Electrical & Electronic Technology (9th Edition) - Edward Hughes - Solutions PDF

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Electrical & Electronic Technology (9th Edition) - Edward Hughes - Solutions...

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Solutions to Exercises Section One: Electrical Principles

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

Exercises 1 1.

F = ma ∴a=

80 F = = 0.4 m/s2 m 200

2. 2 kN 3.

F = ma Here the acceleration is that of gravity, i.e. 9.81 m/s2 ∴ F = 10 × 9.81 = 98.1 N

4. 9.81 m/s, 19.62 m/s, 29.43 m/s 5.

F = ma ∴a=

F 1 × 10 3 = 0.1 m/s2 = m 10 × 103

v = at ∴ t=

5 v = = 50 s a 0 .1

6. 625 kN 7. The velocity has no effect on the weight, hence F = ma = 10 × 9.81 = 98.1 N The acceleration requires extra force, hence F = ma = 10 × (9.81 + 3) = 128.1 N 8. 0.98 m/s2 downwards 9. (a) F = 5 × 10−3 × 300 × 103 × 9.81 = 14 715 N = 14.7 kN (b) W = Fd = 14 715 ×

90 × 103 × 10 60

= 221 × 106 J = 221 MJ =

221 × 10 6 = 61.3 kWh 3 .6 × 10 6

(c) P = Fu = 14 715 ×

90 × 10 3 = 14 715 × 25 60 × 60

= 368 000 W = 368 kW (d) W = 12 mu 2 =

1 2

× 300 × 10 3 × 25 2

= 93.75 × 106 J = 26 kWh 10. 16 370 Nm

Exercises 2 1. 0.000 005 A = 5 × 10−6 A = 5 µA 2. 3 MV 3. R =

6 V = = 10 Ω I 0. 6

4. 40 V 8 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

5. I =

V 240 = = 0.25 A R 960

6. 2.9 kΩ 7. I =

240 V = = 28. 2 × 10 3 A = 28.2 kA R 8.5 × 10 −3

8. 5.1 mA 9. V = IR = 22 × 5 = 110 V 10. 1 mA, 0.67 mA 11.

I= R2 =

10 V1 = = 0.1 A R1 100 V2 100 = = 1000 Ω = 1 kΩ I 0.1

12. 1 kΩ 13.

V/V 5 4 3 2 1 I/A 0.2

14.

0.4

0.6

0.8

1.0

1.2

R/Ω 1000

800 600

10

20

30

40

50

60

I/mA

15. Q = It = 2.5 × 8 = 20 C 16. 8 C 17. V = IR = 2 × 40 = 80 V 18. 212 Nm 19. (a)

Pi = ∴ I=

Po 20 × 10 3 = = 22 727 W = VI η 0 .88 22 727 = 94.7 A 240

(b) W = Pt = 22.727 × 6 = 136.4 kWh Cost = 136.4 × 8 = 1091 p 20. 83.8%, 146 kJ; 67.2 Ω 21. 1000 cm3 has a mass of 1 kg, hence, for the pump (a) Po = 1.5 ×

10 6 40 × 9. 81× = 9810 W 3 60 10 9 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

For motor, Po = Pump input =

Pump output 9810 = ηp 0. 9

= 10 900 W and

Pi =

10 900 10 900 = = 12 824 W = 12.82 kW η 0.85

12 824 = 26.7 A 480 (c) W = Pt = 12.82 × 8 = 102.6 kWh (b) P = VI ∴ I =

22. 59.5 Ω, 0.65 p 23. Energy to raise 1 kg of aluminium from 12 °C to 660 °C = 950 × (660 − 12) = 615 600 J Latent heat for 1 kg = 450 000 J ∴ Heat energy for 1 kg is 615 600 + 450 000 = 1 065 600 J Po =

(a)

∴ Pi =

40 × 1 065 600 = 11 840 W = 11.84 kW 3600 Po 11. 84 = = 13.93 kW η 0 .85

(b) Cost = 13.93 × 8 × 20 = 2229 p

Exercises 3 1. Across 2 Ω resistor, V2 = IR2 = 2 × 2 = 4 V Across 3 Ω resistor, V3 = IR3 = 2 × 3 = 6 V Across 5 Ω resistor, V5 = IR5 = 2 × 5 = 10 V Total voltage V = V2 + V3 + V5 = 4 + 6 + 10 = 20 V 2. 500 Ω, 0.48 A 3.

