

’ = –17.143 V

For v x3


10 Ω

+ x3

−

20 Ω 10 Ω

+ x3

− i o

”

30 Ω 60 Ω 30 Ω

40V

+

−

20 Ω 12 Ω

4A i o

” = [12/(12 + 30)]2 = 24/42, v x3

= -10i o

” = -5.714= [12/(12 + 30)]2 = 24/42, v x3

= -10i o

” = -5.714

= [12/(12 + 30)]2 = 24/42, v x3

= -10i o

” = -5.714 v x

= 14.286 – 17.143 – 5.714 = -8.571 V



4 .

18

Let V o find V

1

= V

1

+ V

2, where V

1

and V

, we use the circuit below.

2

are due to 10-V and 2-A sources respectively. To

1 Ω

10 V

+

_

2 Ω

0.5 V

1

+

V

1

_

2 Ω 1 Ω

0.5 V

- +

1

+

10 V

+

_ i

4 Ω

V

1

_

-10 + 7i – 0.5V

1

= 0

But V

1

= 4i

` i i = 5 i → i = 2, V

1

= 8 V





To find V

2


2 Ω

2 A

2 Ω 1 Ω

1 Ω

0.5 V

2

0.5 V

- +

2

4 V

+

_ i

4 Ω

4 Ω

- 4 + 7i – 0.5V

2

=0

But V

2

= 4i

4 7 2 i = 5 i → i = 0.8,

V o

= V

1

+ V

2

V

2

= 4 i =

= 8 +3.2 = 11.2 V

3.2

+

V

2

_

+

V

2

_





4 .

19

Let v x

= v

1

+ v

2

, where v

1

and v

2

are due to the 4-A and 6-A sources respectively.

2 i

Ω x

−

4i

+ x

4 A v

1

8 Ω

+ v

−

1

2 i

Ω x v

2

6 A

−

8

+

4i

Ω x

+ v

−

2

(a)

To find v

1

, consider the circuit in Fig. (a).

(b) v

1

/8 – 4 + (v

1

– (–4i x

))/2 = 0 or (0.125+0.5)v

1

= 4 – 2i x

or v

1

= 6.4 – 3.2i

x

But, i x

= (v

1

– (–4i x

))/2 or i x

= –0.5v

1

. Thus, v

1

= 6.4 + 3.2(0.5v

1

), which leads to v

1

= –6.4/0.6 = –10.667

To find v

2

, consider the circuit shown in Fig. (b). v

2

/8 – 6 + (v

2

– (–4i x

))/2 = 0 or v

2

+ 3.2i

x

= 9.6

But i x

= –0.5v

2

. Therefore, v

2

+ 3.2(–0.5v

2

) = 9.6 which leads to v

2

= –16

Hence,

Checking, v x

= –10.667 – 16 = –26.67V

. i x

= –0.5v

x

= 13.333A

Now all we need to do now is sum the currents flowing out of the top node.

13.333 – 6 – 4 + (–26.67)/8 = 3.333 – 3.333 = 0





4 .

20

Convert the voltage sources to current sources and obtain the circuit shown below.

3 A

10 Ω 0.6 20 Ω 0.4

R

1 eq

=

1 1 1

10 20 40

= + →

R

R eq

=

4 A

I eq

= 3 + 0.6 + 0.4 = 4

Thus, the circuit is reduced as shown below. Please note, we that this is merely an exercise in combining sources and resistors. The circuit we have is an equivalent circuit which has no real purpose other than to demonstrate source transformation. In a practical situation, this would need some kind of reference and a use to an external circuit to be of real value.

5.714 Ω

18.285 V

+

_

5.714 Ω


40 Ω




4 .

21

To get i o i o

, transform the current sources as shown in Fig. (a).

6 Ω 3 Ω

+

−

12V

(a)

−

+

6V 2 A

6 Ω 3 Ω

(b) i

From Fig. (a), -12 + 9i o

+ 6 = 0, therefore i o

= 666.7 mA

To get v o

, transform the voltage sources as shown in Fig. (b). i = [6/(3 + 6)](2 + 2) = 8/3 v o

= 3i = 8 V

+

− v o

2 A





4 .

22

We transform the two sources to get the circuit shown in Fig. (a).

5 Ω 5 Ω

−

+

10V

4 Ω 10 Ω

2A

1A 10 Ω 4 Ω i

(a)

10 Ω

2A

(b)

We now transform only the voltage source to obtain the circuit in Fig. (b).

10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA





4 .

23

If we transform the voltage source, we obtain the circuit below.

8 Ω

3A

10 Ω 6 Ω 3 Ω

3//6 = 2-ohm. Convert the current sources to voltages sources as shown below.

10 Ω 8 Ω 2 Ω

+

30V

-

Applying KVL to the loop gives

5A

+

10V

-

− 30 + 10 + I ( 10 + 8 + 2 ) = 0 I = 1 A p = VI = I 2 R = 8 W





4 .

24

Transform the two current sources in parallel with the resistors into their voltage source equivalents yield, a 30-V source in series with a 10Ω resistor and a 20V x series with a 10Ω resistor.

-V sources in

We now have the following circuit,

+

8 Ω

V x

–

– +

10 Ω

40 V

+

_

30 V

10 Ω

I 20V x

+

–

We now write the following mesh equation and constraint equation which will lead to a solution for V x

,

28I – 70 + 20V x

= 0 or 28I + 20V x

= 70, but V x

= 8I which leads to

28I + 160I = 70 or I = 0.3723 A or V x

= 2.978 V .





4 .

25

Transforming only the current source gives the circuit below.

9 Ω

18 V

− +

12V

–

+

5 Ω

4 Ω

+

2 Ω

− i

+ −

30 V

−

+

30 V

Applying KVL to the loop gives,

–(4 + 9 + 5 + 2)i + 12 – 18 – 30 – 30 = 0

20i = –66 which leads to i = –3.3 v o

= 2i = –6.6 V





4 .

26

Transforming the current sources gives the circuit below.

2 Ω

15 V

– +

5 Ω

12 V

+

_

–12 + 11i o

–15 +20 = 0 or 11i o

= 7 or i o

= 636.4 mA . i o

4 Ω

+

_ 20 V





4 .

27

Transforming the voltage sources to current sources gives the circuit in Fig. (a).

10||40 = 8 ohms

Transforming the current sources to voltage sources yields the circuit in Fig. (b).

Applying KVL to the loop,

-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4 v x

12i = -48 V

5A

10 Ω 40 Ω

12

+ v

Ω x

−

8A

20 Ω

2A

+

−

8 Ω

40V

(a)

12

+ v i

Ω x

−

(b)

20 Ω

+

− 200V





4.28

Convert the dependent current source to a dependent voltage source as shown below.

1

Ω i o

4

Ω

3

Ω

+ V o

_

8 V

+

_

–

+

V o

Applying KVL,

8 i o

(1 4 3) V o

= 0

But

V o

= 4 i o

− + i o

− 4 i o

= 0 ⎯⎯→ i o

= 2 A





4 .

29

Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 =

(4/3) k ohms

4 k Ω

2 k Ω

(4/3) k Ω

2v o

1.5v

o

− +

+ i

+ 3 mA

1 k Ω v o

3 mA

1 k Ω v o

−

−

(a) (b)

It is clear that i = 3 mA which leads to v o

= 1000i = 3 V

If the use of source transformations was not required for this problem, the actual answer could have been determined by inspection right away since the only current that could have flowed through the 1 k ohm resistor is 3 mA.




