PART ONE. Solutions to End-of-Chapter Problems

Transcription

1 PAT ONE Solutions to End-of-Chapter Problems

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3 CHAPTE QUANTITIES AND UNITS SECTION - Scientific and Engineering Notation. (a) (b) 75, (c),000, (a) (b) (c) 5,000, (a) (b) 99, (c) (a) (b) (c) (already in scientific notation) 5. (a) (b) (c) (a) (b) (c) (a) (b) (c) (a) (b) (c)

4 9. (a) (5 0 3 )(4 0 5 ) (b) (. 0 )(3 0 ) (c) (. 0 9 )(7 0 6 ) (a) (b) ( 8) (c) ( 5) (a) 89, (b) 450, (c),040,000,000, (a) (b) (c) (already in engineering notation) 3. (a) (b) (c) (a) (b) (c) (a) ( ) (b) ( ) (c) ( ) (a) (3 0 3 )( ) 79 0 ( 3 + 3) (b) (. 0 6 )(. 0 6 ).44 0 ( 6 6)

5 (c) (00)( ) (a) (b) (c) (3 ( 6)) (3 3) SECTION - Units and Metric Prefixes 8. (a) 89,000 Ω kω (b) (c) 450,000 Ω kω,040,000,000,000 Ω TΩ 9. (a) A A 345 μa (b) (c) 0.05 A A 5 ma A A.9 na 0. (a) A 3 ma (b) k (c) 0 0 F 0 pf. (a) F 3 μf (b) Ω 3.3 MΩ (c) A 350 na. (a) 5 μa A (b) 43 m (c) 75 kω Ω (d) 0 MW W SECTION -3 Metric Unit Conversions 3. (a) (5 ma) ( 0 3 μa/ma) μa 5000 μa (b) (c) (d) (300 μw)( 0 3 W/μW) 3. mw (5000 k)( 0 3 ) M/k 5 M (0 MW)( 0 3 kw/mw) kw 0,000 kw 5

6 4. (a) ma μa A A (b) 0.05 k m , (c) 0.0 kω Ω 6 MΩ 0 Ω (d) 55 mw kw W 3 0 W (a) 50 ma μa 50 ma ma ma (b) (c) 0 kω +. MΩ 0. MΩ +. MΩ.3 MΩ 0.0 μf pf 0.0 μf μf μf 6. (a) 0 kω 0 kω. kω + 0 kω. kω m 50 0 (b) 6 50 μ MW 0 (c) kw 0 6 SECTION -4 Measured Numbers 7. (a) has 3 significant digits. (b) has significant digits. (c) 50.0 has 5 significant digits. (d) has significant digits. (e) 0.05 has 3 significant digits. (f).6 0 has significant digits. 8. (a) 50, (b) (c) (d) (e)

7 CHAPTE OLTAGE, CUENT, AND ESISTANCE BASIC POBLEMS SECTION - Electrical Charge. Q (charge per electron)(number of electrons) ( C/e)( e) 80 0 C. ( e/c)( C) e 3. The magnitude of the charge on a proton (p) is equal to the magnitude of the charge on the electron (e). Therefore, ( C/p)(9 p) C 4. ( C/p)(7 p) C SECTION -3 oltage 5. (a) W 0 J 0 (b) Q C W 5 J.5 (c) Q C W 00 J 4 Q 5 C W 500 J 5 Q 00 C W 800 J 0 Q 40 C 8. W Q ( )(.5 C) 30 J 9. W.5 J.5 Q 0. C SECTION -4 Current 0. Q 0. C I 0 ma t 0 s 7

8 . (a) Q 75 C I 75 A (b) t s Q 0 C I 0 A (c) t 0.5 s Q 5 C I.5 A t s. Q 0.6 C I 0. A t 3s Q 3. I ; t t Q I 0 C 5 A 4. Q I t (.5 A)(0. s) 0.5 C s SECTION -5 esistance 5. A: Blue, gray, red, silver: 6800 Ω ± 0% B: Orange, orange, black, silver: 33 Ω ± 0% C: Yellow, violet, orange, gold: 47,000 Ω ± 5% 6. A: min 6800 Ω 0.(6800 Ω) 6800 Ω 680 Ω 60 Ω max 6800 Ω Ω 7480 Ω B: min 33 Ω 0.(33 Ω) 33 Ω 3.3 Ω 9.7 Ω max 33 Ω Ω 36.3 Ω C: min 47,000 Ω (0.05)(47,000 Ω) 47,000 Ω 350 Ω 44,650 Ω max 47,000 Ω Ω 49,350 Ω 7. (a) st band red, nd band violet, 3rd band brown, 4th band gold (b) 330 Ω; orange, orange, brown, (B). kω: red, red, red (D) 39 kω: orange, white, orange (A) 56 kω: green, blue, orange (L) 00 kω: brown, black, yellow (F) 8. (a) 36.5 Ω ± % (b).74 kω ± 0.5% (c) 8.5 kω ± % 9. (a) Brown, black, black, gold: 0 Ω ± 5% (b) Green, brown, green, silver: 5,00,000 Ω ± 0% 5. MΩ ± 0% (c) Blue, gray, black, gold: 68 Ω ± 5% 0. (a) 0.47 Ω ± 5%: yellow, violet, silver, gold (b) 70 kω ± 5%: red, violet, yellow, gold 8

9 (c) 5. MΩ ± 5%: green, brown, green, gold. (a) ed, gray, violet, red, brown: 8,700 Ω ± % 8.7 kω ± % (b) Blue, black, yellow, gold, brown: 60.4 Ω ± % (c) White, orange, brown, brown, brown: 930 ± % 9.3 kω ± %. (a) 4.7 kω ± %: brown, yellow, violet, red, brown (b) (c) 39. Ω ± %: orange, white, red, gold, brown 9.76 kω ± %: white, violet, blue, brown, brown 3. (a) 0 Ω (b) kω (c) 83 8 kω (d) 3K3 3.3 kω (e) Ω (f) 0M 0 MΩ Ω, equal resistance on each side of the contact. SECTION -6 The Electric Circuit 5. There is current through Lamp. 6. See Figure -. Figure - SECTION -7 Basic Circuit Measurements 7. See Figure -(a). 9

10 Figure - Figure See Figure -(b). 9. Position : 0, S Position : S, See Figure On the 600 DC scale: (0)(0 Ω) 00 Ω 33. (a) (00 Ω) 00 Ω (b) 5(0 MΩ) 50 MΩ (c) 45(00 Ω) 4500 Ω 34. See Figure -4. Figure -4 0

11 ADANCED POBLEMS 35. I t Q Q I t ( A)(5 s) 30 C W 000 J 33.3 Q 30 C 36. I t Q Q (number of electrons) / (number of electrons/coulomb) e Q C e/c Q C I A 3 t 50 0 s 37. Total wire length 00 ft esistance per 000 ft (000 ft)(6 Ω/00 ft) 60 Ω Smallest wire size is AWG 7 which has 5.47 Ω/000 ft 38. (a) 47J 4.7 Ω ± 5% (b) 560KF 560 kω ± % (c) M5G.5 MΩ ± % 39. The circuit in (b) can have both lamps on at the same time. 40. There is always current through See Figure -5. Figure -5

12 4. See Figure See Figure -6. Figure -6

13 CHAPTE 3 OHM S LAW, ENEGY, AND POWE BASIC POBLEMS SECTION 3- Ohm s Law. I is directly proportional to and will change the same percentage as. (a) I 3( A) 3 A (b) I A (0.8)( A) A 0.8 A 0. A (c) I A + (0.5)( A) A A.5 A. (a) When the resistance doubles, the current is halved from 00 ma to 50 ma. (b) When the resistance is reduced by 30%, the current increases from 00 ma to I /0.7.49(/) (.49)(00 ma) 43 ma (c) When the resistance is quadrupled, the current decreases from 00 ma to 5 ma. 3. Tripling the voltage triples the current from 0 ma to 30 ma, but doubling the resistance halves the current to 5 ma. SECTION 3- Application of Ohm s Law 4. (a) I 5 Ω 5 A (b) I 5 0 Ω.5 A (c) I Ω 0.5 A (d) I 30 5 kω ma (e) I 5. (a) I MΩ 9.7 kω 53. μa 3.33 ma (b) I kω 550 μa (c) I (e) I kω 66 k 0 MΩ 588 μa (d) I 6.60 ma k kω 500 ma 3