R=

V 120 = = 80 Ω I 1 .5

= R1 + R2 + R3 = 30 + 30 + R3 ∴ R3 = 20 Ω 4. 60 Ω 5. VAB =

30 × 100 = 30 V 30 + 70

VBC = VLN − VAB = 100 − 30 = 70 V VAC = VLN = 100 V VBN = VBC = 70 V 6. −35 V, −5 V, 35 V, 50 V, 10 V 7. (a) I6 =

V 9 = = 1.5 A R6 6

I9 =

9 = 1.0 A 9

I15 =

9 = 0.6 A 15 10 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

(b) I = I6 + I9 + I15 = 1.5 + 1.0 + 0.6 = 3.1 A (c) R =

V 9 = = 2.9 Ω I 3. 1

8. 24 Ω, 36 Ω 9. V = I8 R8 = 3 × 8 = 24 V I4 = I16 =

24 V = =6A R4 4 V 24 = = 1.5 A R16 16

10. (a), (d) 11. (a) I1 = I2 + I3: incorrect – with link removed, I1 = I2 (b) I1 = I4 + I5: correct – I1 = I2 = I4 + I5 (c) I1 = I2: correct (d) I1 = I3: incorrect – I3 = 0 (e) I1 = I3 + I4 + I5: correct 12. 4.04 V, 780 J 13. 30 V 12 Ω

20 Ω

2Ω 8Ω

For the load, R = 8 + (a) For load V =

12 × 20 = 15. 5 Ω 12 + 20

15 .5 × 30 = 26.57 V 2 + 15 .5

(b) Battery current =

E 30 = = 1.71 A Rt 2 + 15.5

Current in 12 Ω resistor =

20 × 1.71 = 1.07 A 20 + 12

∴ P = I 2R = 1.072 × 12 = 13.8 W 14. 5 Ω 15. For circuit A, RA = 6 + 4 = 10 Ω For circuit B, RB = 10 + 5 = 15 Ω For parallel network, R = Battery current =

10 × 15 = 6Ω 10 + 15

25 = 4.0 A 6 + 0.25 11 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

Current in circuit A =

15 RB I= × 4 = 2.4 A RA + R B 10 + 15

Current in circuit B = 4.0 − 2.4 = 1.6 A V6 = 2.4 × 6 = 14.4 V V4 = 2.4 × 4 = 9.6 V V10 = 1.6 × 10 = 16.0 V V5 = 1.6 × 5 = 8.0 V P = VI = (16 + 8)4 = 96 W 16. 10.96 V; 121.7 A, 78.3 A; 1.33 kW, 0.86 kW 17. Voltage across 12 Ω resistance; P=

V2 R

hence

∴ V = √{36 × 12} = 20.78 V I12 =

20. 78

I18 =

20. 78 = 1.15 A 18

I36 =

20. 78 = 0.58 A 36

12

= 1.73 A

∴ I = 1.73 + 1.15 + 0.58 = 3.46 A Voltage across fourth resistance = 60 − 20.78 = 39.22 V ∴ Resistance =

39. 22 = 11.33 Ω 3 .46

P = VI = 60 × 3.46 = 208 W 18. 40 Ω, 302.5 W 19. For aluminium wire,

ρ l 28 × 10 − 9 × 7.5 = = 0. 267 Ω π −6 A 4 × 10 V = IRa = 3 × 0.267 = 0.802 V

Ra =

For copper wire, Rc = = A=

V I

0 .802 = 5 −3

=0 401 . Ω

ρ l 17 × 10 − 9 × 6 = A A 17 × 10 −9 × 6 π = 0.254 mm2 = × d2 0 .401 4

d = 0.569 mm 20. 70 °C 21.

R1 (1 + α 0θ1) = R2 (1 + α0 θ2 ) 250 (1 + 0. 004 28 × 15) 1 .0642 = = R2 ( 1 + 0. 004 28 × 45) 1 .1926 ∴ R2 = 280.16 Ω Hence increase in resistance = 280 − 250 = 30 Ω 12 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

22. 0.0173 µΩ m, 0.398 Ω 23.