4 .

30

Transform the dependent current source as shown below.

i x

24 Ω 60 Ω

+

12V

-

30 Ω

10 Ω

+

7i

- x

Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below.

+

12V

i x

24 Ω

30 Ω 70 Ω 0.1i

x

-

Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below.

i x

24 Ω 21 Ω

+

12V

+

2.1i

x

- -

Applying KVL to the loop gives

45 i x

− 12 + 2 .

1 i x

= 0 i x

=

12

47 .

1

= 254 .

8 mA



4 .

31

Transform the dependent source so that we have the circuit in

Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in

Fig. (b).

3 Ω

12V

+

+ x

−

−

8 Ω 6 Ω v x

/3

From Fig. (b),

12V

+

−

3 Ω

+ x

−

(a)

(b) i

(24/7) Ω

+

−

(8/7)v x v x

= 3i, or i = v x

/3.

Applying KVL,

-12 + (3 + 24/7)i + (24/21)v x

= 0

12 = [(21 + 24)/7]v x

/3 + (8/7)v x

, leads to v x

= 84/23 = 3.652 V





4 .

32

As shown in Fig. (a), we transform the dependent current source to a voltage source,

15 Ω 10 Ω

5i x

− +

60V

+

−

50 Ω 40 Ω

60V

+

−

15 Ω

(a)

50 Ω

(b)

50 Ω

0.1i

x

60V

− i

+ x

15 Ω i x

25 Ω

−

+

(c)

In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c),

-60 + 40i x

– 2.5i

x

= 0, or i x

= 1.6 A

2.5i

x





4 .

33

(a) R

Th

= 10||40 = 400/50 = 8 ohms

V

Th

= (40/(40 + 10))20 = 16 V

(b) R

Th

= 30||60 = 1800/90 = 20 ohms

2 + (30 – v1)/60 = v

1

/30, and v

1

= V

Th

120 + 30 – v

1

= 2v

1

, or v

1

= 50 V

V

Th

= 50 V





4 .

34

To find R

Th


10 Ω

40 Ω

20 Ω

R

Th

+

−

(a)

10

40V

Ω

40 v

1

Ω

(b)

3 A

20 Ω v

2

+

V

Th

R

Th

= 20 + 10||40 = 20 + 400/50 = 28 ohms

To find V

Th

, consider the circuit in Fig. (b).

At node 1, (40 – v

1

)/10 = 3 + [(v

1

– v

2

)/20] + v

1

/40, 40 = 7v

1

– 2v

2

(1)

At node 2, 3 + (v1- v2)/20 = 0, or v1 = v2 – 60 (2)

Solving (1) and (2), v

1

= 32 V, v

2

= 92 V, and V

Th

= v

2

= 92 V





4 .

35

To find R

Th


R

Th

= R ab

= 6||3 + 12||4 = 2 + 3 =5 ohms

To find V

Th

, consider the circuit shown in Fig. (b).

R

Th a b

6 Ω 3 Ω 12 Ω 4 Ω

6 Ω v

1

(a)

2 A

V

Th v

2

+

−

12V v

1

+

−

3

+

Ω 12 Ω

+ v

−

2

4 Ω

+

−

19V

(b)

At node 1, 2 + (12 – v

1

)/6 = v

1

/3, or v

1

= 8

At node 2, (19 – v

2

)/4 = 2 + v

2

/12, or v

2

= 33/4

But, -v

1

+ V

Th

+ v

2

= 0, or V

Th

= v

1

– v

2

= 8 – 33/4 = -0.25




COSMOS: Complete Online Solutions Manual Organization System a + −

R

Th

= 5 Ω

V

Th

10

+

Ω

−

= (-1/4)V b v o

= V

Th

/2 = -0.25/2 = –125 mV





4 .

36

Remove the 30-V voltage source and the 20-ohm resistor.

10 Ω

R

Th a b

40 Ω

10 Ω

+

−

50V

+

V a

Th b

40

From Fig. (a),

From Fig. (b),

(a)

R

Th

= 10||40 = 8 ohms

(b)

V

Th

= (40/(10 + 40))50 = 40V

8 Ω i a

+

−

12 Ω

40V

+

− b

30V

(c)

The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL,

30 – 40 + (8 + 12)i = 0, which leads to i = 500mA

Ω




4 .

37

R

N

is found from the circuit below.

20 Ω

a

40 Ω

12 Ω

b

R

N

= 12 //( 20 + 40 ) = 10 Ω

I

N

is found from the circuit below.

2A

20 Ω

+ 40 Ω

a

120V

-

12 Ω

I

N b

Applying source transformation to the current source yields the circuit below.

20 Ω 40 Ω + 80 V -

+

120V

-

I

N

Applying KVL to the loop yields



− 120 + 80 + 60 I

N

= 0 I

N

= 40 / 60 = 666.7 mA .





4 .

38

We find Thevenin equivalent at the terminals of the 10-ohm resistor. For R

Th the circuit below.

1 Ω

4 Ω

, consider

5 Ω

R

Th

16 Ω

R

Th

= 1 + 5 //( 4 + 16 ) = 1 + 4 = 5 Ω

For V

Th


1 Ω

V

1

4 Ω V

2

5 Ω +

3A 16 Ω V

Th

+

-

12 V

-

At node 1,

3 =

V

1 +

V

1

−

16

At node 2,

V

1

−

4

V

2 +

12

4

−

5

V

2

V

2 = 0

Solving (1) and (2) leads to

V

Th

= V

2

= 19 .

2

48 = 5 V

1

48 =

− 4 V

2

− 5 V

1

+ 9 V

2

(1)

(2)





Thus, the given circuit can be replaced as shown below.

5 Ω

+ +

19.2V V o

10 Ω

- -

Using voltage division,

V o

=

10

10

+ 5

( 19 .

2 ) = 12 .

8 V




4 .

39

We obtain R

Th


10 Ω

10 Ω 5 Ω

16

R

Th

= = +

25

= 20

To find V

Th


Ω

10

24

Ω

At node 1,

24 −

10

At node 2,

V

1

+

_

V

1

10

V V

10

2

+

V

2

_

3 A

→

V

2

5

16

54 2

1

−

R

Th

V

+

Th

_

2

(1)



V V

10

2

V

5

2 →

Substracting (1) from (2) gives

6 = − 5 V

1

→

=

V

2

=

V

1

− 6 V

1.2 V

2

(2)

But

V 16 3 + V

Th

= 0 → V

Th

= − 49.2 V





4 .

40

To obtain V

− + +

Th

, we apply KVL to the loop.

70 (10 20) kI + 4 V o

= 0

But V o

= 10

70 70 kI kI

→ I = 1 mA

− + +

Th

=

I

2

I

R

Th

0 → V

Th

= 60 V

To find R

Th

, we remove the 70-V source and apply a 1-V source at terminals a-b, as shown in the circuit below. a

I f

1

–

V o

10 k Ω

I

2 c

20 Ω

+

+

_

+

–

4 V o b

We notice that V o

− + kI

1

+ 4 V o

= -1 V.

= 0 → I

1

= 0.25 mA

1 V

10 k

=

1 V

I

2

=

= 0.35 mA

1

0.35

k Ω = 2.857 k Ω





4 .