14 6. I 0 Ω. A 7. (a) I (c) I 5 0 kω 5.8 kω.50 ma (b) I 8.33 ma 5. MΩ.7 μa 8. Orange, violet, yellow, gold, brown 37.4 Ω ± % S I 0.3 A 37.4 Ω 4 9. I 37.4 Ω 0.64 A 0.64 A is greater than 0.5 A, so the fuse will blow. 0. (a) I ( A)(8 Ω) 36 (b) I (5 A)(47 Ω) 35 (c) I (.5 A)(60 Ω) 550 (d) I (0.6 A)(47 Ω) 8. (e) I (0. A)(470 Ω) 47. (a) I ( ma)(0 Ω) 0 m (b) I (50 ma)(33 Ω).65 (c) I (3 A)(4.7 kω) 4. k (d) I (.6 ma)(. kω) 3.5 (e) I (50 μa)( kω) 50 m (f) I (500 ma)(.5 MΩ) 750 k (g) I (850 μa)(0 MΩ) 8.5 k (h) I (75 μa)(47 Ω) 3.53 m. S I (3 A)(7 Ω) 8 3. (a) I (3 ma)(7 kω) 8 (b) I (5 μa)(00 MΩ) 500 (c) I (.5 A)(47 Ω) (a) (c) (e) 5. (a) (c) I I I I I 0 A 50 5 A A 0 k 5 A ma 5 Ω (b) 0 Ω (d) 300 Ω kω (b) kω (d) I I I I A A 7 ma Ω 0.55 Ω 3.5 kω kω 500 μa 4

15 (e) I k ma MΩ 6. I 6 ma 3 kω 7. (a) I 8 A 4 Ω (b) I 4 ma 3 kω (c) I μa 0. MΩ 00 kω 8. I Ω 0.8 A SECTION 3-3 Energy and Power 9. P W t 6 J.6 W 0 s 0. Since watt joule, P 350 J/s 350 W. P W t 7500 J 5 h 7500 J 5 h h 3600 s 7500 J 8,000 s 0.47 J/s 47 mw. (a) 000 W 0 3 W kw (b) 3750 W W 3.75 kw (c) 60 W W 0.60 kw (d) 50,000 W W 50 kw 3. (a),000,000 W 0 6 W MW (b) W 3 MW (c) W W 50 MW (d) 8700 kw W 8.7 MW 4. (a) W W 000 mw (b) 0.4 W W 400 mw (c) 0.00 W 0 3 W mw (d) 0.05 W W.5 mw 5. (a) W,000,000 μw (b) W 500 μw (c) 0.5 mw 50 μw (d) mw 6.67 μw 5

16 6. (a).5 kw W 500 W (b) 0.5 MW W 500,000 W (c) 350 mw W W (d) 9000 μw W W 7. W P in watts t W Q Q I t W P I t So, ( )( A) W 8. W J P W t s kw 000 W 000 J s kw-second 000 J kwh J kwh J SECTION 3-4 Power in an Electric Circuit 9. P I (5.5 )(3 ma) 6.5 mw 30. P I (5 )(3 A) 345 W 3. P I (500 ma) (4.7 kω).8 kw 3. P I ( A) (0 0 3 Ω) 0 4 W 00 μw 33. P (60 ) 60 Ω 5.8 W 34. P (.5 ) 56 Ω W 40. mw 35. P I P 00 W I ( A) 5 Ω 6

17 watts for minute kwmin kwmin 60 min/ hr 83.3 kwh W/s (000 W/kW)(3600 s/h) kwh 38. (50 W)( h) 600 Wh 50 W 0.05 kw (0.05 kw)( h) 0.6 kwh I 0 Ω 0.5 A L P I (.5 )(0.5 A) 0.56 W 56 mw 40. P W t 56 mj 56 mw s W (56 mj/s)(90 h)(3600 s/h) 50,544 J tot SECTION 3-5 The Power ating of esistors 4. P I (0 ma) (6.8 kω) 0.68 W Use the next highest standard power rating of W. 4. If the 8 W resistor is used, it will be operating in a marginal condition. To allow for a safety margin of 0%, use a W resistor. SECTION 3-6 Energy Conversion and oltage Drop in a esistance 43. (a) + at top, at bottom of resistor (b) + at bottom, at top of resistor (c) + on right, on left of resistor SECTION 3-7 Power Supplies and Batteries 44. OUT P ( W)(50 Ω) 7.07 L L 7

18 45. Ampere-hour rating (.5 A)(4 h) 36 Ah 80 Ah 46. I 8 A 0 h 47. I 650 mah 48 h 3.5 ma 48. P LOST P IN P OUT 500 mw 400 mw 00 mw P OUT 400 mw % efficiency 00% 00% 80% PIN 500 mw 49. P OUT (efficiency)p IN (0.85)(5 W) 4.5 W SECTION 3-8 Introduction to Troubleshooting 50. The 4th bulb from the left is open. 5. If should take five (maximum) resistance measurements. ADANCED POBLEMS 5. Assume that the total consumption of the power supply is the input power plus the power lost. P OUT W P OUT % efficiency 00% PIN POUT W P IN 00% 00% % efficiency 3.33 W 60% The power supply itself uses P IN P OUT 3.33 W W.33 W Energy W Pt (.33 W)(4 h) 3.9 Wh 0.03 kwh 53. f I A 50 Ω 8

19 54. Measure the current with an ammeter connected as shown in Figure 3-. Then calculate the unknown resistance with the formula, /I. Figure Calculate I for each value of : I 0 00 Ω 0 A I 0 00 Ω 00 ma 0 30 I 3 00 ma I ma 00 Ω 00 Ω I ma I ma 00 Ω 00 Ω I 7 I 9 I Ω Ω Ω 600 ma I ma I 0 A Ω Ω 700 ma 900 ma The graph is a straight line as shown in Figure 3-. This indicates a linear relationship between I and. Figure 3-9

20 56. S I 5 ma 00 Ω (a) I (c) I (e) I S S S.5 00 Ω 3 00 Ω 0 00 Ω 7.5 ma (b) I 5 ma (d) I 50 ma S S 00 Ω 4 00 Ω 0 ma 0 ma 57. I A 0.5 Ω I A Ω 3 I 0.5 A Ω ma 0 50 ma (0 )(30 ma) 50 ma 6 new value The voltage decreased by 4, from 0 to The current increase is 50%, so the voltage increase must be the same; that is, the voltage must be increased by (0.5)(0 ) 0. The new value of voltage is 0 + (0.5)(0 ) (0.4 CM Ω/ft)(4ft) Wire resistance: W 0.54 Ω 64.3 CM (a) 6 I 59.9 ma + W Ω (b) (59.9 ma)(00 Ω) 5.99 (c) W For one length of wire, W 0.3 kw 30 days (30 days)(4 h/day) 70 h Energy (0.3 kw)(70 h) 6 kwh kwh 3days P kwh/day kwh/day 4 h/day.0 kw 0

21 63. The minimum power rating you should use is W so that the power dissipation does not exceed the rating. 64. (a) ( ) P 0 Ω 4.4 W (b) W Pt (4.4 W)( min)(/60 h/min) 0.48 Wh (c) Neither, the power is the same because it is not time dependent. 65. (max) I max (max) 0.5 A min 8 Ω A fuse with a rating of less than.5 A must be used. A A fuse is recommended. Multisim Troubleshooting Problems 66. is open. 67. No fault 68. is shorted. 69. Lamp 4 is shorted. 70. Lamp 6 is open.

22 CHAPTE 4 SEIES CICUITS BASIC POBLEMS SECTION 4- esistors in Series. See Figure 4-. Figure 4-. The groups of series resistors are,, 3, 9 4 ; 3, 7, 4, 6 ; 6, 8, ; 0,, 5, 5 See Figure 4-. Figure k + 33 k + 47 k + k 70 k Ω+ 8 Ω + Ω 50

23 SECTION 4- Total Series esistance 5. T 8 Ω + 56 Ω 38 Ω 6. (a) T 560 Ω +.0 kω 560 Ω (b) T 47 Ω + 33 Ω 80 Ω (c) T.5 kω +. kω + 0 kω 3.7 kω (d) T.0 kω +.8 kω + 00 kω +.0 MΩ,0,800 Ω (round to.0 MΩ) 7. (a) T.0 kω kω +. kω 7.9 kω (b) T 0 Ω + 0 Ω + Ω +.0 Ω 33 Ω (c) T.0 MΩ kω +.0 MΩ kω + 0 MΩ 3.4 MΩ See Figure 4-3. Figure T (5.6 kω) 67. kω 9. T 6(47 Ω) + 8(00 Ω) + ( Ω) 8 Ω Ω + 44 Ω 6 Ω 0. T T ( ) 0 kω (4.7 kω +.0 kω +. kω kω) 0 kω.8 kω 8. kω. (a) kω + 33 kω + 47 kω + kω 70 kω (b) Ω + 8 Ω + Ω 50 Ω (c) kω + 8. kω +.0 kω +.0 kω.4 kω (d) Ω Ω Ω Ω Ω.97 kω 3

24 . T kω + 50 Ω +.4 kω +.97 kω 84.4 kω SECTION 4-3 Current in a Series Circuit 3. I S T 0 Ω 0. A 4. I 5 ma at all points in the circuit. SECTION 4-4 Application of Ohm s Law 5. (a) T. kω kω +.0 kω 8.8 kω 5.5 I 65 μa 8.8 kω T (b) T.0 MΩ +. MΩ kω 3.76 MΩ 6 I 4.6 μa 3.76 MΩ The ammeters are connected in series. See Figure 4-4. Figure (a). kω S T 8.8 kω 5.6 kω S T 8.8 kω 3.0 kω 3 S m T 8.8 kω 4