R2 = R1(1 + α 20θ ) 230 = 150(1 + 0. 0039θ ) 1 .25 1.23 = 1 + 0.0039 θ ∴ θ=

0. 23 = 58 .1 ° 0 .0039

∴ Temperature = 58.1 + 20 = 78.1 °C

Exercises 4 1.

A

IA

B

I

IB

100 V

95 V 80 Ω

5Ω

(a)

3Ω

100 = IA 5 + (IA + IB) 80 = 85IA + 80IB

¨

95 = IB 3 + (IA + IB) 80 = 80IA + 83IB

≠

8300 = 7055IA + 6640IB

¨ × 83 = Æ

7600 = 6400IA + 6640IB

≠ × 80 = Ø

700 = 655IA

Æ−Ø

∴ IA = 1.069 A discharge ∴ 100 = 90.84 + 80IB ∴ IB = 0.115 A discharge (b) V = 80(1.069 + 0.115) = 94.66 V 2. 5; if A is positive with respect to B, load current is from D to C. 3.

I1 I1 − I2

100 Ω

2V V

50 Ω

1Ω

1.5 V

I2

2 = I1 + 50(I1 − I2) = 51I1 − 50I2

¨

1.5 = I2 × 100 − 50(I1 − I2) = −50I1 + 150I2 6 = 153I1 − 150I2

≠ ¨×3=Æ

7.5 = 103I1

≠+Æ

∴ I1 = 72.8 mA 2 = (51 × 72.8 × 10−3) − 50I2 ∴ I2 = 34.3 mA V = 50(72.8 − 34.3) × 10−3 = 1.93 V 13 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

4. 11.55 A, 3.86 A 5.

A

B

108 V

120 V

3Ω

25 Ω

2Ω

IA

30 V C

IB IA + IB

(a) 108 + 30 = 3IA + 25( IA + IB) 138 = 28IA + 25IB

¨

120 + 30 = 2IB + 25(IA + IB) 150 = 25IA + 27IB

≠

3726 = 756IA + 675IB

¨ × 27 = Æ

3750 = 625IA + 675IB

≠ × 25 = Ø

−24 = 131IA

Æ−Ø

∴ IA = − 0.183 A charge 150 = (25 × −0.183) + 27IB ∴ IB = 5.726 A discharge ∴ IC = IA + IB = 5.543 A (b) V = 120 − (2 × 5.726) = 108.5 V 6. 0.32 A 7. (a) Nodal analysis: The series branch containing the 2 V source and 2 Ω resistor is replaced by its Norton equivalent, a 1 A current source and a parallel 2 Ω resistor: V1

V2

Node 1

Node 2

4Ω

2Ω

1A

At Node 1: 1 =

8Ω

2A

V1 V1 − V2 + 2 4

4 = 2V1 + V1 − V2 4 = 3V1 − V2 At Node 2:

¨

V1 − V2 V + 2= 2 4 8 2V1 − 2V2 + 16 = V2 16 = −2V1 + 3V2

Eq.≠ ×

3 2

:

24 = −3V1 +

Eq.¨ + Eq.Æ: 28 =

9 2

≠ Æ

V2

7 V2 2

Hence V2 = 8 V and voltage across 4 Ω resistor, V1 = 4 V 14 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

(b) Superposition theorem: Suppress 2 A source: I1 4Ω 2Ω

8Ω

By Ohm’s Law, current through 4 Ω resistor: I1 = Suppress 2 V source:

2 A 14

4Ω 2Ω

8Ω

2A

Current through 4 Ω resistor: I2 = 2 ×

8 16 A= A 14 6+8

Combining the currents through the 4 Ω resistor due to the two sources: Itotal = I2 − I1 =

16 2 − A=1A 14 14

Hence voltage across 4 Ω resistor = 4 V (c) Thévenin’s theorem: Replace 4 Ω resistor by open circuit between A and B as shown: A

