Chapter 4 - Alexander`s Fundamentals of Electric circuits Ch.4 Solution PDF

Preview

Summary

Alexander`s Fundamentals of Electric circuits Ch.4 Solution...

Description

Chapter 4, Solution 1. 5

30V

+ 

25  i o

i

40 

15 

40 (25  15)  20 , i = [30/(5+20)] = 1.2 and i o = i20/40 = 600 mA.

Since the resistance remains the same we get can use linearity to find the new value of the voltage source = (30/0.6)5 = 250 V.

4.2 Using Fig. 4.70, design a problem to help other students better understand linearity.

Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find v o in the circuit of Fig. 4.70. If the source current is reduced to 1 A, what is v o ?

Figure 4.70

Solution 6 (4  2)  3, i1  i 2 

io 

1 A 2

1 1 i1  , v o  2i o  0.5V 2 4

5

4

i1

io

i2 1A

If i s = 1A, then v o = 0.5 V

8

6

2

Chapter 4, Solution 3. R 3R io 3R Vs

3R

+ 

+ R

vo

1V

(a) We transform the Y sub-circuit to the equivalent  . 3R 2 3 3 3 3  R, R  R  R 4R 4 4 4 2

vs independent of R 2 i o = v o /(R) vo 

When v s = 1V, v o = 0.5V, i o = 0.5A (b) (c)

3R

 (a)

R 3R 

+ 

When v s = 10V, v o = 5V, i o = 5A When v s = 10V and R = 10, v o = 5V, i o = 10/(10) = 500mA

(b)

1.5R

Chapter 4, Solution 4. If I o = 1, the voltage across the 6 resistor is 6V so that the current through the 3 resistor is 2A. 2

2

2A 1A

3A

3A

i1 +

3

6

4

Is

2

4

v1  (b)

(a) 3 6  2 , v o = 3(4) = 12V, i1 

vo 4

 3A.

Hence I s = 3 + 3 = 6A If

I s = 6A I s = 9A

Io = 1 I o = 9/6 = 1.5A

Is

Chapter 4, Solution 5. 2

Vs

If v o = 1V,

If v s =

10 3

Then v s = 15

3

v1

+ 

6

1 V1     1  2V 3 10 2 Vs  2    v1  3 3

vo = 1 vo =

3 x15  4.5V 10

vo

6

6

Chapter 4, Solution 6. Due to linearity, from the first experiment, 1 Vo  V s 3 Applying this to other experiments, we obtain:

Experiment 2 3 4

Vs

Vo

48 1V -6 V

16 V 0.333 V -2V

Chapter 4, Solution 7. If V o = 1V, then the current through the 2- and 4- resistors is ½ = 0.5. The voltage across the 3- resistor is ½ (4 + 2) = 3 V. The total current through the 1- resistor is 0.5 +3/3 = 1.5 A. Hence the source voltage v s  1x1.5  3  4.5 V If v s  4.5   1V Then v s  4



1 x 4  0.8889 V = 888.9 mV. 4.5

1 Chapter 4, Solution 8. Let V o = V 1 + V 2 , where V 1 and V 2 are due to 9-V and 3-V sources respectively. To find V 1 , consider the circuit below.

V1

3 9

1 + _

9  V1 V1 V1   3 9 1

9V

  V1  27/ 13  2.0769

To find V 2 , consider the circuit below. V1

9

V2 V2 3  V2   9 3 1

3

  V2  27/ 13  2.0769

V o = V 1 + V 2 = 4.1538 V

+ _

3V

Chapter 4. Solution 9. Given that I = 4 amps when V s = 40 volts and I s = 4 amps and I = 1 amp when V s = 20 volts and I s = 0, determine the value of I when V s = 60 volts and I s = –2 amps.

VS

+ 

I

IS

At first this appears to be a difficult problem. However, if you take it one step at a time then it is not as hard as it seems. The important thing to keep in mind is that it is linear! If I = 1 amp when V s = 20 and I s = 0 then I = 2 amps when V s = 40 volts and I s = 0 (linearity). This means that I is equal to 2 amps (4–2) when I s = 4 amps and V s = 0 (superposition). Thus, I = (60/20)1 + (–2/4)2 = 3–1 = 2 amps.

Chapter 4, Solution 10. Using Fig. 4.78, design a problem to help other students better understand superposition. Note, the letter k is a gain you can specify to make the problem easier to solve but must not be zero. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem For the circuit in Fig. 4.78, find the terminal voltage V ab using superposition.

Figure 4.78 For Prob. 4.10. Solution Let v ab = v ab1 + v ab2 where v ab1 and v ab2 are due to the 4-V and the 2-A sources respectively.

3vab1

10 

10 

+

3vab2 +

+ 4V

+ 

v ab1

+ 2A

 (a)

 (b)

For vab1 , consider Fig. (a). Applying KVL gives, - v ab1 – 3 v ab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2 , consider Fig. (b). Applying KVL gives,

v ab2

–vab2 – 3vab2 + 10x2 = 0, which leads to v ab2 = 5 v ab = 1 + 5 = 6 V

Chapter 4, Solution 11. Let vo = v 1 + v 2 , where v 1 and v 2 are due to the 6-A and 80-V sources respectively. To find v1 , consider the circuit below.