41

To find R

Th

, consider the circuit below

14 Ω

6 Ω 5 Ω

a

b

R

Th

= 5 //( 14 + 6 ) = 4 Ω = R

N

Applying source transformation to the 1-A current source, we obtain the circuit below.

6 Ω - 14V + 14 Ω V

Th

a

+

6V 3A

-

5 Ω

b

At node a,

14 + 6 − V

Th

6 + 14

I

N

=

V

Th

R

Th

=

Thus,

R

Th

= R

N

=

= 3 +

V

Th

5

( − 8 ) / 4

4 Ω ,

=

V

Th

− 2 A

= − 8 V,

V

Th

I

N

= − 8 V

= − 2 A




4 .

42

To find R

Th


10 Ω

20

20

Ω

Ω a

10 Ω

10 Ω

10 Ω b

(a) a

10 Ω

30

30

Ω

Ω

(b)

30 Ω

10

20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b).

10||30 = 7.5 ohms. R

Th

= R ab

= 30||(7.5 + 7.5) = 10 ohms . b

Ω

To find V

Th

, we transform the 20-V and the 5-V sources. We obtain the circuit shown in

Fig. (c).

10 Ω 10 Ω

10 V

− +

10 Ω 10 Ω

30V

+

+

− i

1

50V

+

− i

2

10 Ω

(c)

For loop 1, -30 + 50 + 30i

1

– 10i

2

Solving (1) and (2),

= 0, or -2 = 3i i

1

= 0, i

2

= 2 A

1

– i

For loop 2, -50 – 10 + 30i

2

– 10i

1

= 0, or 6 = -i

1

+ 3i

2

2

(1)

(2)

Applying KVL to the output loop, -v ab

– 10i

V

Th

= v

1

+ 30 – 10i ab

= 10 volts

2

= 0, v ab

= 10 V



4 .

43

To find R

Th


10 Ω 10 Ω

R

Th a b

(a)

5 Ω

+

−

10 Ω

20V v a

+

− a

10

+

Ω

V

Th

(b)

+ v

− b b

5 Ω

R

Th

= 10||10 + 5 = 10 ohms

To find V

Th

, consider the circuit in Fig. (b). v b

= 2x5 = 10 V, v a

= 20/2 = 10 V

But, -v a

+ V

Th

+ v b

= 0, or V

Th

= v a

– v b

= 0 volts

2 A





4 .

44

(a) For , consider the circuit in Fig. (a).

R

Th

= 1 + 4||(3 + 2 + 5) = 3.857 ohms

For V

Th

, consider the circuit in Fig. (b). Applying KVL gives,

10 – 24 + i(3 + 4 + 5 + 2), or i = 1

V

Th

= 4i = 4 V

3 Ω 1 Ω a

2 Ω

4 Ω b

R

Th

5 Ω

(a)

(b) For , consider the circuit in Fig. (c).

+

− 24V

2

3

Ω

Ω i

+

−

4

10V

5

1

Ω

Ω

(b)

Ω

2

3

Ω

Ω

4

5

(c)

1

Ω

Ω

Ω b c

R

Th

24V

+

−

2

3

Ω

Ω

5 Ω v

4 o

1

Ω

(d)

Ω

2A a b b c

+

+

V

V

Th

Th





R

Th

= 5||(2 + 3 + 4) = 3.214 ohms

To get V

Th

, consider the circuit in Fig. (d). At the node, KCL gives,

[(24 – vo)/9] + 2 = vo/5, or vo = 15

V

Th

= vo = 15 V





4 .

45

For R

N


6 Ω

6 Ω

4 Ω R

N

4A

6

6

Ω

Ω

4 Ω

I

N

(a) (b)

R

N

= (6 + 6)||4 = 3 ohms

For I

N

, consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors. Hence,

I

N

= 4/2 = 2 A





4 .

46

R

N

is found using the circuit below.

10 Ω

10 Ω

20 Ω c

R

N

= 20//(10+10) = 10 Ω

To find I

N

, consider the circuit below. c

10 Ω a b

4 A

10 Ω

The 20Ω resistor is short-circuited and can be ignored.

I

N

= ½ x 4 = 2 A

R

N

20 Ω b I

N





4 .

47

Since V

Th

= V ab

30 −

12

V

Th

= V

= x

V

Th

60

, we apply KCL at the node a and obtain

+ 2 V

Th

V

Th

= 150 / 126 = 1 .

19 V

To find R

Th


12 Ω V x a

At node a, KCL gives

1 = 2 V x

+

V x

60

+

V x

12

R

Th

=

V x

1

= 0 .

4762 Ω ,

Thus,

V

Th

= 1 .

19 V , R

Th

= R

N

V x

I

N

=

V

Th

R

Th

= 0 .

4762 Ω ,

60 Ω

= 60 / 126 = 0 .

4762

= 1 .

19 / 0 .

4762 =

I

N

= 2 .

5 A

2V

2 .

5 x

1A





4 .

48

To get R

Th


I o

4

+

Ω

−

10I o 2 Ω

V

+

−

1A

From Fig. (a),

(a)

I o

= 1,

I

2A o

4

+

Ω

−

10I

6 – 10 – V = 0, or V = -4

(b) o

R

N

= R

Th

= V/1 = -4 ohms

To get V

Th

, consider the circuit in Fig. (b),

I o

= 2, V

Th

= -10I o

+ 4I o

= -12 V

I

N

= V

Th

/R

Th

= 3A

2 Ω

+

V

Th

−





4 .

49

R

N

= R

Th

= 28 ohms

To find I

N

, consider the circuit below,

3A

10 Ω v o

20 Ω

40V

+

−

40 Ω i o

I sc

= I

N

At the node, (40 – v o

)/10 = 3 + (v o

/40) + (v o

/20), or v o

= 40/7 i o

= v o

/20 = 2/7, but I

N

= I sc

= i o

+ 3 = 3.286 A





4 .

50

From Fig. (a), R

N

= 6 + 4 = 10 ohms

4

6

Ω

Ω

2A

4

6

Ω

Ω

12V

+

−

I sc

= I

N

From Fig. (b),

(a) (b)

2 + (12 – v)/6 = v/4, or v = 9.6 V

-I

N

= (12 – v)/6 = 0.4, which leads to I

N

= -0.4 A

Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c).

0.4A

10 Ω

(c) i

5 Ω

4A i = [10/(10 + 5)] (4 – 0.4) = 2.4 A





4 .

51

(a) From the circuit in Fig. (a),

R

N

= 4||(2 + 6||3) = 4||4 = 2 ohms

R

Th +

V

Th

6 Ω 4 Ω 6 Ω 4 Ω

3 Ω 2 Ω

120V

+

−

3 Ω 6A

2

(a) (b)

For I

N

or V

Th

, consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c).

Ω

2 Ω

+ V

4

Th

Ω 2 Ω

40V

+

− i

(c)

12V

+

−

Applying KVL to the circuit in Fig. (c),

-40 + 8i + 12 = 0 which gives i = 7/2

V

Th

= 4i = 14 therefore I

N

= V

Th

/R

N

= 14/2 = 7 A





(b) To get R

N

, consider the circuit in Fig. (d).