B

2Ω

8Ω

2V

16 V

Note that the 2 A current source and parallel 8 Ω resistor have been replaced by their Thévenin equivalent, i.e. a 16 V source in series with an 8 Ω resistor. No current flows in this circuit. Thus, open circuit voltage, ET = 14 V (B higher voltage than A). Resistance between A and B (voltage sources replaced by short circuits): RT = 10 Ω Thévenin circuit now: A

B 4Ω

10 Ω

14 V

14 A=1A 10 + 4 Hence voltage across 4 Ω resistor = 4 V Current in circuit: I =

15 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

8. 2.84 Ω, 1.45 Ω 9. Apply the delta/star transformation to CDE RC = (10 × 20)/35 = 5.71 Ω RD = (5 × 20)/35 = 2.86 Ω RE = (5 × 10)/35 = 1.43 Ω A

20 Ω 10 Ω

D 2.86 Ω

2V

1.43 Ω

40 Ω B

E 5.71 Ω

30 Ω

C

Apply Thevenin’s Theorem. For EB open circuit: VAE = 2 ×

20 + 2. 86 = 1.60 V 20 + 2.86 + 5. 71

VAB = 2 ×

10 = 0.5 V 10 + 30

note no volt drop across 1.43 Ω since open circuit

∴ VBE = 1.6 − 0.5 = 1.10 V Rin = 1.43 + ∴ IBE =

22. 86 × 5. 71 10 × 30 + = 13. 5 Ω 22. 86 + 5. 71 10 + 30

1 .10 = 20.6 mA 13. 5 + 40

10. 0.047 A 11. For 10 Ω open circuit, the voltage across 20 Ω resistor is Vo/c = 6 ×

10 =4V 10 + 5

where 10 Ω is result of 20 Ω in parallel with 5 and 15 Ω resistors ∴ voltage across 15 Ω resistor is Vo/c = 4 ×

15 =3V 15 + 20

Looking into network with battery short circuited 5 Ω and 20 Ω in parallel gives

5 × 20 = 4Ω 5 + 20

15 Ω in parallel with 4 and 5 Ω resistors Rin =

9 × 15 = 5. 625 Ω 9 + 15

hence I10 =

3 = 0.192 A 10 + 5.625 16 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

12. 40 Ω; 6 V, 0.9 W; 0.8 W 13. Solution by Thevenin’s Theorem Vo/c = 105 +

110 − 105 × 1 = 108. 33 V 1 + 0. 5

Rin =

1 × 0 .5 = 0.333 Ω 1 + 0 .5

I8 =

108 .33 = 13.0 A 8 + 0. 333

∴ V = 8 × 13 = 104 V For battery, current is 105 = I × 1 + 104 ∴

I=1A

For generator, current is I = 13 − 1 = 12 A 14. 2.295 A, 0.53 A, 1.765 A, both batteries discharging; 4.875 A 15. RAB = 50 + 100 +

50 × 100 = 183.3 Ω 150

RBC = 100 + 150 + RCA = 50 + 150 + 16. A, 4.615 Ω;

50 × 150 = 275 Ω 100

B, 12.31 Ω; C, 18.46 Ω

17. RAB = 70 + 100 + RBC = 100 + 90 + RAC = 70 + 90 + R=

100 × 150 = 550 Ω 50

70 + 100 = 247.8 Ω 90 100 × 90 70

= 318.6 Ω

70 × 90 = 223.0 Ω 100

160 × 160 = 80 Ω 160 + 160

18. 0.8 A 19. 10 V 30 V

0

10 Ω

i1 N

20 V

20 Ω

20 Ω i2

This can be solved by application of Maxwell’s circulating currents. 10 − 30 = 10i1 + 20(i1 − i2) −20 = 30i1 − 20i2

¨

30 − 20 = 20(i2 − i1) + 20i2 10 = −20i1 + 40i2

≠

− 40 = 60i1 − 40i2

¨×2

−30 = 40i1 17 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

∴

i1 = −0.75 A VNO = 10 − i1 10 = 17.5 V

This is the open circuit voltage hence it is possible to apply Thevenin’s Theorem 1 1 1 1 = + + Rin 10 20 20

∴ R in = 5 Ω

V0 = I(R + Rin) 17.5 = 1(5 + R) ∴ R = 12.5 Ω This could also have been solved by the Superposition Theorem.