I1

va

20 

10  vb + V1 _

6A

40 

4 i1

At node a, 6

va va  v b  40 10

  240  5va  4 v b

(1)

At node b, –I 1 – 4I1 + (v b – 0)/20 = 0 or v b = 100I 1 But

i1 

va  vb 10

which leads to 100(va –v b )10 = v b or v b = 0.9091v a

Substituting (2) into (1), 5va – 3.636v a = 240 or v a = 175.95 and vb = 159.96 However,

v1 = v a – v b = 15.99 V.

To find v2 , consider the circuit below.

(2)

io

10  + v2 _

40 

20 

vc  

4 io

– +

30 V

0  vc (30  vc )  4 io  0 50 20 (0  vc ) But io  50 5 v c (30  v c )  0  50 20 0  v c 0  10 1   i2  50 50 5 v 2  10 i2  2 V



v c   10 V

v o = v1 + v 2 =15.99 + 2 = 17.99 V and i o = v o /10= 1.799 A.

Chapter 4, Solution 12. Let v o = 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. For v o1 , consider the circuit below. 2A

2A

5

4

io 5 

+ vo1  6

3

+ v o1 

12 

5

6||3 = 2 ohms, 4||12 = 3 ohms. Hence, i o = 2/2 = 1, vo1 = 5io = 5 V For vo2 , consider the circuit below. 6

5

4

+ vo2  12V

+ 

3

5

6

+ v o2  3

+ 12 

12V

+ 

v1

3

 3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5 v o2 = (5/8)v1 = (5/8)(16/5) = 2 V

For vo3 , consider the circuit shown below. 5 + vo3  6 

3

5

4

12 

+ 19V 

 2

+ vo3  12 

4 + v2

+ 

 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

19V

Chapter 4, Solution 13. Let vo  v1  v2  v 3 , where v 1 , v2 , and v 3 are due to the independent sources. To find v1 , consider the circuit below. 8

2A

+

5

10 

v1 _

10 x2  4.3478  10 8  5 To find v2 , consider the circuit below.

v1  5 x

4A

8

10 

5

+ v2 _

8 x4  6.9565  8 10  5 To find v3 , consider the circuit below.

v2  5x

8

12 V + –

10 

5

+ v3 _

5   v 3  12   2.6087   5  10  8 

vo  v1  v 2  v 3  8.6956 V =8.696V.

Chapter 4, Solution 14. Let v o = 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. For vo1 , consider the circuit below. 6 4

2 +

+ 

3

v o1

20V

 6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V For vo2 , consider the circuit below. 6 4

6 4V

2

2

+

+ 1A

4

+

3

v o2

v o2





3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3 , consider the circuit below. 6 2A 4

2A

2

3 +

v o3

3



3  vo3 +

6||(4 + 2) = 3, v o3 = (-1)3 = –3 v o = 10 + 1 – 3 = 8 V

3

Chapter 4, Solution 15. Let i = i1 + i 2 + i3 , where i1 , i 2 , and i 3 are due to the 20-V, 2-A, and 16-V sources. For i1 , consider the circuit below. io

20V

+ 

1 i1

2

 4

 3

4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i 1 = io /2 = 2.5 A For i3 , consider the circuit below. + 1 2

v o’

 4

i3  +

 3

16V

 2||(1 + 3) = 4/3, v o ’ = [(4/3)/((4/3) + 4)](-16) = -4 i 3 = vo ’/4 = -1 For i2 , consider the circuit below.

2

1

1

2A

2A

 (4/3) i2

 4

3 

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

i2 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 = i2R = (1.875)23 = 10.55 watts

Chapter 4, Solution 16. Let io = i o1 + i o2 + i o3 , where i o1 , i o2 , and i o3 are due to the 12-V, 4-A, and 2-A sources. For i o1 , consider the circuit below.

12V

3

4

i o1 + 

10 

2

 5

10||(3 + 2 + 5) = 5 ohms, i o1 = 12/(5 + 4) = (12/9) A 4A

For io2 , consider the circuit below.

3

i o2 4

2

5

10 

i1 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 io3 , consider the circuit below.