R

N

= 2||(4 + 6||3) = 2||6 = 1.5 ohms

6 Ω 4 Ω 2 Ω

+

3 Ω 2 Ω R

N V

Th 12V

+

− i

(d) (e)

To get I

N

, the circuit in Fig. (c) applies except that it needs slight modification as in

Fig. (e). i = 7/2, V

Th

= 12 + 2i = 19, I

N

= V

Th

/R

N

= 19/1.5 = 12.667 A





4 .

52

For R

Th


3 k Ω

I o

20I o

2 k Ω

(a)

6V

+

−

3 k Ω

I o

2 k Ω

20I o

(b)

For Fig. (a), I o

= 0, hence the current source is inactive and a b a b

+

R

V

R

Th

= 2 k ohms

For V

Th


V

Th

= (-20I o

I o

= 6/3k = 2 mA

)(2k) = -20x2x10 -3 x2x10 3 = -80 V

Th

Th





4 .

53

To get R

Th


6 Ω 3 Ω

From Fig. (b),

+ v

− o

0.25v

2

(a)

Ω o a

1A b

2 Ω

+ v

− o

0.25v

2 Ω o

1/2

(b) v

1/2 ab

+

− a b

1A v o

= 2x1 = 2V, -v ab

+ 2x(1/2) +v o

= 0 v ab

= 3V

R

N

= v ab

/1 = 3 ohms

To get I

N

, consider the circuit in Fig. (c).

0.25v

o

6 Ω

+

2 Ω a

18V

+

−

3 Ω v

− o I sc

= I

N b

(c)

[(18 – v o

)/6] + 0.25v

o

= (v o

/2) + (v o

/3) or v o

= 4V

But, (v o

/2) = 0.25v

o

+ I

N

, which leads to I

N

= 1 A





4 .

54

To find V

Th

− 3 +

=V

1000 i o x

, consider the left loop.

+ 2 V x

= 0 3 = 1000 i o

For the right loop,

V x

= − 50 x 40 i o

= − 2000 i o

Combining (1) and (2),

3 = 1000 i o

− 4000 i o

= − 3000 i o i o

V x

= − 2000 i o

= 2 V

Th

= 2

+

=

2 V x

− 1

(1) mA

(2)

.

To find R

Th

, insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below.

1 k Ω i x i o

+

2V x

-

40i o

+

V x

-

+

50 Ω 1V

-

V x

= 1 , i o

= −

2 V x

1000

= − 2 mA i x

= 40 i o

+

V x

50

= − 80 mA +

1

50

A = -60mA

R

Th

= i

1 x

= − 1 / 0 .

060 = − 16 .

67 Ω





4 .

55

To get R

N

, apply a 1 mA source at the terminals a and b as shown in Fig. (a).

8 k Ω

I

+ a

2V

+

− v ab

/1000

+

−

80I

50 k Ω v ab

−

I

N b

(b)

We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At node a,

(v ab

/50) + 80I = 1 (1)

Also,

(2) -8I = (v ab

/1000), or I = -v ab

/8000

From (1) and (2), (v ab

/50) – (80v ab

/8000) = 1, or v ab

= 100

R

N

= v ab

/1 = 100 k ohms

To get I

N


I

8 k Ω v ab

/1000

+

−

80I

50 k Ω v

Since the 50-k ohm resistor is shorted,

(a)

I

N

= -80I, v ab

= 0

Hence, 8i = 2 which leads to I = (1/4) mA

I

N

= -20 mA ab

+

− a b

1mA




4 .

56

We remove the 1-k Ω resistor temporarily and find Norton equivalent across its terminals.

R

N

is obtained from the circuit below.

12 k Ω 2 k Ω 10 k Ω c

24 k Ω c

R

N

R

N

= 10 + 2 + 12//24 = 12+8 = 20 k Ω

I

N

is obtained from the circuit below.

12 k

36 V

+

_

24 k Ω

2 k 10 k

3 mA b I

N

We can use superposition theorem to find I

N

16-V and 3-mA sources respectively. We find I

12 k

. Let I

N

= I

1

1


2 k

+ I

2

, where I

1

and I

10 k

2

are due to

36 V

+

_

24 k Ω b I

1



Using source transformation, we obtain the circuit below.

12//24 = 8 k Ω

I

1

=

8

3 mA

(3 mA =

12 k 24 k

12 k b

To find I

2


12 k 24 k

2 k 10 k

3 mA b

2k + 12k//24 k = 10 k Ω

I

2

=0.5(-3mA) = -1.5 mA

I

N

= 1.2 –1.5 = -0.3 mA




I

2

I

1


The Norton equivalent with the 1-k Ω resistor is shown below a

+

I n

20 k Ω

V

– o

1 k Ω b

20

+





V o

= 1 k








4 .

57

To find R

Th

, remove the 50V source and insert a 1-V source at a – b, as shown in Fig. (a).

+

B

2 Ω

A a i

3 Ω v x

−

6 Ω

(a)

0.5v

x

10 Ω b

+

−

1V

We apply nodal analysis. At node A, i + 0.5v

x

= (1/10) + (1 – v x

)/2, or i + v x

= 0.6

At node B,

(1)

(2) (1 – v o

)/2 = (v x

/3) + (v x

/6), and v x

= 0.5

From (1) and (2), i = 0.1 and

R

Th

= 1/i = 10 ohms

To get V

Th


50V

−

+

3 Ω v x

+

− v

1

6 Ω

2 Ω v

(b)

2

0.5v

x

10 a

Ω b

+

V

−

Th

At node 1, (50 – v

1

)/3 = (v

1

/6) + (v

1

– v

2

)/2, or 100 = 6v

1

– 3v

2

(3)

At node 2, 0.5v

x

+ (v

1

– v

2

)/2 = v

2

/10, v x

= v

1

, and v

1

= 0.6v

2

(4)

From (3) and (4), v

2

= V

Th

= 166.67 V

I

N

= V

Th

/R

Th

= 16.667 A

R

N

= R

Th

= 10 ohms



4 .

58

This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of R

N infinity . I

N

can be found by solving for I sc

.

= i b

R

1 v o

β i b

V

S

+

−

R

2

Writing the node equation at node vo, i b

+ β i b

= v o

/R

2

= (1 + β )i b

But i b

= (V s

– v o

)/R

1

I sc v o

= V s

– i b

R

1

V s

– i b

R

1

= (1 + β )R

2 i b

, or i b

= V s

/(R

1

+ (1 + β )R

2

)

I sc

= I

N

= β i b

= β V s

/(R

1

+ (1 + β )R

2

)





4 .

59

R

Th

= (10 + 20)||(50 + 40) 30||90 = 22.5 ohms

To find V

Th

, consider the circuit below. i

1

10 Ω

+

8A 50 Ω

V

Th i

2

20 Ω

40 Ω i

1

= i

2

= 8/2 = 4, 10i

1

+ V

Th

– 20i

2

= 0, or V

Th

= 20i

2

–10i

1

= 10i

1

= 10x4

V

Th

= 40V , and I

N

= V

Th

/R

Th

= 40/22.5 = 1.7778 A




4 .

60

The circuit can be reduced by source transformations.

2A

18 V 12 V

10 Ω

+

−

+

−

10 V

+

−

2A

5 Ω a a

3A

3.333

Ω

Norton Equivalent Circuit

10 b

5

Ω

3A

Ω

2A a

3.333

Ω b

10 V

+

−

Thevenin Equivalent Circuit b



4 .