25 .0 MΩ (b) S T 3.76 MΩ. MΩ S T 3.76 MΩ kω 3 S 6.38 T 3.76 MΩ 7. (a) T 3(470 Ω) 40 Ω 48 I 34.0 ma 40 Ω (b) I (34.0 ma)(470 Ω) 6 (c) P min I (34.0 ma) 470 Ω W 8. T I each S T 5 ma 5 kω 5 kω.5 kω 4 SECTION 4-5 oltage Sources in Series 9. See Figure 4-5. Figure The total voltage is SECTION 4-6 Kirchhoff s oltage Law. S S S ( ) 0 ( )

26 3. (a) By Kirchhoff s voltage law: ( ) See Figure 4-6(a). (b) 8 ; 6 ; 3 4 ; 4 3 See Figure 4-6(b). Figure 4-6 SECTION 4-7 oltage Dividers Ω Ω 4.4% 5. (a) 47 Ω AB Ω (b). kω kω 5.5 kω AB k. k 3.3 k Ω + Ω + Ω 6.5 kω A S kω kω kω S kω 3 B S 3 C 6

27 3 680 Ω 7. min S Ω Ω max S Ω 8. T k 0 (by measurement); I.79 ma; 5.6 kω k (.79 ma)( kω).79 ; 560 (.79 ma)(560 Ω) ; 0k (.79 ma)(0 kω) 7.9 SECTION 4-8 Power in Series Circuits 30. P T 5(50 mw) 50 mw 3. T 5.6 kω + kω Ω + 0 kω 7.6 kω P I T (.79 ma) (7.6 kω) W 55 mw SECTION 4-9 oltage Measurements 3. oltage from point A to ground (G): AG 0 esistance between A and G: AG 5.6 kω kω +.0 kω +.0 kω 3. kω esistance between B and G: BG 5.6 kω +.0 kω +.0 kω 7.6 kω esistance between C and G: CG.0 kω +.0 kω kω BG CG DG BG AG CG AG DG AG 7.6 kω k Ω kω k Ω.0 kω k Ω

28 33. Measure the voltage at point A with respect to ground and the voltage at point B with respect to ground. The difference of these two voltages is. A B 34. T kω kω + 00 kω +.0 MΩ + 00 kω.3 MΩ T 5.76 MΩ A AG T 5 T.3 M.4 Ω B C D. MΩ BG T 5 T.3 M 7.76 Ω. MΩ CG T 5 T.3 M 7. Ω 00 kω DG T 5 T.3 M 647 m Ω AC A C CA C A SECTION 4-0 Troubleshooting 37. (a) Zero current indicates an open. 4 is open since all the voltage is dropped across it. (b) S ma Ω 4 and 5 have no effect on the current. There is a short from A to B. 38. T 0 kω + 8. kω + kω +. kω kω 38 kω The meter reads about 8 kω. It should read 38 kω. The 0 kω resistor is shorted. ADANCED POBLEMS 39. I (0 ma)(680 Ω) 6.8 I (0 ma)(.0 kω) 0 4 I 4 (0 ma)(70 Ω).7 5 I 5 (0 ma)(70 Ω).7 3 S ( ) 3 30 ( ) I kω 780 Ω 0 ma 8

29 40. T 3(5.6 kω) +.0 kω + (00 Ω) 8 kω Three 5.6 kω resistors, one kω resistor, and two 00 Ω resistors 4. A 0, T kω + 0 kω + 47 kω + kω kω 96.6 kω kω B A k k Ω 0 kω C B 0k k Ω 47 kω D C 47k k Ω kω E D k k Ω F 0 4. I (0 ma)(00 Ω) 5 S Ω I 0 ma P 6 6 mw 80 Ω I (0 ma) 6 I 6 (0 ma)(80 Ω) 5.6 S (0 + 6 ) 30 ( ) Ω I 0 ma Ω I 0 ma 43. S I T (50 ma)(.5 kω) 375 I new 50 ma 0.5(50 ma) 50 ma 6.5 ma 88 ma S 375 new 000 Ω I 88 ma new 500 Ω must be added to the existing 500 Ω to reduce I by 5%. 44. P I I max P 0.5 W 0 Ω A 64.5 ma Since all resistors in series have the same current, use the largest to determine the maximum current allowable because the largest has the greatest power. Thus, the 0 Ω resistor burns out first. 9

30 45. (a) P T W + W + W 0.5 W W W W 8 4 I PT W T 400 Ω 9. ma (b) S I T T (9. ma)(400 Ω) 45.8 (c) P I P 0.5 W /8 I (9.mA) 343 Ω 46. See Figure W /4 (9.mA) 0.5 W / (9.mA) 686 Ω 37 Ω Figure 4-7 Figure See Figure 4-8. When the potentiometer is at minimum setting (0 Ω), OUT 0 : 0 + kω 0 ma ( kω ).0 kω 0 kω.0 kω kω 30

31 When the potentiometer is at maximum setting, OUT 00 : kω + 0 k 00 Ω + (.0 kω + )0 ( kω + )00 0 kω kω kω 54 kω 48. See Figure 4-9. ( ) 5.4 kω ma ( ) 9.9 kω ma ( ) kω ma kω ma A series of standard value resistors must be used to approximately achieve each resistance as follows: 4700 Ω Ω + Ω 5.4 kω 800 Ω Ω + 0 Ω 9.9 kω Ω + 80 Ω + 00 Ω 6.5 kω Ω Ω + 80 Ω + 00 Ω + 00 Ω 8.8 kω The highest power dissipation is in the 800 Ω resistor. P ( ma) 800 Ω 8. mw All resistors must be at least /8 W. Figure 4-9 3

32 49. The groups are:, 7, 8, and 0 ;, 4, 6, and ; 3, 5, 9, and See Figure kω Ω Ω +.0 kω 4.3 kω kω kω kω + 0 kω 3.6 kω kω kω + 8. kω kω 9.9 kω Problem 56: There is a short between these two points. Figure Position : T Ω + 80 Ω Ω.0 kω Position : T Ω + 90 Ω + 80 Ω Ω Ω 3.67 kω 5. Position A: T Ω Ω + 50 Ω +.0 kω. kω I 5.45 ma T. kω Position B: T Ω + 50 Ω +.0 kω.98 kω I 6.06 ma T.98 kω Position C: T Ω +.0 kω.5 kω I 7.95 ma T.5 kω Position D: T 4.0 kω I ma.0 kω T 3

33 5. Position A: T.0 kω 9 I.0 kω T 9 ma Position B: T kω + 33 kω + kω 56 kω 9 I 6 μa 56 kω T Position C: T kω + 33 kω + 68 kω + 7 kω + kω 5 kω 9 I 59.6 μa 5 kω T 53. First, find the value of 5 with the switch in Position D. 8 6 ma +.8 kω ma.8 kω. kω Position A: T 5.38 kω I 8 /5.38 kω 3.35 ma (3.35 ma)(.8 kω) 6.03 (3.35 ma)(.0 kω) (3.35 ma)(80 Ω).75 4 (3.35 ma)(560 Ω) Position B: T 4.8 kω I 8 /4.8 kω 3.73 ma (3.73 ma)(.8 kω) 6.7 (3.73 ma)(.0 kω) (3.73 ma)(80 Ω) Position C: T 4 kω I 8 /4 kω 4.5 ma (4.5 ma)(.8 kω) 8. (4.5 ma)(.0 kω) Position D: T 3 kω I 8 /3 kω 6 ma (6 ma)(.8 kω) Note: The voltage approach can also be used. 54. See Figure 4-0. The results in the table are correct. 33

34 55. Yes, 3 and 5 are each shorted. efer to Figure Yes, there is a short between the points indicated in Figure (a) burned open due to excessive power because it had the largest value in ohms. (b) eplace (0 kω). (c) T 47.7 kω I max P 0.5 W 0 kω 7.07 ma TOTAL I max T (7.07 ma)(47.7 kω) 338 Multisim Troubleshooting Problems 58. is open is shorted. 60. is open. 6. Lamp 4 is open. 6. No fault 63. The 8 Ω resistor is shorted. 34

35 CHAPTE 5 PAALLEL CICUITS BASIC POBLEMS SECTION 5- esistors in Parallel. See Figure 5-.. See Figure 5-. Figure 5- Figure 5- SECTION 5- Total Parallel esistance 3. From Problem : ( + + ) T kω 0 kω kω 68 kω 56 kω 08 kω 3.43 kω 4. T MΩ. MΩ 4.7 MΩ MΩ MΩ 557 kω 5. (a) 47 Ω 6 kω 5.6 Ω (b) 560 Ω.0 kω 359 Ω (c).5 kω. kω 0 kω 89 Ω (d).0 kω. MΩ.0 MΩ 470 kω 996 Ω 35