Exercises 5 1. Q = It = 5 × 10−6 × 10 × 60 = 3.0 mC V=

Q 3 × 10 −3 = 150 V = 20 × 10 −6 C

2. 45 µF, 4.615 µF 3.

Cp = 4 + 2 = 6 µF Cs =

6×9 = 3.6 µF 6+9

Q = CV = 3.6 × 10−6 × 20 = 72 µC = C9V9 ∴ V9 =

72 =8V 9

∴ V4 = 12 V W=

1 1 4 CV 2 = × × 122 = 288 µJ 2 2 10 6

4. 1200 µC; 120 V, 80 V; 6 µF 5.

1 1 1 1 3+ 2 + 1 = + + = = 1 ∴ C = 1 µF 6 C 2 3 6 Q = CV = 1 × 500 = 500 µC V1 =

500 500 500 = 250 V V2 = = 167 V V3 = = 83 V 2 3 6

W = 12 CV 2 =

1 2

× 6 × 832 = 20.83 mJ

6. 15 µF in series 7. Cp = 3 + 6 = 9 µF Cs =

4× 9 = 2.77 µF 4+ 9

Q = CV = 2.77 × 20 = 55.4 µC Q r = 55.4 × 39 = 18.47 µC 8. 4.57 µF, 3.56 µF 9. CA120 = CB80 ∴ CA = 23 CB

and

CB = 32 CA

CA140 = (CB + 3)60 = 60CB + 180 = 90CA + 180 ∴ CA = 3.6 µF

and

CB = 5.4 µF 18 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

10. 200 V; 1.2 mC, 2 mC, 3.2 mC on A, B, C respectively 11.

C1 V2 V2 20 = = = C2 V1 30 600 − V2 ∴ 1200 − 2V2 = 3V2 C 400 = 30 200

∴ V2 = 240 V

and

V1 = 360 V

∴ C = 60 µF

∴ CC = 60 − 20 = 40 µF W = 12 CV 2 =

1 2

× 40 × 10−6 × 2002 = 0.8 J

12. 267 V; 0.32 J; 0.213 J 13. C =

4 ×6 = 2.4 µF 4+6

Q = CV = 2.4 × 250 = 600 µC 600 = 150 V 4

V1 =

V2 = 100 V

and

V = (600 + 600)/(4 + 6) = 120 V Q 1 = 120 × 4 = 480 µC and

Q 2 = 720 µC

Reverse connection would cause complete discharge. 14. 62.5 V 15. Q = CV1 = 3 × 200 = 600 µC W1 = 12 CV 2 = V2 = W2 =

1 2

× 3 × 2002 = 60 mJ

600 = 120 V 5 1 2

× 5 × 1202 = 36 mJ

16. 590 pF, 0.354 µC, 200 kV/m, 8.85 µC/m2 17. (a) C =

ε 0 ε r A 8. 854 × 10 −12 × 6 × 900 × 10− 4 = 3 × 10 −3 d = 1594 pF

(b) Q = CV = 1594 × 10−12 × 500 = 0.797 µC (c) E =

500 3 × 10 −3

= 167 kV/m −6

(d) D =

Q 0 .797 × 10 = 8.86 µ C/m2 = A 900 × 10 −4

18. 664 pF, 0.2656 µC, 100 kV/m, 4.43 µC/m2 19. (a) C =

ε 0 ε r A 8. 854 × 10−12 × 1 × 2002 × 10−6 = 177 pF = d 2 × 10 −3

(b) C = 177 × 6 = 1062 pF (c) Q = 177 × 200 = 1062 × V

∴ V = 33.3 V

(d) Q = 177 × 200 = 0.0354 µC 20. 619.5 pF, 31 ms, 0.062 µC 21. C =

ε 0ε r A 8. 854 × 10− 12 × 5 × 1000 × 10− 6 × ( 11− )1 = = 2212 pF 0 .2 × 10 −3 d

22. 1.416 mm 19 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

23. (a)