61

To find R

Th


Let R = 2||18 = 1.8 ohms , R

Th

= 2R||R = (2/3)R = 1.2 ohms .

To get V

Th

, we apply mesh analysis to the circuit in Fig. (d).

6 Ω

2 Ω

6 Ω a

2 Ω 2 Ω

6 Ω

2 Ω

18 Ω

(a) a b

2 Ω

18 Ω 18

(b)

Ω

2 Ω b

1.8 Ω

1.8 Ω

1.8 Ω

(c) a

R b


Th




2 Ω

12V

+

−

6 Ω i

3

2 Ω i

1

−

+

6 Ω

12V

6 i

2

Ω

12V

+

−

2 Ω a b

+

V

Th

(d)

-12 – 12 + 14i

1

– 6i

2

– 6i

3

= 0, and 7 i

1

– 3 i

2

– 3i

3

= 12 (1)

12 + 12 + 14 i

2

– 6 i

1

– 6 i

3

= 0, and -3 i

1

+ 7 i

2

– 3 i

3

= -12 (2)

– 6 i

1

– 6 i

2

= 0, and -3 i

1

– 3 i

2

+ 7 i

3

= 0 (3)

This leads to the following matrix form for (1), (2) and (3),









−

−

7

3

3

−

−

7

3

3

−

−

7

3

3















 i i i

1

2

3









=









−

12

12

0









7 − 3 − 3 7 12 − 3

∆ = − 3

− 3

7

− 3

− 3

7

= 100 , ∆

2

= − 3

− 3

− 12

0

− 3

7

= − 120 i

2

= ∆ / ∆

2

= -120/100 = -1.2 A

V

Th

= 12 + 2i

2

= 9.6 V , and I

N

= V

Th

/R

Th

= 8 A





4 .

62

Since there are no independent sources, V

Th

= 0 V

To obtain R

Th


0.1i

o

+ v o

−

1

40 Ω

2v o

+

−

At node 2,

2 v

10 i o

1

20

Ω

Ω i x

V

S

+

− i x

+ 0.1i

o

= (1 – v

1

)/10, or 10i x

+ i o

= 1 – v

1

(1)

At node 1,

(v

1

/20) + 0.1i

o

= [(2v o

– v

1

)/40] + [(1 – v

But i o

= (v

1

/20) and v o

= 1 – v

1

, then (2) becomes,

1

)/10] (2)

1.1v

1

/20 = [(2 – 3v

1

)/40] + [(1 – v

1

)/10]

2.2v

1

= 2 – 3v

1

+ 4 – 4v

1

= 6 – 7v

1 or

From (1) and (3), v

1

= 6/9.2

10i x

+ v

1

/20 = 1 – v

1

10i x

= 1 – v

1

– v

1

/20 = 1 – (21/20)v

1

= 1 – (21/20)(6/9.2) i x

= 31.52 mA, R

Th

= 1/i x

= 31.73 ohms.

(3)





4 .

63

Because there are no independent sources, I

N

= I sc

= 0 A

R

N

can be found using the circuit below. v o

+

−

10

20

Ω

Ω v

1

0.5v

o i o

+

−

1V

Applying KCL at node 1, v

1

= 1, and v o

= (20/30)v

1

= 2/3 i o

= (v

1

/30) – 0.5v

o

= (1/30) – 0.5x2/3 = 0.03333 –

0.33333 = – 0.3 A.

Hence,

R

N

= 1/(–0.3) = –3.333 ohms





Chapter 4, Solution 64.

With no independent sources, V

Th below.

= 0 V . To obtain R

Th

, consider the circuit shown

10i x

+

–

4 Ω v o i x

2

1

Ω

Ω i o

+

−

1V i x

= [(1 – v o

)/1] + [(10i x

– v o

)/4], or 5v o

= 4 + 6i x

But i x

= v o

/2. Hence,

5v o

= 4 + 3v o

, or v o

= 2, i o

= (1 – v o

)/1 = -1

Thus, R

Th

= 1/i o

= –1 ohm

(1)





4 .

65

At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent.

R

Th

= 2 + 4 // 12 = 2 + 3 = 5 Ω , V

Th

=

12

12

+ 4

( 32 ) =

Thus, the circuit can be replaced by that shown below.

5 Ω

24 V

I o

+

24 V

-

+

V o

-

Applying KVL to the loop,

− 24 + 5 I o

+ V o

= 0 V o

= 24 − 5 I o





4 .

66

We first find the Thevenin equivalent at terminals a and b. We find R

Th in Fig. (a).

using the circuit

3 Ω a

2 Ω b

3 Ω a

2 Ω

V

Th

− +

10V b

5 Ω

(a)

R

Th

5 Ω

+

−

20V

+ i

(b)

30V

−

+

R

Th

= 2||(3 + 5) = 2||8 = 1.6 ohms

By performing source transformation on the given circuit, we obatin the circuit in (b).

We now use this to find V

Th

.

10i + 30 + 20 + 10 = 0, or i = –6

V

Th

+ 10 + 2i = 0, or V

Th

= 2 V p = V

Th

2 /(4R

Th

) = (2) 2 /[4(1.6)] = 625 m watts




4 .

67

We first find the Thevenin equivalent. We find R

Th


80 Ω

20 Ω

10 Ω 90 Ω

R

Th

R

Th

= = + = Ω

We find V

Th

using the circuit below. We apply mesh analysis.

80

10

Ω

Ω

I

I

1

40 V

+ –

2

20

90 Ω

Ω

V

+

TH

_

+ i

1

− = → i

1

= 0.4

− 90 i

2

+ i

2

+ =

− 20 +

Th

= 0

→

→ i

2

= −

V

Th

0.4

= − 28 V



(a) R = R

Th

= 25 Ω

(b) P max

=

4

V

R

2

Th

Th

=

(28) 2

100

= 7.84 W





4 .

68

This is a challenging problem in that the load is already specified. This now becomes a

"minimize losses" style problem. When a load is specified and internal losses can be adjusted, then the objective becomes, reduce R

Thev in maximum power transfer to the load.

as much as possible, which will result

R

12 V

+

-

10 Ω

+

-

8V

20 Ω

Removing the 10 ohm resistor and solving for the Thevenin Circuit results in:

R

Th

= (Rx20/(R+20)) and a V oc

= V

Th

= 12x(20/(R +20)) + (-8)

As R goes to zero, R

Th

goes to zero and V

Th

goes to 4 volts, which produces the maximum power delivered to the 10-ohm resistor.

P = vi = v 2 /R = 4x4/10 = 1.6 watts

Notice that if R = 20 ohms which gives an R

Th

= 10 ohms, then V

Th

becomes -2 volts and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts.

It is also interesting to note that the internal losses for the first case are 12 2 /20 = 7.2 watts and for the second case are = to 12 watts. This is a significant difference.




4 .

69

We need the Thevenin equivalent across the resistor R. To find R

Th


22 k Ω v

1

10 k Ω v o

+

40 k Ω

0.003v

o

30 k Ω

−

Assume that all resistances are in k ohms and all currents are in mA.