36 6. (a) T (4.7 kω)(. kω) k Ω +. kω.5 kω (b) T (7 Ω)(56 Ω) + 7 Ω + 56 Ω 8. Ω (c) T (.5 k Ω)(. kω) +.5 k Ω +. kω 89 Ω 7. T kω kω 5 Ω 8. T 3 Ω 5 00 Ω T 0 Ω 0 0 Ω T3 5 Ω T Ω 0 Ω 5 Ω.58 Ω SECTION 5-3 oltage in a Parallel Circuit T I T 0 ma T 600 Ω The total current divides equally among the four equal resistors. 0 ma I I I 3 I 4 5 ma 4 0. The resistors are all in parallel across the source. The voltmeters are each measuring the voltage across a resistor, so each meter indicates 00. SECTION 5-4 Application of Ohm s Law. (a) T 33 kω 33 kω 33 kω kω I T μa kω 36

37 (b) T.0 kω 3.9 kω 560 Ω 39 Ω 5 I T 76 ma 39 Ω. I T I + I + I A 3. (a) I 0 56 k I 0 k 4. T 79 μa 455 μa 8 (b) I 444 μa 8 k 8 I 80 μa 00 k I S T 5.5 ma each 4( kω) 8 kω kω SECTION 5-5 Kirchhoff s Current Law 5. I T 50 ma ma ma 350 ma 6. I T I + I + I 3 + I 4 + I 5 I 5 I T (I + I + I 3 + I 4 ) I ma (50 ma + 50 ma + 5 ma + 00 ma) 500 ma 35 ma 75 ma 7. I 3 I T (I + I 4 ) 50 ma 35 ma 5 ma I I ma See Figure 5-3. Figure

38 8. I T I + I + I 3 + I 4 + I 5 + I A A A A +. A +. A 4.4 A 9. (a) I T I + I + I 3 + I 4 + I 5 + I 6 + I 7 + I A A A A +. A +. A +.0 A +.0 A 6.4 A (b) I ground 6.4 A SECTION 5-6 Current Dividers 0. The 0 kω resistor has the highest current..7 kω. I IT 3 A 3.7 k + Ω.0 kω I IT 3 A 3.7 k + Ω.9 A 8 ma. (a). MΩ I IT 0 μa 3. M 6.88 μa + Ω.0 MΩ I IT 0 μa 3. M 3.3 μa + Ω (b) T 56 Ω kω. kω 3.3 kω 5.6 kω T I I T I I T I 3 I 3 T I 4 I 4 T T T T 56.0 k Ω Ω 0 ma 5.6 ma 56 Ω 0 ma. k.35 ma Ω 56 Ω 0 ma 3.3 k.56 ma Ω 56 Ω 0 ma 5.6 k 9 μa Ω SECTION 5-7 Power in Parallel Circuits 3. P T 5(40 mw) 00 mw 4. (a) T.0 MΩ. MΩ 688 kω P T T T (688 kω) 68.8 μw (b) T.0 kω. kω 3.3 kω 5.6 kω 56 Ω P T (0 ma) (56 Ω) 5.6 mw T T 38

39 5. P I P 75 W I each A I T 6(0.65 A) 3.75 A SECTION 5-8 Troubleshooting P 75 W 6. I each A The current in each bulb is independent of the number of parallel bulbs. I T 3.75 A 0.65 A 3.5 A 7. First determine what the total current should be: T 0 Ω 00 Ω.0 kω 560 Ω 70 Ω Ω 0 I T 0.4 ma Ω The measured current is 00.4 ma which is 0 ma less than it should be. Therefore, one of the resistors must be open. 0 open.0 kω I 0 ma The.0 kω resistor is open. 8. T kω 0 kω 8. kω.3 kω 5 5 I T 0.87 ma T.3 kω The meter indicates 7.8 ma. Therefore, a resistor must be open. 5 I ma 8. kω I I T I M 0.87 ma 7.8 ma 3.05 ma This shows that I 3 is missing from the total current as read on the meter. Therefore, 3 is open. 9. G T Ω 70 Ω 330 Ω 8.5 ms G meas 07.6 Ω 4.8 ms G open G T G meas 8.5 ms 4.8 ms 3.70 ms So, open 70 Ω G 3.70 ms is open. open 39

40 30. G T Ω 00 Ω 0 Ω 0 Ω G meas 4.6 ms 40.7 Ω There is a resistor open. 9. ms G open G T G meas 9. ms 4.6 ms 4.5 ms So, open Ω 4.5 ms One of the 0 Ω resistors is open, but identification requires more information. ADANCED POBLEMS 3. S I ( ma)(50 Ω) 50 m S 50 m 5 Ω I ma 3 I S 3 50 m 0.5 ma 00 Ω I 4 I T (I + I + I 3 ) 7.5 ma 3.5 ma 4 ma 4 S 50 m.5 Ω I 4 ma 4 3. T I T T (00 ma)(5 Ω) 500 m.5 T.5 I 0.4 ma 0 Ω 33. T T 0.48 I 0 A 0 A 4.8 A T 0.48 I 0 A 0 A.4 A T 0.48 I 3 0 A 0 A.6 A 3 3 T 0.48 I 4 0 A 0 A. A

41 34. (a) P T I T T (50 ma) (.0 kω).5 W PT.5 W Number of resistors n 0 Peach 0.5 W All resistors are equal because each has the same power. (b) T n n T 0(.0 kω) 0 kω I 50 ma (c) I T 5 ma n 0 (d) S I T T (50 ma)(.0 kω) (a) I I T I 50 ma 00 ma 50 ma 0 00 ma 00 Ω 0 50 ma 00 Ω (b) P P T P W 0.75 W.5 W S I.5 W.5 W S I S I 0.75 W 0.75 W S I Thus,.5 W 0.75 W I I.5I 0.75I.5 I I I I + I 00 ma.67i 00 ma 00 ma I ma I.67(74.9 ma) 5 ma 0.75 W S 74.9 ma 0 0 I 5 ma 80 Ω I S ma 34 Ω 4

42 00 (c) I 3 00 ma.0 kω 00 I 47 ma 680 Ω I I T I I 0.5 A 47 ma 00 ma 53 ma Ω 53 ma 36. (a) T 50 kω (b) T + 50 kω 470 kω (c) T 50 kω (d) T 3 45 kω kω 470 kω 90 kω 93 kω 37. I max 0.5 A 5 T(min) I (68 Ω) 68 Ω + max A T(min) 30 Ω (68 Ω) (30 Ω)(68 Ω + ) Ω 38. Position A: T kω 0 kω 70 kω 99.7 kω 4 4 I T 4 μa T 99.7 kω 4 I 4.9 μa 560 kω 4 I 09 μa 0 kω I μa 70 kω 4

43 Position B: T kω 0 kω 70 kω.0 MΩ 80 kω. MΩ 78.7 kω 4 4 I T 305 μa T 78.7 kω I 4.9 μa I 09 μa I μa 4 I μa.0 MΩ 4 I μa 80 kω I 6 4. MΩ 0.9 μa Position C: T MΩ 80 kω. MΩ 374 kω 4 4 I T 64. μa T 374 kω I μa I μa I μa 39. From Kirchhoff s current law, the total current into the room is equal to the current in the appliances: I HEATE + I LAMPS + I ACUUM 8.0 A + (0.833 A) A 4.7 A. The current into the room is less than the capacity of the breaker, so the vacuum cleaner can be plugged in without exceeding the capacity of the breaker. 40. T I 4 4 (00 ma)(5 Ω) 500 m.5 T.5 I 0 Ω.4 ma 0 Ω kω 0 kω 560 kω 390 kω. MΩ 00 kω 33.6 kω kω.0 MΩ 80 kω 680 kω 35. kω kω.8 MΩ 78.9 kω 4. T + T ( + ) T + T T T T ( T ) T ( T ) (680 Ω)(00 Ω) 680 Ω 00 Ω 83 Ω 43

44 43. I.5 ma. ma 0.3 ma S 3 4 I (0.3 ma)(.0 kω) 0.3 I. ma 0.8 ma 0.4 ma kω 750 Ω I 0.4 ma I μa 3.3 kω I ma 90 μa 709 μa Ω I 709 μa S I T T (50 ma)(.5 kω) 375 I new 50 ma + (0.5)(50 ma) 50 ma ma 33 ma S 375 T(new).0 kω I 33 ma new T(new) T new + T(new) new new T T T(new) T T(new) new T(new) T T T T(new) T(new) (.0 kω)(.50 kω).50 kω.0 kω 6 kω 45. T 4.7 kω 0 kω 0 kω.4 kω 4 I T 0.3 ma.4 kω With the 4.7 kω resistor open, 5 I 5 ma 5 kω Therefore, the 4.7 kω resistor is open. 46. Pins - T.0 kω 3.3 kω 767 Ω (correct reading) When one resistor is open, the reading is either.0 kω or 3.3 kω. Pins 3-4 T 70 Ω 390 Ω 60 Ω (correct reading) When one resistor is open, the reading is either 70 Ω or 390 Ω. Pins 5-6 T.0 MΩ.8 MΩ 680 kω 50 kω 0 kω (correct reading) 5 open: T.8 MΩ 680 kω 50 kω 5 kω 6 open: T.0 MΩ 680 kω 50 kω 6 kω 7 open: T.0 MΩ.8 MΩ 50 kω 84 kω 8 open: T.0 MΩ.8 MΩ 680 kω 330 kω 44