3

i o3

2

i2 4

10 

5

2A

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

Chapter 4, Solution 17. Let v x = v x1 + v x2 + v x3 , where v x1 ,v x2 , and v x3 are due to the 90-V, 6-A, and 40-V sources. For vx1 , consider the circuit below. 30  + 90V

+ 

20 

10 

60 

v x1

 i o 10  +  v x1

30  20 

3A

12 

20||30 = 12 ohms, 60||30 = 20 ohms By using current division, i o = [20/(22 + 20)]3 = 60/42, vx1 = 10i o = 600/42 = 14.286 V For vx2 , consider the circuit below. 10  i ’ o + 30 

10  i ’ o + vx2 

v x2 

60  6A

30 

20 

6A

20 

12 

i o ’ = [12/(12 + 30)]6 = 72/42, vx2 = –10i o ’ = –17.143 V For vx3 , consider the circuit below. 10  + 30 

60 

v x3

20 

10 



30 

+ 40V

+ 

20 

vx3

io  12 

4A

i o ” = [12/(12 + 30)]2 = 24/42, vx3 = -10io ” = -5.714= [12/(12 + 30)]2 = 24/42, v x 3 = -10i o ” = -5.714 = [12/(12 + 30)]2 = 24/42, vx3 = -10i o ” = -5.714 v x = 14.286 – 17.143 – 5.714 = -8.571 V

Chapter 4, Solution 18. Let V o = V 1 + V 2, where V 1 and V 2 are due to 10-V and 2-A sources respectively. To find V1 , we use the circuit below. 1 0.5 V 1

2

+ 10 V

+ _

V1 _

2

1

0.5 V 1 - + +

10 V

i

+ _

4

-10 + 7i – 0.5V 1 = 0 But V 1 = 4i `10  7 i  2 i  5 i

  i  2,

V1  8 V

V1 _

To find V 2 , we use the circuit below. 1 0.5 V 2

2

+ 4

2A

2

V2 _

1

0.5 V 2 - + +

4V

+ _

- 4 + 7i – 0.5V 2 =0 But V 2 = 4i 4  7i  2 i  5 i   i  0.8,

i

4

V2  4 i  3.2

V o = V 1 + V 2 = 8 +3.2 =11.2 V

V2 _

Chapter 4, Solution 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. ix

v1

ix

v2

+ 2

4A

8

v1

+ 6A 8

2



+

+

v2 

4i x

4i x (a)

(b)

To find v1 , consider the circuit in Fig. (a). v1 /8 – 4 + (v1 – (–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 ix = –0.5v 1 . Thus, v 1 = 6.4 + 3.2(0.5v1 ), which leads to v 1 = –6.4/0.6 = –10.667

To find v2 , consider the circuit shown in Fig. (b). v 2 /8 – 6 + (v 2 – (–4ix ))/2 = 0 or v 2 + 3.2i x = 9.6 But ix = –0.5v 2 . Therefore, v 2 + 3.2(–0.5v2 ) = 9.6 which leads to v 2 = –16 Hence,

vx = –10.667 – 16 = –26.67V.

Checking, i x = –0.5vx = 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

Chapter 4, Solution 20. Convert the voltage sources to current sources and obtain the circuit shown below.

3A

10 

0.6

20 

1 1 1 1     0.1 0.05 0.025  0.175 Re q 10 20 40

 

0.4

40 

R eq = 5.714 Ω

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 

18.285 V 4A

5.714 

+ _

4.21 Using Fig. 4.89, design a problem to help other students to better understand source transformation. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Apply source transformation to determine v o and i o in the circuit in Fig. 4.89.

Figure 4.89

Solution To get io , transform the current sources as shown in Fig. (a). io

6

3 i

+  12V

+ 

6V 2 A

6

3

+ vo 2 A 

(b)

(a) From Fig. (a),

-12 + 9io + 6 = 0, therefore i o = 666.7 mA

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

Chapter 4, Solution 22. We transform the two sources to get the circuit shown in Fig. (a). 5  + 10V

5

4

 10

2A

(a)

i 1A

10 

 4

 10

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

Chapter 4, Solution 23 If we transform the voltage source, we obtain the circuit below.

8

10 

6

3

5A

3A

3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 2 10  8

+

+ 10V -

30V -

Applying KVL to the loop gives  30  10  I (10  8  2)  0   I = 1 A p  VI  I 2 R  8 W

Chapter 4, Solution 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 10-Ω resistor and a 20V x -V sources in series with a 10-Ω resistor. We now have the following circuit,

8 +

10 

Vx –

– + 30 V

40 V

10 

+ _ I

20Vx

+ –

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.

Chapter 4, Solution 25. Transforming only the current source gives the circuit below. 18 V

9

+ – +

12V

5 i

4 +

vo 2



 +

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

Chapter 4, Solution 26. Transforming the current sources gives the circuit below. 2

15 V

5

io

4

– +

12 V

+ _

–12 + 11i o –15 +20 = 0 or 11i o = 7 or i o = 636.4 mA.

+ _

20 V

Chapter 4, Solution 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

12  + vx  5A

10 

 40

8A

 20

2A

(a)

8 + 

12  + vx 

40V

i

(b)

20  + 

200V

Chapter 4, Solution 28. Convert the dependent current source to a dependent voltage source as shown below. 1

4

io

3

+ Vo _

8V

+ _

– +

Applying KVL,  8  io (1 4  3)  Vo  0 But Vo  4io  8  8io  4io  0

  io  2 A

Vo

Chapter 4, Solution 29. Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 = (4/3) k ohms  4 k  2 k

2vo

 (4/3) k

+

1.5vo + 3 mA

 1 k

3 mA

i

 1 k

+ vo 

vo 

(a)

(b)

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 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.

Chapter 4, Sol...