1mA

10||40 = 8, and 8 + 22 = 30

1 + 3v o

= (v

1

/30) + (v

1

/30) = (v

1

/15)

15 + 45v o

= v

1

But v o

= (8/30)v

1

, hence,

15 + 45x(8v

1

/30) v

1

, which leads to v

1

= 1.3636

R

Th

= v

1

/1 = –1.3636 k ohms

R

Th

being negative indicates an active circuit and if you now make R equal to 1.3636 k ohms, then the active circuit will actually try to supply infinite power to the resistor. The correct answer is therefore: p

R

=



 − 1363 .

V

Th

6 + 1363 .

6





2

1363 .

6 =





V

Th

0





2

1363 .

6 = ∞

It may still be instructive to find V

Th

. Consider the circuit below.

10 k Ω v o

22 k Ω v

1

+

100V

+

− v o

−

40 k Ω

0.003v

o

30 k Ω

+

V

−

Th



(100 – v o

)/10 = (v o

/40) + (v o

– v

1

)/22 (1)

[(v o

– v

1

)/22] + 3v o

= (v

1

/30) (2)

Solving (1) and (2), v

1

= V

Th

= -243.6 volts





4 .

70

We find the Thevenin equivalent across the 10-ohm resistor. To find V

Th circuit below.

, consider the

3V x

5 Ω 5 Ω

+

4V

-

15 Ω

6 Ω

+ V x -

+

V

-

Th

From the figure,

V x

=

To find R

Th,

0 , V

Th

=

15

15

+ 5

( 4 ) = 3

consider the circuit below:

V

3V x

+

4V

-

5 Ω

V

1

15 Ω

5

6

Ω

Ω

+ V x

-

At node 1,

V

2

1A





4 − V

1

5

= 3 V x

+

V

1

15

+

V

1

−

5

V

2 , V x

= 6 x 1 = 6

At node 2,

1 + 3 V x

+

V

1

− V

2 = 0 V

1

= V

2

−

5

Solving (1) and (2) leads to V

2

= 101.75 V

95

R

Th

=

V

1

2 = 101 .

75 Ω , p max

=

V

Th

2

4 R

Th

258 = 3 V

2

− 7 V

1

(1)

(2)

=

9

4 x 101 .

75

= 22 .

11 mW





4 .

71

We need R

Th

and V

Th

at terminals a and b. To find R

Th terminals a and b as shown below.

, we insert a 1-mA source at the

3 k Ω v o

+

−

1 k Ω

−

+

10 k

120v

Ω o a

40 k b

Ω

1mA

Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in volts. At node a,

1 = (v a

/40) + [(v a

+ 120v o

)/10], or 40 = 5v a

+ 480v o

(1)

The loop on the left side has no voltage source. Hence, v o

= 0. From (1), v a

= 8 V.

R

Th

= v a

/1 mA = 8 kohms

To get V

Th

, consider the original circuit. For the left loop, v o

= (1/4)8 = 2 V

For the right loop, v

R

= V

Th

= (40/50)(-120v o

) = -192

The resistance at the required resistor is

R = R

Th

= 8 kohms p = V

Th

2 /(4R

Th

) = (-192) 2 /(4x8x10 3 ) = 1.152 watts





4 .

72

(a) R

Th

and V respectively.

Th

are calculated using the circuits shown in Fig. (a) and (b)

From Fig. (a),

From Fig. (b),

4 Ω

(b)

2 Ω

6

(a)

Ω

R

Th

-V

= 2 + 4 + 6 =

Th

+ 12 + 8 + 20 = 0, or V

R i = V

Th

Th

/(R

Th

+

−

12 ohms

2 Ω

8V

Th

4

= 40 V

Ω 6

(b)

+ R) = 40/(12 + 8) = 2A

Ω

20V

+ −

12V

− +

(c)

(d)

For maximum power transfer, p = V

Th

2 /(4R

Th

) = (40) 2

R

L

= R

Th

= 12 ohms

/(4x12) = 33.33 watts .

+

V

Th

−





4 .

73

Find the Thevenin’s equivalent circuit across the terminals of R.

10 Ω 25 Ω

R

Th

20 Ω 5 Ω

R

Th

= 10 // 20 + 25 // 5 = 325 / 30 = 10 .

833 Ω

10 Ω 25 Ω

+ + V

Th

-

60 V + +

-

V a

V

20 Ω 5 Ω b

- -

V a

− V a

=

20

( 60 )

+

30

V

Th

+

=

V b

40

= 0

, V b

=

5

30

(

V

60

Th

)

=

=

V

10 a

− V b

= 40 − 10 = 30 V p max

=

V

Th

2

4 R

Th

=

4

30 x 10

2

.

833

= 20 .

77 W





4 .

74

When R

L

is removed and V s

is short-circuited,

R

Th

= R

1

||R

2

+ R

3

||R

4

= [R

1

R

2

/( R

1

+ R

2

)] + [R

3

R

4

/( R

3

+ R

4

)]

R

L

= R

Th

=

(R

1

R

2

R

3

+ R

1

R

2

R

4

+ R

1

R

3

R

4

+ R

2

R

3

R

4

)/[( R

1

+ R

2

)( R

3

+ R

4

)]

When R

L

is removed and we apply the voltage division principle,

V oc

= V

Th

= v

R2

– v

R4

= ([R

2

/(R

1

+ R

2

)] – [R

4

/(R

3

+ R

4

)])V s

= {[(R

2

R

3

) – (R

1

R

4

)]/[(R

1

+ R

2

)(R

3

+ R

4

)]}V s p max

= V

Th

2

/(4R

Th

)

= {[(R

2

R

3

) – (R

1

R

4

)]

2

/[(R

1

+ R

2

)(R

3

+ R

4

)]

2

}V s

2

[( R

1

+ R

2

)( R

3

+ R

4

)]/[4(a)] where a = (R

1

R

2

R

3

+ R

1

R

2

R

4

+ R

1

R

3

R

4

+ R

2

R

3

R

4

) p max

=

[(R

2

R

3

) – (R

1

R

4

)] 2 V s

2 /[4(R

1

+ R

2

)(R

3

+ R

4

) (R

1

R

2

R

3

+ R

1

R

2

R

4

+ R

1

R

3

R

4

+ R

2

R

3

R

4

)]





4 .

75

We need to first find R

Th

and V

Th

.

R R

(a)

R

Consider the circuit in Fig. (a).

R

Th

1V

+

−

2V

+

−

3V

(1/R

Th

) = (1/R) + (1/R) + (1/R) = 3/R

(b)

+

−

R

Th

= R/3

From the circuit in Fig. (b),

((1 – v o

)/R) + ((2 – v o

)/R) + ((3 – v o

)/R) = 0 v o

= 2 = V

Th

For maximum power transfer,

R

Th

= [(V

P max

R

L

= R

Th

= R/3

= [(V

Th

) 2 /(4R

Th

)] = 3 mW

Th

) 2 /(4P max

)] = 4/(4xP max

) = 1/P max

R = 3/(3x10 -3 ) = 1 k ohms

= R/3

R

R

R v o

+

V

Th

−




4.76

V

Applicando il teorema di Millmann

V =

1

5

+

30

10 40

+

1

+

1

10 40 12

= 6 Volt

Da cui si ottiene i =

V

12

= 0.5 Ampere


4.77

V

Applicando il teorema di Millmann si ha

V =

3

2 v k

1

0 +

+

3 mA

1

2 k 4 k

=

6 v

0

+ 12

3 dove il valore della tensione dipende dalla grandezza v

0

= − × 6 mA = − 6 Volt

Da cui

V =

6 v

0

+ 12

3

= − 8 Volt


4.78

Per poter applicare Millmann il circuito va modifcato come segue

10k

100 +

vo

27k

15

+

-

15vo

Da cui si ottiene che v

0

= 10

+ k

+ v

0

1

27 k

1

10 k 27 k

+

1

40 k

Da cui si ottiene che v

0

= − 25,42 Volt



4 .