45 47. (a) One of the resistors has burned open because the power exceeded 0.5 W. Since each resistor has the same voltage, the smallest value will reach the maximum power dissipation first, as per the formula P /. (b) P, P (0.5 W)(.8 kω) 30 (c) eplace the.8 kω resistor and operate the circuit at less than 30 or use a higher wattage resistor to replace the existing.8 kω. 48. (a) kω. kω. kω 3.3 kω 8 kω.0 kω 4 Ω (b) kω 4.7 kω 6.8 kω 5.6 kω 5.6 kω.0 kω 58 Ω (c) Ω (d) Ω 49. (a) ( 3 4 ) + ( ) 4 Ω + 58 Ω 940 Ω (b) Ω (c) Ω (d) Ω Multisim Troubleshooting Problems 50. is open is open. 5. No fault 53. (a) The measured resistance between pin and pin 4 agrees with the calculated value. (b) The measured resistance between pin and pin 3 agrees with the calculated value. 45

46 CHAPTE 6 SEIES-PAALLEL CICUITS BASIC POBLEMS SECTION 6- Identifying Series-Parallel elationships., 3, and 4 are in parallel and this parallel combination is in series with both and 5. T ( 3 4 ) (a) in series with the parallel combination of and 3. See Figure 6-(a). (b) in parallel with the series combination of and 3. See Figure 6-(b). (c) in parallel with a branch containing in series with a parallel combination of four other resistors. See Figure 6-(c). Figure 6-3. See Figure 6-. Figure 6-46

47 4. (a) and 4 are in series with the parallel combination of and 3. T ( 3 ) (b) is in series with the parallel combination of, 3, and 4. T + ( 3 4 ) SECTION 6- Analysis of Series-Parallel esistive Circuits 5. T + T (.0 kω)(667 Ω).0 kω 667 Ω T 003 Ω 6. Brown, black, black, gold 0 Ω ± 5% Orange, orange, black, gold 33 Ω ± 5% Two 0 Ω resistors are in series with three 33 Ω resistors that are in parallel. 33 Ω AB 0 Ω + 0 Ω + 0 Ω + Ω 3 Ω 3 7. (a) T 56 Ω + Ω + 00 Ω 00 Ω 56 Ω + Ω + 50 Ω 8 Ω (b) T 680 Ω 330 Ω 0 Ω Ω Ω Ω 79 Ω 8. T Ω + 0 Ω + Ω 3 Ω 3 I T I I ma 3Ω 96.8 ma I I 3 I ma 3 5 (96.8 ma)(0 Ω) 968 m 3 4 (3.3 ma)(33 Ω) (a) T 8 Ω.5 I T.7 ma 8 Ω I I 4 I T.7 ma.7 ma I I ma I (.7 ma)(56 Ω) 655 m 3 I T ( T 3 ) (.7 ma)(50 Ω) 585 m 4 I 4 4 (.7 ma)( Ω) 57 m (b) T(p) Ω 330 Ω 0 Ω Ω T T(p) + 79 Ω 3 I T 3.8 ma 79Ω I I T 3.8 ma 47

48 I T(p) I T 0.5 Ω 3.8 ma 680 Ω 68 μa I 3 T(p) 3 I T 0.5 Ω 3.8 ma 330 Ω.7 ma I 4 T(p) 4 I T 0.5 Ω 3.8 ma 0 Ω.9 ma I (3.8 ma)(680 Ω) (3.8 ma)( Ω) 40 m 0. (a) 4.7 kω kω kω kω kω 5.65 kω 5.65 kω.7 kω.83 kω AB.83 kω 0 kω.55 kω (b) I T S T 6.55 kω 3.87 ma (c) The resistance to the right of AB is.83 kω. The current through this part of the circuit is I 6 /.83 kω 3.8 ma. I 3 I 5 Ω.7 k 3.8 ma 3.8 ma.06 ma kω 8.35 kω I.06 ma μa (d) S 6. From Problem 0, 6. 6 I. ma.7 kω. From Problem 0, I 3.06 ma I µa I 4 I 3 I 5.06 ma 530 µa 530 μa 48

49 SECTION 6-3 oltage Dividers with esistive Loads 56 kω 3. OUT 5 k 7.5 unloaded Ω L.0 MΩ 56 kω 53 kω 56 kω OUT 5 09 k 7.9 loaded Ω 4. With no load: 6.6 kω A 9.9 k 8 Ω 3.3 kω B 9.9 k 4 Ω With a 0 kω resistor connected from output A to ground: (6.6 kω)(0 kω) AG 3.98 kω 6.6 kω + 0 kω 3.98 kω A(loaded) 7.8 k 6.56 Ω With a 0 kω resistor connected from output B to ground: (3.3 kω)(0 kω) BG.48 kω 3.3 kω.48 kω B(loaded) 9.08 k 3.8 Ω efer to Figure 6-3. Figure

50 5. The 56 kω load will cause a smaller decrease in output voltage for a given voltage divider because it has less effect on the circuit resistance than the 0 kω load does. 6. With no load: T 0 kω kω +.7 kω 8.3 kω I. ma 8.3 kω With a 0 kω load: (8.3 kω)(0 kω) T 0 kω kω 8.3 kω + 0 kω I.5 ma 4.54 kω SECTION 6-4 Loading Effect of a oltmeter 7. The voltmeter presents the least loading across the kω load MΩ.0 MΩ 909 kω 909 kω 909 kω M MΩ kω +.0 MΩ.909 MΩ 3.3 MΩ 9. ACT MΩ M 3.3 Δ ACT M ACT 0. % M 00% M % %. The total resistance of the meter is M (0 k / )(0 ) 00 k. M 00 k00 k 50 k 33% 00 k00 k00 k 50 k M. M 0 M00 k 99 k 0 M00 k00 k 99 k M 49.8% 50

51 SECTION 6-5 The Wheatstone Bridge 3. UNK (8 kω)(0.0) 360 Ω 4 4. UNK ; UNK 390 Ω Ω kω 8.4 Ω 5. X 5 kω. kω 4.5 kω 7.33 kω 6. Change in thermistor resistance from 5C to 65C. Δ therm 5 Ω(65C 5C) 5 Ω(40C) 00 Ω At 65C: therm kω + 00 Ω. kω 3.0 kω A S kω B.0 kω kω 4 S 3 4 OUT B A SECTION 6-6 Thevenin s Theorem kω 7. TH 00 kω kω 8 kω TH 5 k Ω.7 8. (a) TH Ω + 78 Ω 47 Ω Ω + 5 Ω 73 Ω 78 Ω TH m Ω + Ω + Ω (b) (c) TH 00 Ω 70 Ω 73 Ω 00 Ω TH m Ω TH 00 kω 56 kω 35.9 kω 56 kω TH.5 56 kω 538 m 5

52 9. TH kω. kω +. kω.5 kω.58 kω 3 4. kω.5 kω TH A B S S k 3.7 k + + Ω Ω kω L TH.4 TH + L.06 L 6.8 kω L.06 I 6 μa L 4.7 kω L SECTION 6-7 The Maximum Power Transfer Theorem 30. TH 00 kω kω 8 kω L TH 8 kω 3. L TH 75 Ω 3. TH 73 Ω Therefore, L TH 73 Ω for maximum power transfer. SECTION 6-8 The Superposition Theorem 33. For the source: T Ω + 56 Ω 7 Ω 8. Ω I T 8.46 ma 8. Ω Ω I 3 56 IT 8.46 ma 5.7 ma (up) Ω For the.5 source: T Ω + 00 Ω 7 Ω 77.3 Ω.5 I T 9.4 ma 77.3 Ω Ω I 3 00 IT 9.4 ma 5.3 ma (up) Ω I 3(total) 5.7 ma ma.0 ma (up) 5

53 34. For the source: Ω I 3 7 IT 8.46 ma.75 ma (up) Ω For the.5 source: I I T 9.4 ma (down) I (total) 9.4 ma.75 ma 6.7 ma (down) SECTION 6-9 Troubleshooting (680 Ω)(4.7 kω) 35. eq 594 Ω 680 Ω kω T 560 Ω Ω Ω 64 Ω The voltmeter indicates Ω The voltmeter should read: Ω The meter reading is incorrect, indicating that the 680 Ω resistor is open. 36. If opens: A 5, B 0, and C 0 4 ( 3 + ) kΩ.6 kω ( 3 ) k 6.8 Ω The 7.6 reading is incorrect.. kω.kω k 4.5 Ω The 5.4 reading is incorrect. The 3.3 kω resistor is open. 38. (a) open: 5 ; (b) (c) 3 open: 3 5 ; open: T kω Ω Ω +. kω 4.3 kω.0 kω 5 5 T 4.3 k Ω 560 Ω 5 5 T 4.3 k.99 Ω 53

