Electric Circuit Chap 4

7/30/2019 Electric Circuit Chap 4

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Chapter 4, Problem 16.

Given the circuit in Fig. 4.84, use superposition to get io.


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Solution 4.16

Let io = io1 + io2 + io3, where io1, io2, and io3 are due tothe 12-V, 4-A, and 2-A sources. For io1, consider the circuit below.

10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A

For io2, consider the circuit below.

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

i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9

For io3, consider the circuit below.

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

i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9

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

12V

+

4

10

io1

5

2 3

io2 3

5 4

2

10

4A

i1

3

4 5 10

io3

2 A

i2

2


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Problem 4.19 Use superposition to solve for vx and ix in the circuit of Fig. 4.87.


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Solution 4.19: Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources

respectively.

To find v1, consider the circuit in Fig. (a).

v1/8 = 4 + (-4ix v1)/2

But, -ix = (-4ix v1)/2 and we have -2ix = v1. Thus,

v1/8 = 4 + (2v1 v1)/8, which leads to v1 = -32/3

To find v2, consider the circuit shown in Fig. (b).

v2/2 = 6 + (4ix v2)/8

But ix

= v2/2 and 2i

x= v

2. Therefore,

v2/2 = 6 + (2v2 v2)/8 which leads to v2 = -16

Hence, vx = (32/3) 16

Writing a mesh equation around the outside loop we get,

2ix + vx + 4ix = 0 or 2ix = vx

Thus, we get, ix = 13.333 A.

+

4 A

4ix

2

ix v1

8

(a)

+

6 A

4ix

2

ix v2

8

(b)

+

v1

+

v2


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Apply source transformation to find vx in the circuit of Fig. 4.95.


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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

vx = 12i = -48 V

12

10

(a)

8A40

+ vx

2A5A 20

40V

+

(b)

8

i 200V

+

12 20

+ vx


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Use source transformation to find vo in the circuit of Fig. 4.97.

4 k

1 k

3vo

3 mA

+

2 k

+

vo


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Chapter 4, Solution 29.

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

(4/3) k ohms

It is clear that i = 3 mA which leads to vo = 1000i = 3 V

If the use of source transformations was not required for this problem, the actual answercould have been determined by inspection right away since the only current that could

have flowed through the 1 k ohm resistor is 3 mA.

+

vo

(a)

2 k

4 k

1 k

1.5vo

3 mA

+

i

(b)

(4/3) k2vo

3 mA

+

vo

1 k


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Find the Thvenin and Norton equivalents at terminals a-b of the circuit shown in

Fig. 4.108.


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Chapter 4, Solution 41

To find RTh, consider the circuit below

14

a

6 5

b

NThRR ==+=

4)614//(5Applying source transformation to the 1-A current source, we obtain the circuit below.

6 - 14V + 14 VTha

+

6V 3A 5

-

b

At node a,

V85

3146

614=+=

+

+

Th

ThThV

VV

A24/)8( ===Th

Th

N

R

VI

Thus,

A2V,8,4 ====NThNTh

IVRR


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Find the Thvenin equivalent between terminals a-b of the circuit in Fig. 4.120

+-


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Chapter 4, Solution 54

To find VTh =Vx, consider the left loop.

xoxoViVi 2100030210003 +==++ (1)

For the right loop,

ooxiixV 20004050 == (2)

Combining (1) and (2),

mA13000400010003 ===ooooiiii

222000 ===Thox

ViV

To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independentsource, as shown below.

1 k ix.

io + + +2Vx 40io Vx 50 1V

- - -

mA21000

2,1 === x

ox

ViV

-60mAA

50

1mA80

50

40 =+=+= xox

Vii

=== 67.16060.0/11

x

Th

iR


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Find the maximum power transferred to resistorR in the circuit of Fig. 4.135

0.003vo


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We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit

below.

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

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

1 + 3vo = (v1/30) + (v1/30) = (v1/15)

15 + 45vo = v1

But vo = (8/30)v1, hence,

15 + 45x(8v1/30) v1, which leads to v1 = 1.3636

RTh = v1/1 = 1.3636 k ohms

RTh 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. Thecorrect answer is therefore:

pR = 6.13630

V6.1363

6.13636.1363

V2

Th2

Th

=

+=

It may still be instructive to find VTh. Consider the circuit below.

(100 vo)/10 = (vo/40) + (vo v1)/22 (1)

22 k

10 k

v1

+

vo

40 k 30 k0.003vo

1mA

100V

+

22 k v1

+

vo

40 k 30 k0.003vo

10 k vo

+

VTh


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[(vo v1)/22] + 3vo = (v1/30) (2)

Solving (1) and (2),

v1 = VTh = -243.6 volts


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Chapter 5, Problem 67

Obtain the output voin the circuit of Fig. 5.95.


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Solution 5.67Voltage follower , inv amplifier , summer

Voltage follower, summer

vo = )2.0(

20

80)4.0(

20

80

40

80

== 8.02.3 2.4V


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Find the load voltage vLin the circuit of Fig. 5.98.


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Since no current flows into the input terminals of ideal op amp, there is no voltage drop

across the 20 k resistor. As a voltage summer, the output of the first op amp is

v01 = 0.4

The second stage is an inverter

012 v100

250v =

== )4.0(5.2 -1V


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Determine the load voltage vLin the circuit of Fig. 5.99.


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The first stage is a noninverting circuit. The output is

V8.10)8.1(

10

1050v01 =

+=

The second stage is a voltage follower,

== 012 vv 10.8V

Electric Circuit Chap 4

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