7 9

Follow the steps in Example 4.14. The schematic and the output plots are shown below.

From the plot, we obtain,

V = 92 V [i = 0, voltage axis intercept]

R = Slope = (120 – 92)/1 = 28 ohms









4 .80

(a) The schematic is shown below. We perform a dc sweep on a current source, I1, connected between terminals a and b. We label the top and bottom of source I1 as 2 and

1 respectively. We plot V(2) – V(1) as shown.

V

Th

= 4 V [zero intercept]

R

Th

= (7.8 – 4)/1 = 3.8 ohms





(b) Everything remains the same as in part (a) except that the current source, I1, is connected between terminals b and c as shown below. We perform a dc sweep on

I1 and obtain the plot shown below. From the plot, we obtain,

V = 15 V [zero intercept]

R = (18.2 – 15)/1 = 3.2 ohms









4 .81

The schematic is shown below. We perform a dc sweep on the current source, I1, connected between terminals a and b. The plot is shown. From the plot we obtain,

V

Th

=

-80 V

[zero intercept]

R

Th

= (1920 – (-80))/1 =

2 k ohms



4 .82

After drawing and saving the schematic as shown below, we perform a dc sweep on I1 connected across a and b. The plot is shown. From the plot, we get,

V = 167 V [zero intercept]

R = (177 – 167)/1 = 10 ohms





4 .

8 3

The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We perform dc sweep on I1. In the Trace/Add menu, type v(1) – v(2) which will result in the plot below. From the plot,

V

Th


R

Th

= (40 – 17.5)/1 = 22.5 ohms [slope]









4 .

8 4

The schematic is shown below. We perform a dc sweep on the current source, I2, connected between terminals a and b. The plot of the voltage across I2 is shown below.

From the plot,

V

Th


R

Th

= (10 – 6.7)/1 = 3.3 ohms . Note that this is in good agreement with the exact value of 3.333 ohms.









4 .

8 5

V

Th

= V oc

= 12 V, I sc

= 20 A

R

Th

= V oc

/I sc

= 12/20 = 0.6 ohm.

0.6 Ω i

12V

+

−

2 Ω i = 12/2.6 , p = i 2 R = (12/2.6) 2 (2) = 42.6 watts





4 .

8 6

V

Th

= V oc

= 12 V, I sc

= I

N

= 1.5 A

R

Th

= V

Th

/I

N

= 8 ohms, V

Th

= 12 V , R

Th

= 8 ohms





4 .

8 7

Let the equivalent circuit of the battery terminated by a load be as shown below.

R

Th

I

L

+ +

V

Th

-

V

L

R

L

-

For open circuit,

R

L

= ∞ ,

When R

L

= 4 ohm, V

L

=10.5,

I

L

=

V

L

R

L

= 10 .

8 / 4 = 2 .

7

But

V

Th

= V oc

= V

L

V

Th

= V

L

+ I

L

R

Th

R

Th

= 10 .

8 V

=

V

Th

− V

L

I

L

=

12 − 10 .

8

2 .

7

= 0 .

4444 Ω





4 .

8 8

(a) Consider the equivalent circuit terminated with R as shown below.

R

Th

a

+ +

V

Th

V ab

R

- -

b

V ab

=

R +

R

R

Th

V

Th

6 =

10

10

+ R

Th

V

Th or

60 + 6 R

Th

= 10 V

Th

(1) where R

Th

is in k-ohm.

Similarly,

12 =

30

30

+ R

Th

V

Th

360 + 12 R

Th

= 30 V

Th

(2)

Solving (1) and (2) leads to

V

Th

= 24 V, R

Th

= 30 k Ω

(b) V ab

=

20

20

+ 30

( 24 ) = 9 .

6 V





4 .

8 9

We replace the box with the Thevenin equivalent.

R

Th

V

Th

+

− v

+

− i

R

V

Th

= v + iR

Th

When i = 1.5, v = 3, which implies that V

Th

= 3 + 1.5R

Th

(1)

When i = 1, v = 8, which implies that V

Th

= 8 + 1xR

Th

(2)

From (1) and (2), R

Th

= 10 ohms and V

Th

= 18 V.

(a) When R = 4, i = V

Th

/(R + R

Th

) = 18/(4 + 10) = 1.2857 A

(b) For maximum power, R = R

TH

Pmax = (V

Th

) 2 /4R

Th

= 18 2 /(4x10) = 8.1 watts




(b)


4 .90

(a)

+ i m

= 9.975 mA

I s v m

From Fig. (a),

−

R s

(a)

R m

I s

R s

R s

(b) i

R m m

= 9.876 mA v m

= R m i m

= 9.975 mA x 20 = 0.1995 V

I s

= 9.975 mA + (0.1995/R s

)

From Fig. (b), v m

= R m i m

= 20x9.876 = 0.19752 V

I s

= 9.876 mA + (0.19752/2k) + (0.19752/R s

)

= 9.975 mA + (0.19752/R s

Solving (1) and (2) gives,

R s

= 8 k ohms , I s

= 10 mA

I s

R s

R s

(b) i

R m m

’ = 9.876 mA

8k||4k = 2.667 k ohms i m

’ = [2667/(2667 + 20)](10 mA) = 9.926 mA





4 .91

To find R

Th,

consider the circuit below.

R

Th

5k Ω

A B

30k Ω 20k Ω

10k Ω

R

Th

= 30 + 10 + 20 // 5 = 44 k Ω

To find V

Th


5k Ω

A B

i o

+

30k Ω 20k Ω

4mA 60 V

-

10k Ω

V

A

= 30 x 4 = 120 , V

B

=

20

25

( 60 ) = 48 , V

Th

= V

A

− V

B

= 72 V





The Thevenin equivalent circuit is shown below.

44k Ω

I

R i

+

72 V

-

2k Ω

I =

44 +

72

2 + R i mA assuming R i

is in k-ohm.

(a) When R i

=500 Ω ,

I =

44 +

72

2 + 0 .

5

= 1.548

mA

(b) When R i

= 0 Ω ,

I =

44 +

72

2 + 0

= 1.565

mA





4 .92

It is easy to solve this problem using Pspice.

(a) The schematic is shown below. We insert IPROBE to measure the desired ammeter reading. We insert a very small resistance in series IPROBE to avoid problem. After the circuit is saved and simulated, the current is displaced on IPROBE as 99 .

99 µ A .

(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown below. We obtain exactly the same result as in part (a).





4 .

9 3

R x

= (R

3

/R

1

)R

2

= (4/2)R

2

= 42.6, R

2

= 21.3 which is (21.3ohms/100ohms)% = 21.3%





4 .

9 4

R x

= (R

3

/R

1

)R

2

(a) Since 0 < R

2 ohms, R x

< 50 ohms, to make 0 < R x

< 10 ohms requires that when R

= 10 ohms.