54 3 470 Ω T 4.3 k.67 Ω 5. kω T 4.3 k 7.80 Ω (d) (e) 5 open: T kω Ω Ω kω 5.33 kω.0 kω 5 5 T 5.33 k.8 Ω 560 Ω 5 5 T 5.33 k.58 Ω Ω T 5.33 k.3 Ω kω T 5.33 k 9.9 Ω Point C shorted to ground: T kω Ω Ω.03 kω kω 5 5 T.03 k 7.39 Ω 560 Ω 5 5 T.03 k 4.4 Ω Ω T.03 k 3.47 Ω 39. (a) open: 0, (b) open: T kω kω 4.3 kω.0 kω 0 0 T 4.3 k.33 Ω kω T 4.3 k 7.67 Ω 54

55 (c) 3 open: T kω kω 4.3 kω.0 kω 0 0 T 4.3 k.33 Ω kω 0 0 T 4.3 k 7.67 Ω (d) 4 shorted: T kω Ω Ω kω 5.33 kω ADANCED POBLEMS 40. (a) The parallel combination of and 3 is in series with the parallel combination of 4 and 5. This is all in parallel with. (b) and are in series with the parallel combination of 3 and 4. Also, 5 and 8 are in series with the parallel combination of 6 and 7. These two series-parallel combinations are in parallel with each other. 4. esistors 8, 9, and can be removed with no effect on the circuit because they are shorted by the pc connection. See Figure 6-4. Figure

56 4. The circuit is redrawn and simplified as shown in Figure 6-5. ( ( Ω.0 kω) + (.0 kω +.0 kω) 560 Ω) 90 Ω) 56 Ω) 560 Ω ( 58 Ω 90 Ω) + 56 Ω) 560 Ω 7 Ω (a) T ( Ω) ( Ω) (b) I T 60 7 Ω ma 560 Ω (c) I 56 Ω IT (0.56) ma 4 ma Ω + Ω (d) 966 Ω I 90 Ω I 56 Ω (0.55)4 ma 58.7 ma 90 Ω Ω 90 Ω I 90 Ω (90 Ω) (58.7 ma) 90 Ω 53.4 The voltage from point B to the negative side of the battery is Ω Ω B Ω 4. Ω + Ω Ω 4. AB. Figure B. kω kω.05 kω A. kω.05 kω.06 kω T 4.7 kω +.06 kω 5.76 kω.06 kω A kω.05 kω.05 kω B A kω.05 kω.0 kω C B kω (0.5)(.7 ) 850 m 56

57 44. The circuit is simplified in Figure 6-6 step-by-step to determine T. T 6 Ω 0 I T I I 9 6. ma 6 Ω 864 Ω I 6.mA ma Ω 80 Ω I 3 I 8 6.mA ma Ω 880 Ω I ma ma Ω 80 Ω I ma ma Ω Figure Using the currents found in Problem 44: I T (6. ma)(00 Ω).6 I (8.6 ma)(80 Ω) I 3 3 (7.84 ma)(0 Ω).7 4 I 4 4 (4.06 ma)(80 Ω) I 5 5 (3.78 ma)(00 Ω) 378 m 6 I 5 6 (3.78 ma)(680 Ω) m 8 I 8 8 (7.84 ma)(0 Ω).7 9 I T 9 (6. ma)(00 Ω).6 57

58 46. esistance of the right branch: Ω Ω Ω + 00 Ω 70 Ω esistance of the left branch: L Ω Ω 030 Ω Total resistance: T + L.0 kω Ω 643 Ω 00 I T 60.9 ma 643 Ω Current in right branch: L 030 Ω I IT 60.9 ma L ma + Ω Current in left branch: 70 Ω I L IT 60.9 ma L ma + Ω oltages with respect to the negative terminal of the source: A I 4 (38.0 ma)(560 Ω).3 B I ( ) (.9 ma)(780 Ω) 7.9 AB A B Writing KL around outside loop, and substituting I (I T ma): 0 + (I T ma) 47 kω + I T (33 kω) 0 (47 kω + 33 kω) I T I T ma 80 k 3 I T 3 (3.34 ma)(33 kω) 0 I (I T ma).34 ma S kω I.0 ma 48. C GND 6.0 kω B GND ( ) 4 kω. kω.05 kω A GND ( 3 + B GND ).05 kω. kω.06 kω T + A GND 5.6 kω +.06 kω 6.66 kω.06 kω A k.86 Ω.05 kω B k.47 Ω.0 kω C.47 k 735 m Ω 58

59 49. I max 00 ma 4 T 40 Ω 00 ma 4 6 T 4 6 T 6(40 Ω ) 60 Ω 4 40 Ω 60 Ω 80 Ω With load: L 60 Ω 000 Ω 56.6 Ω 56.6 Ω OUT Ω + Ω 50. efer to Figure T kω 5 ma kω + kω 4 kω Ω Ω With kω loads across the.5 and the 5 outputs: A ( 3L + ) L S ( 3L + ) L + ( ) ( ) 500 Ω k Ω Ω kω 500 Ω k Ω Ω kω+ kω Ω Ω + kω 59

60 3.3.5 Ω kω 333 Ω kω Ω B A 3 Figure efer to Figure 6-8(a). With the source acting alone: T.96 kω I T.0 ma.96 kω. kω I.0 ma. k.69 k 577 μa Ω + Ω.0 kω I μa.0 k. k 80 μa up Ω + Ω efer to Figure 6-8(b). With the 3 source acting alone: T.96 kω 3 I T.53 ma.96 kω.69 kω I 5.53 ma 665 μa up. kω +.69 kω I 5(total) 665 μa + 80 μa 845 μa 60

61 Figure Using Superposition: Current from the source: T 0 Ω + 00 Ω 560 Ω 00 Ω 80 Ω 0 Ω Ω 64 Ω I T 45.5 ma 64 Ω 43.5 Ω I L 45.5 ma 80 Ω.4 ma (up) Current from the 6 source: T 00 Ω + 0 Ω 560 Ω 00 Ω 80 Ω 00 Ω + 57 Ω 57 Ω 6 I T 38. ma 57 Ω 57 Ω I L 38. ma 80 Ω.66 ma (up) Current from the 0 source: T 560 Ω + 0 Ω 00 Ω 00 Ω 80 Ω 560 Ω Ω 599 Ω I T ma 599 Ω 38.8 Ω I L 6.7 ma 80 Ω 0.79 ma (down) Current from the 5 source: T 00 Ω + 0 Ω 00 Ω 560 Ω 80 Ω 00 Ω + 57 Ω 57 Ω 5 I T 3.8 ma 57 Ω 57 Ω I L 3.8 ma 80 Ω. ma (down) I L(total ).4 ma +.66 ma 0.79 ma. ma.08 ma (up) 6

62 53. efer to Figure 6-9(a). TH.65 kω.35 kω.5 kω efer to Figure 6-9(b)..69 kω TH k 3. Ω efer to Figure 6-9(c). 0 kω k.7 Ω kω Figure (a) When SW is connected to +, the voltage at the junction of 3, 4, and 5 is 4 kω 6.4 kω.6 k Ω k ( 4 k 6.4 k ) Ω + Ω Ω 36.6 kω The voltage at the junction of 5, 6, and 7 is 4 kω 36 kω 4.4 k 3 Ω k ( 4 k 36 k ) Ω + Ω Ω 36.4 kω 4 kω 4 kω OUT k 36 k.5 Ω Ω 6

63 (b) When SW is connected to +, the voltage at the junction of,, and 3 is 4 kω 4.6 kω.k Ω k ( 4 k 4.6 k ) Ω + Ω Ω 36.kΩ The voltage at the junction of 3, 4, and 5 is 4 kω 4.4 kω kω + 4 kω 4.4 kω ( ).6 kω kω 4 kω 4.4 kω 4.4 k 3 Ω.08.3 k ( 4 k 4.4 k ) Ω + Ω Ω 6.4 kω 4 kω 4 kω OUT k 36 k 0.75 Ω Ω 55. See Figure kω 3.9 kω kω. kω 4.7 kω 68 kω 9. kω 47 kω 7 kω 33 kω. kω 6. kω.0 kω Figure See Figure 6-. Figure 6-63

64 57. Position : T 0 kω + 30 kω 330 kω 0 kω kω 37.5 kω 7.5 kω k 88.0 Ω 0 kω k 58.7 Ω 0 kω k 9.3 Ω Position : T 0 kω + 0 kω 330 kω 0 kω kω 38.9 kω 0 kω kω k 89. Ω 8.9 kω k 58. Ω 0 kω k 9. Ω Position 3: T 30 kω + 0 kω 330 kω 30 kω kω 39.7 kω 0 kω kω k 89.8 Ω 0 kω kω k 59.6 Ω 9.7 kω k 9.3 Ω 58. (a) 560 kω G S DD +. M Ω kω (b) I I DD G MΩ I 5.80 μa S.75 I S.7 ma.5 kω I D S I.7 ma S 5.80 μa (c) D DD I D D 6 (67 μa)(4.7 kω) DS D S DG D G