C1 =

ε0 εr A1 8 .854 × 2 × 10 −12 × 2000 × 10 −4 = 1770 .8 pF = d1 2 × 10− 3

C2 =

8. 854 × 8 × 10−12 × 2000 × 10− 4 = 4722 pF 3 × 10−3

∴C=

1770. 8 × 4722 = 1288 pF 1770. 8 + 4722

Q = CV = 1288 × 5000 × 10−12 = 6.44 × 10−6 C

(b)

∴ V1 =

6 .44 × 10 −6 = 3636 V 1771 × 10−12

∴ E1 =

3636 = 1.82 MV/m 2 × 10 − 3

V2 = 5000 − 3636 = 1364 V E2 =

1364 = 0.454 MV/m 3 × 10 −3

(c) W = 12 CV 2 =

1 2

× 1288 × 10−12 × 52 × 106 = 0.0161 J

24. 0.245 m2, 2.3 MV/m in air-gap Q = CV = k × 1 × V1 = k × 6 × V2

25. ∴

V1 6 6 = ∴ V1 = × 5000 = 4284 V 1+ 6 1 V2

∴ E1 =

4284 = 857 kV/m × 5 10− 3

and E2 =

714 = 143 kV/m 5 × 10 − 3

Hence maximum electric field strength in air 26. 20 µC, 15 µC; 1000 µJ, 2250 µJ; 35 µC; 140 V; 2450 µJ 27.

C=

ε 0 ε r A ε 0 ε r × 2002 × 10 −6 = 40 ε 0ε r = 1 × 10 − 3 d

=

Q 0.0011 × 32 = = 352 × 10 − 6 × 10 −6 100 V

ε0ε r =

352 × 10−12 = 8.8 × 10− 12 F/m 40

electric field strength 100 kV/m electric flux density 0.88 µC/m2 28. 2.81; 30 kV/m, 0.7425 µC/m2; 0.4455 µJ 29.

i,v 0.2 A 500 V 0.1 A

Voltage

t/s 1

2

3 Current

–0.1 A

20 © Pearson Education Limited 2005

4

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

30. T = CR = 100 × 10−6 × 8 × 103 = 0.8 s i=C

(a)

Initially i = V/R = 100/8000 = 12.5 mA

= 100 × 10−6 ×

∴ ∴

dv dt

dv = 12.5 × 10−3 dt

dv = 125 V/s dt

(b) i0 = 12.5 mA (c) Q = CV = 100 × 10−6 × 100 = 10 mC (d) W = 12 CV 2 = 12 × 100 × 10−6 × 1002 = 0.5 J 31. T = CR = 10 × 10−6 × 50 × 103 = 0.5 s T

P.D. across C

50 V

T T 0

Time 0.5 s

i0 =

1.0 s

1.5 s

2.0 s

50 V = R 50 × 103

= 1 mA 1 0.5 s

Current (mA)

0.5 s

0.5

0

0.5 s

1.0 s

1.5 s

2.0 s

32. 18.4 µA, 4000 µJ 33.

i=

V −t/CR e ∴ e−t/CR = 0.9 ∴ t/CR = 0.1054 R

∴ t = 0.1054 × 0.01 × 10−6 × 100 × 103 = 105 µs i0 = V/R =

1000 = 10 mA ∴ VR = 10 × 10−3 × 0.9 × 100 × 103 = 900 V 100 × 10−3

∴ Vc = V − VR = 100 V 21 © Pearson Education Limited 2005

HUGHES: ELECTRICAL AND ELECTRONIC TECHNOLOGY, 9E INSTRUCTOR’S MANUAL

34. 50 mA, 500 mA, 1.25 J 35. v = V(e−t/CR) ∴ 60 = 230e−t/CR ∴

t = 1.344 CR

∴ t = 1.344 × 20 × 10−6 × 50 × 106 = 1344 s or 22 min 24 s

36. 100 MΩ 37. W = 12 CV 2 ∴ 8.5 = 12 × C × 22 × 106 ∴ C = 4.25 µF C=

ε0εr A 4 .25 × 10 −6 × 0. 1 × 10 −3 ∴A= = 8.727 m2 d 8. 854 × 10 −12 × 5. 5

= l × b = l × 0.11 ∴ l = 79.34 m For this length, the plates would probably be wound spirally in ...