10 = (R

3

/R

1

)50 or R

3

= R

1

/5

2

= 50 so we select R

1

= 100 ohms and R

3

= 20 ohms

(b) For 0 < R x

< 100 ohms

100 = (R

3

/R

1

)50, or R

3

= 2R

1

So we can select R

1

= 100 ohms and R

3

= 200 ohms





4 .

9 5

For a balanced bridge, v ab circuit in Fig. (a), where i

1

= 0. We can use mesh analysis to find v ab

and i

2

are assumed to be in mA.

. Consider the

2 k Ω

220V

+

− i

1

3 k a

Ω

+ v

5 k Ω i

2 ab

− b

6 k Ω

10 k Ω

0

(a)

From (1) and (2),

220 = 2i

1

+ 8(i

1

– i

2

) or 220 = 10i

1

– 8i

2

(1)

0 = 24i

2

– 8i

1

or i

2

= (1/3)i

1

(2) i

1

= 30 mA and i

2

= 10 mA

Applying KVL to loop 0ab0 gives

5(i

2

– i

1

) + v ab

+ 10i

2

= 0 V

Since v ab

= 0, the bridge is balanced.

When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes unbalanced. (1) remains the same but (2) becomes

0 = 32i

2

– 8i

1

, or i

2

= (1/4)i

1

(3)

Solving (1) and (3), i

1

= 27.5 mA, i

2

= 6.875 mA v ab

= 5(i

1

– i

2

) – 18i

2

= -20.625 V

V

Th

= v ab

= -20.625 V





To obtain R

Th

Fig. (c).

, we convert the delta connection in Fig. (b) to a wye connection shown in

2 k Ω

3 k Ω 6 k Ω R

2

R

1

6 k Ω

5 k Ω 18

(b)

R

3

(c)

18 k Ω

R

1

= 3x5/(2 + 3 + 5) = 1.5 k ohms, R

2

= 2x3/10 = 600 ohms,

R

3

= 2x5/10 = 1 k ohm.

R

Th

= R

1

+ (R

2

+ 6)||(R

3

+ 18) = 1.5 + 6.6||9 = 6.398 k ohms

R

L

= R

Th

= 6.398 k ohms

P max

= (V

Th

) 2 /(4R

Th

) = (20.625) 2 /(4x6.398) = 16.622 mWatts





4 .

9 6

V

S i x

+

−

R s i x

R o

+

−

β

-V s

+ (R s

+ R i x

= V s o

)i x

+ β R o i x

= 0

/(R s

+ (1 + β )R o

)

R o i x




4 .

9 7

V o

/V g

= R p

/(R g

+ R s

+ R p

R eq

= R p

||(R g

+ R s

) = R g

R g

= R p

(R g

+ R s

)/(R p

+ R g

+ R s

)

R g

R p

+ R g

2 + R g

R s

= R p

R g

+ R p

R s

R p

R s

= R g

(R g

+ R s

)

(1), p

/ α = R g

+ R s

+ R p

R g

+ R s

= R p

((1/ α ) – 1) = R

Combining (2) and (1a) gives,

R s

= [(1 - α )/ α ]R eq p

(1 - α )/ α (1a)

(3)

= (1 – 0.125)(100)/0.125 = 700 ohms

From (3) and (1a),

R p

R p

(1 - α )/ α = R g

= R g

+ [(1 - α )/ α ]R g

= R g

/ α

/(1 - α ) = 100/(1 – 0.125) = 114.29 ohms

(b)

R

Th

I

V

Th

+

−

R

L

V

Th

= V s

= 0.125V

g

= 1.5 V

R

Th

= R g

= 100 ohms

I = V

Th

/(R

Th

+ R

L

) = 1.5/150 = 10 mA



4 .

9 8

Let 1/sensitivity = 1/(20 k ohms/volt) = 50 µ A

For the 0 – 10 V scale,

R m

= V fs

/I fs

= 10/50 µ A = 200 k ohms

For the 0 – 50 V scale,

R m

= 50(20 k ohms/V) = 1 M ohm

V

Th

+

−

R

Th

R m

V

Th

= I(R

Th

+ R m

)

(a) A 4V reading corresponds to

I = (4/10)I

V

Th fs

= 0.4x50 µ A = 20 µ A

= 20 µ A R

Th

+ 20 µ A 250 k ohms

= 4 + 20 µ A R

Th

(1)

(b) A 5V reading corresponds to

I = (5/50)I fs

= 0.1 x 50 µ A = 5 µ A

V

Th

= 5 µ A x R

Th

+ 5

V

Th

= 5 + 5 µ A R

µ

Th

A x 1 M ohm

(2)

From (1) and (2)

0 = -1 + 15 µ A R

Th

which leads to R

Th

From (1),

= 66.67 k ohms

V

Th

= 4 + 20x10 -6 x(1/(15x10 -6 )) = 5.333 V





4 .

9 9

(a) The resistance network can be redrawn as shown in Fig. (a),

10 Ω 8 Ω 10 Ω

9V

+

− i

1

40 Ω

8 i

Ω

2

(a)

60 Ω

10 Ω

+

V

−

Th

R

V

Th

+

−

R

Th

(b)

V o

+

−

R

Th

= 10 + 10 + 60||(8 + 8 + 10||40) = 20 + 60||24 = 37.14 ohms

Using mesh analysis,

-9 + 50i

1

- 40i

2

= 0

116i

2

– 40i

1

= 0 or i

1

= 2.9i

2

(1)

(2)

From (1) and (2), i

2

= 9/105

V

Th

= 60i

2

= 5.143 V

From Fig. (b),

V o

= [R/(R + R

Th

)]V

Th

= 1.8

R/(R + 37.14) = 1.8/5.143 which leads to R = 20 ohms

(b) R = R

Th

= 37.14 ohms

I max

= V

Th

/(2R

Th

) = 5.143/(2x37.14) = 69.23 mA

R





4 .100

12V

+

−

4 k Ω

6 k Ω

B

E

+

V

−

Th

V

Th

R

Th

= [R

2

/(R

1

= R

1

||R

2

= 6||4 = 2.4 k ohms

+ R

2

)]v s

= [4/(6 + 4)](12) = 4.8 V





4 .101

The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be connected to the wye connection as shown in Fig. (b),

20

60

Ω

Ω

14 Ω a

30 b

R

Th

Ω

R

1

R

R

3

2 b

30

R a

Th

Ω

(a) (b)

R

1

= 20x60/(20 + 60 + 14) = 1200/94 = 12.766 ohms

R

2

= 20x14/94 = 2.979 ohms

R

3

= 60x14/94 = 8.936 ohms

R

Th

= R

3

+ R

1

||(R

2

+ 30) = 8.936 + 12.766||32.98 = 18.139 ohms





To find V

Th

, consider the circuit in Fig. (c).

20 Ω

I

1

I

T

14 Ω

60

I

T

Ω

16 V

+ −

(c) a

+

30 Ω b

V

Th

I

T

= 16/(30 + 15.745) = 349.8 mA

I

1

= [20/(20 + 60 + 14)]I

T

= 74.43 mA

V

Th

= 14I

1

+ 30I

T

= 11.536 V

I

40

= V

Th

/(R

Th

+ 40) = 11.536/(18.139 + 40) = 198.42 mA

P

40

= I

40

2 R = 1.5748 watts




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