65 59. The circuit is redrawn in Figure 6-. The meter reading at point A should be: 6 kω A 50 6 k 56.3 Ω The meter reading of 8.8 is incorrect. The most likely failure is an open kω resistor. This will cause the voltage at point A to be higher than it should be. To verify, let s calculate the voltage assuming that one of the kω resistors is open. kω A 50 k 8.8 This verifies an open kω. Ω Now check B :. kω B k 4.3 Ω This meter reading is correct. Figure The circuit is redrawn in Figure 6-3. (0 kω + 47 kω)(00 kω) BG 36.3 kω 0 kω + 47 kω + 00 kω AG 33 kω + BG 33 kω kω 69.3 kω T AG 7 kω 69.3 kω + 7 kω 96.3 kω 69.3 kω AG AG T 96.3 kω 36.3 kω BG BG T 96.3 kω 47 kω 47 kω CG BG correct 57 kω 57 kω AC AG CG correct 65

66 Figure The.5 reading indicated on one of the meters shows that the series-parallel branch containing the other meter is open. The 0 reading on the other meter shows that there is no current in that branch. Therefore, if only one resistor is open, it must be the. kω. 6. The circuit is redrawn in Figure 6-4. The resistance from point A to ground is A kω 8. kω 4.87 kω 4.87 kω 4.87 kω A k 4.87 k 9.57 k 5.3 Ω + Ω Ω The meter reading of 5.3 at point A is correct. 3.3 kω B k. Ω The meter reading of 30 at point B is incorrect. Either the 5.6 kω resistor is shorted or the 3.3 kω resistor is open. Since resistors tend to fail open, the 3.3 kω is most likely open. Figure A 0 6 B kω kω 66

67 Multisim Troubleshooting Problems is open. 65. is shorted. 66. is open. 67. No fault is open is shorted is open. 7. In fact, 5 is shorted, but it must be removed from the bridge before that can be determined. 67

68 CHAPTE 7 MAGNETISM AND ELECTOMAGNETISM BASIC POBLEMS SECTION 7- The Magnetic Field. Since B A φ, when A increases, B (flux density) decreases. φ 500 μwb. B 3000 μwb/m 3000 μt A 0.5 m φ 3. B A There are 00 centimeters per meter. ( m/00 cm m /0,000 cm A 50 cm m 0,000 cm 0.05 m φ BA ( T)(0.05 m ) 37.5 μwb G T (0.6 G)( T/0 4 G) 60 μt 5. T 0 4 G (00,000 µt)(0 4 G/T) 000 G SECTION 7- Electromagnetism 6. The compass needle turns 80. μ 7. μ r μ 0 μ 0 4π 0 7 Wb/At m Wb/At m μ r π 0 Wb/At m l 0.8 m 8. 7 μ A (50 0 Wb/At m )(0.08 m ) 9. F m NI (500 t)(3 A) 500 At At/Wb 68

69 SECTION 7-3 Electromagnetic Devices 0. When a solenoid is activated, its plunger is retracted.. (a) An electromagnetic force moves the plunger when the solenoid is activated. (b) A spring force returns the plunger to its at-rest position.. The relay connects +9 to pin turning on lamp and turning off lamp. 3. The pointer in a d Arsonval movement is deflected by the electromagnetic force when there is current through the coil. SECTION 7-4 Magnetic Hysteresis 4. F m 500 At H F 500 At m 7500 At/m l 0. m 5. The flux density can be changed by changing the current. Fm NI 500(0.5 A) 6. (a) H 47 At/m l l 0.3 m Fm NI (b) φ l μa μ μ r μ 0 μ μ r μ 0 (50)(4π 0 7 ) Wb/At m A ( cm)( cm) (0.0 m)(0.0 m) m (500 t)(0.5 A) 5 At φ m.39 0 At/Wb -7 (34 0 )( m ) φ 5.3 μwb (c) B 30,750 μwb/m A m 5.3 μwb 7. Material A has the most retentivity. SECTION 7-5 Electromagnetic Induction 8. The induced voltage doubles when the rate of change of the magnetic flux doubles. 00 m 9. I induced ma 00 Ω induced 69

70 0. The magnetic field is not changing; therefore, there is no induced voltage. 3 φ.4 0 Wb. B 0.7 Wb/m 0.7 T A (0.085 m) v 44 m v ind Bl sin θ 3.0 m/s (0.7 T)(0.085 m)(sin 90 ). (a) Positive (with respect to other end). (b) The induced force will oppose the motion; it is downward. SECTION 7-6 DC Generators 3. efficiency Pout 0.80 P in Pout 45 W P in 56.3 W I A I F + I L A + A 3 A 5. (a) P I ( A)(4 ) 68 W (b) P I (.0 A)(4 ) 4 W SECTION 7-7 DC Motors 6. (a) P 0.05Ts 0.05(3.0 N-m)(00 rpm) 378 W (b) hp 378 W 746 W 0.5 hp 7. P in 6 W; P out 50 W. efficiency Pout 50 W P 6 W 80.6% in ADANCED POBLEMS rev/s peaks/rev 0 peaks/s 70

71 9. The output voltage has a 0 dc peak with a 0 Hz ripple. See Figure 7-. Figure 7- Multisim Troubleshooting Problems 30. Upper lamp is open. 3. The design is flawed. is too little voltage to operate two relays in series but 4 is too much to operate a lamp. Install a separate power supply for the lamps and change the to 4 for the relays. 7

72 CHAPTE 8 INTODUCTION TO ALTENATING CUENT AND OLTAGE BASIC POBLEMS SECTION 8- The Sinusoidal Waveform. (a) f (b) f (c) f (d) f (e) f (f) f. (a) T (b) T (c) ) T (d) T (e) T (f) T T s Hz T 0. ms 5 Hz T 50 ms 0 Hz T ms khz T 500 μs khz T 0 μs 00 khz f f f f f f Hz 60 Hz 500 Hz khz s 00 khz 5 MHz 6.7 ms ms ms 5 μs 00 ns 3. T 0 μs 5 cycles μs 7

73 4. T f 0 ms 0.0 ms 50 khz 0 μs 0.0 ms 500 cycles in 0 ms 5. T 00 μs f 0 khz (00 μs/cycle)(00 cycles) 0 ms SECTION 8- oltage and Current alues of Sine Waves 6. (a) rms p 0.707( ) 8.48 (b) pp p ( ) 4 (c) AG p 7.64 π π 7. (a) I p.44i rms.44(5 ma) 7.07 ma (b) I AG 0.637I p 0.637(7.07 ma) 4.5 ma (c) I pp I p (7.07 ma) 4.4 ma 8. p 5 pp p 50 rms p 7.7 AG p (a) 7.7 (b) 5 (c) 0 (d) 7.7 SECTION 8-3 Angular Measurement of a Sine Wave 0. (a) 7.5 (b) 5 (c) 0. θ waveform A leading. With respect to 0 : Sine wave with peak at 75 is shifted 5 leading. Sine wave with peak at 00 is shifted 0 lagging. Phase difference: θ

74 3. See Figure 8-. Figure 8-4. (a) (b) (c) (d) (e) (f) π rad rad 80 π rad rad 80 π rad rad 80 π rad rad 80 π rad rad 80 π rad rad (a) (b) (c) (d) (e) (f) π 57.3 /rad.5 8 rad π 57.3 /rad 60 3 rad π 57.3 /rad 90 rad 3π 57.3 /rad 08 5 rad 6π 57.3 /rad 6 5 rad.8π 57.3 /rad 34 rad SECTION 8-4 The Sine Wave Formula 6. p.44(0 ) 8.8 (a) v p sin θ (8.8 )sin5 7.3 (b) v (8.8 )sin (c) v (8.8 )sin 50.7 (d) v (8.8 )sin

75 (e) (f) (g) (h) v (8.8 )sin v (8.8 )sin v (8.8 )sin v (8.8 )sin (a) i I p sin θ (00 ma)sin ma (b) i (00 ma)sin ma (c) i (00 ma)sin ma (d) i (00 ma)sin ma (e) i (00 ma)sin ma (f) i (00 ma)sin ma 8. p.44 rms.44(6.37 ) 9 π (a).5 8 v (9 )sin π (b) 45 4 v (9 )sin π (c) 90 v (9 )sin π (d) 35 4 v (9 )sin (e) π 80 v (9 )sin π (f) 70 v (9 )sin 70 9 (g) π 360 v (9 )sin v (5 )sin ( ) 3.0 v (5 )sin ( ) 4.5 v (5 )sin ( ) 3.0 v (5 )sin ( ) 7.5 v (5 )sin ( ).5 v (5 )sin ( ) v (5 )sin (30 30 ) 0 v (5 )sin (45 30 ) 3.88 v (5 )sin (90 30 ) 3.0 v (5 )sin (80 30 ) 7.5 v (5 )sin (00 30 ).60 v (5 )sin ( ) 5 75

76 SECTION 8-5 Analysis of AC Circuits. (a) 0 I rms kω (b) I AG 0 A over a full cycle. (c) 0 I p.0 kω 0 ma (d) I pp (0 ma) 0 ma (e) i I p 0 ma p 7.07 ma. (rms) (p).44(35 ) 49.5 (AG) 0.637(49.5 ) 3.5 (rms) s (p).44(55 ) 77.8 (AG) 0.637(77.8 ) I pp ma.0 kω I I rms pp 6 ma ma 4 I rms 4 (5.66 ma)(560 Ω) 3.7 rms Applying Kirchhoff s voltage law: s 0.707(8 ) (30 ) p (.44)(0.6 ) 5 max 4 + p 39 min 4 p 9 5. p (.44)(3 ) 4.4 DC p min DC p 5 6 SECTION 8-6 Alternators (AC Generators) 7. f number of pole pairs rev/s 50 rev/s 50 Hz 8. f number of pole pairs rev/s 3600 rev/min rev/s 60 s/min 60 rev/s f pole pairs 60 rev/s 0 Hz 76

77 9. rev/s f pole pairs 400 Hz 00 rps 30. N 0 f 0(400 Hz) 6 poles s 3000 rpm SECTION 8-7 AC Motors 3. A one-phase motor requires a starting winding or other means to produce torque for starting the motor, whereas a three-phase motor is self-starting. 3. The field is set up by current in the stator windings. As the current reaches a peak in one winding, the other windings have less current and hence less effect on the field. The result is a rotating field. SECTION 8-8 Nonsinusoidal Waveforms 33. t r 3.5 ms 0.5 ms 3.0 ms t f 6.0 ms 3 ms 3.0 ms t W 4.5 ms.5 ms.0 ms Amplitude (a) t μs % duty cycle W 00% 00% 5% T 4 μs (b) t 0 ms % duty cycle W 00% 00% 66.7% T 30 ms 35. (a) AG baseline + (duty cycle)(amplitude) + (0.5)(.5 ) (b) AG + (0.67)(3 ) (a) f (b) f 37. (a) f (b) f 50 khz 4 μ s 33.3 Hz 30 ms 0 μ s 00 μ s 50 khz 0 Hz 77

78 38. f 5 khz T 40 μs 3rd harmonic 75 khz 5th harmonic 5 khz 7th harmonic 75 khz 9th harmonic 5 khz th harmonic 75 khz 3th harmonic 35 khz 39. fundamental frequency 5 khz SECTION 8-9 The Oscilloscope 40. olts/div 0. m; Time/div 50 ms p olts/div Number of divisions 0. /div 3 divisions 0.6 T Time/div Number of divisions 50 ms/div 0 divisions 500 ms 4. p 0.6 rms p 0.707(0.6 ) 0.44 T 500 ms f Hz T 500 ms 4. p olts/div Number of divisions /div. divisions. rms p 0.707(. ).56 T Time/div Number of divisions 0. μs/div 6.8 divisions 0.68 μs f.47 MHz T 0.68 μs 43. Amplitude olts/div Number of divisions 0.5 /div.8 div.4 t W Time/div Number of divisions 0. s. div 0. s 0 ms T Time/div Number of divisions 0. s 4 div 0.4 s 400 ms t 0 ms % duty cycle W 00% 00% (0.3)00% 30% T 400 ms ADANCED POBLEMS 44. t f. khz ms At t 0. ms: 0. ms θ ms v (5 )sin 94.9 (35.36)sin

79 At t 0. ms: 0. ms θ ms v (5 )sin 58. (35.36)sin Δv I max max T A 00 Ω + 47 Ω 47 Ω 00 Ω AG DC L DC 00 T Ω See Figure 8-. Figure Average value area under curve/period ( )(ms) ms AG 7 ms 7 ms (a).5 cycles are displayed. (b) p /div.8 div 5.6, rms 0.707(5.6 ) 3.96 (c) T 4 div 0 µs/div 80 μs, f T 80 μs.5 khz 48. See Figure 8-3. Figure 8-3 Figure See Figure

80 50. p(in) ( div)(5 /div) 5 T in ( div)(0. ms/div) 00 μs f in 00 μ s 5 khz tot 560 Ω + (470 Ω (560 Ω Ω)) 560 Ω + 33 Ω 883 Ω 470 Ω 33 Ω 470 Ω 33 Ω p(out) ( ) p in 835 m Ω + Ω Ω Ω Ω f out f in 5 khz The scope display for channel shows five cycles of the output waveform with the peak being division high relative to the zero crossing of the sine wave. 5. p(out) (0. /div)(3 div) 0.6 T out (0 div)(50 ms/div) 500 ms f out Hz 500 ms.0 kω.0 kω (. kω +.0 kω) p(out).0 k. k.0 k.0 k (. k.0 k ) p Ω + Ω Ω + Ω Ω + Ω 76 Ω ( 0.33) ( in).76 k p 0.35 p(in) Ω p ( out) 0.6 p(in) f in f out Hz ( in) Multisim Troubleshooting Problems 5. p 35.3 ; T ms is open. 54. is open. 55. Amplitude 5 ; T ms 56. No fault 80

81 CHAPTE 9 CAPACITOS BASIC POBLEMS SECTION 9- The Basic Capacitor Q 50 μc. (a) C 5 μf 0 (b) Q C (0.00 μf)( k) μc Q mc (c) 0 C 00 μf. (a) (0. μf)(0 6 pf/μf) 00,000 pf (b) (0.005 μf)(0 6 pf/μf) 500 pf (c) (5 μf)(0 6 pf/μf) 5,000,000 pf 3. (a) (000 pf)(0 6 μf/pf) 0.00 μf (b) (3500 pf)(0 6 μf/pf) μf (c) (50 pf)(0 6 μf/pf) μf 4. (a) ( F)(0 6 μf/f) 0. μf (b) (0.00 F)(0 6 μf/f) 00 μf (c) ( F)(0 6 μf/f) μf 5. W C W (0 mj) C (00 ) μf 6. C Aε r ( F/m) (0.00 m )(5)( F/m).39 nf d 63.5 µm 7. C Aε r ( F/m) d (0. m )(.006)( m F/m) F 88.5 pf 8

82 Aε r ( ) 8. C d 5 Cd ()( 8 0 ) A ε r ( ) ( )( ) m l A m.9 0 m (almost. miles on a side) The capacitor is too large to be practical and, of course, will not fit in the Astrodome. 9. C Aε r ( ) d ( 0.09)(.5)( ) nf μf 0. ΔT 50 C ( 00 ppm/ C)50 C 0,000 ppm ΔC ( 0 0 ppm) pf C pf 0 pf 990 pf. ΔT 5 C (500 ppm/ C)5 C,500 ppm ( 0 6 pf/μf)(0.00 μf) 000 pf 000 ΔC,500 ppm.5 pf 0 6 SECTION 9- Types of Capacitors. The plate area is increased by increasing the number of layers of plate and dielectric materials. 3. Ceramic has a higher dielectric constant than mica. 4. See Figure 9-. Figure 9-5. (a) 0.0 μf (b) μf (c) 0.00 μf (d) pf 6. Aluminum, tantalum; electrolytics are polarized, others are not. 8

83 7. (a) Encapsulation (b) Dielectric (ceramic disk) (c) Plate (metal disk) (d) Conductive leads SECTION 9-3 Series Capacitors 8. C T 000 pf 00 pf 5 9. (a) C T + μf. μf 0.69 μf (b) C T 00 pf pf 390 pf 69.7 pf (c) C T μf 4.7 μf 47 μf μf.6 μf 0. (a) C T 0.69 μf CT 0.69 μf μf 0 0 F F μ μ 0.69 μf.μf 0. F 3.3 μ 6.9 (b) (c) C T 69.7 pf 69.7 pf 00pF pf pf 560pF pf pf 390pF pf 7.9 C T.6 μf.6 μf 0μF 30 0 F 7.8 μ 83

84 .6 μf 4.7μF F 6.8 μ.6 μf 47μF F.68 μ.6 μf μf 30 F 3.59 μ. Q T Q Q Q 3 Q 4 0 μc Q 0 μc.3 C 4.7 μf Q C Q 3 C Q 4 C μc μf 0 0 μc. μf μc 0 μf SECTION 9-4 Parallel Capacitors. (a) C T 47 pf + 0 pf pf 057 pf (b) C T 0. μf μf μf μf 0. μf 3. C T C + C. μf μf 5.5 μf Q T C T (5 5 μf)(5 ) 7.5 μc 4. Use four 0.47 μf capacitors and one 0. μf capacitor in parallel: C T 4(0.47 μf) + 0. μf. μf 84

85 SECTION 9-5 Capacitors in DC Circuits 5. (a) τ C (00 Ω)( μf) 00 μs (b) τ C (0 MΩ)(56 pf) 560 μs (c) τ C (4.7 kω)( μf). μs (d) τ C (.5 MΩ)(0.0 μf) 5 ms 6. (a) 5τ 5C 5(47 Ω)(47 μf).04 ms (b) 5τ 5C 5(3300 Ω)(0.05 μf) 48 μs (c) 5τ 5C 5( kω)(00 pf) μs (d) 5τ 5C 5(4.7 MΩ)(0 pf) 35 μs 7. τ C (00 Ω)( μf) 0 μs (a) v C 5 ( e t/c ) 5 ( e 0μs/0μs ) 5 ( e ) 9.48 (b) v C 5 ( e ) 3.0 (c) v C 5 ( e 3 ) 4.3 (d) v C 5 ( e 4 ) 4.7 (e) v C 5 ( e 5 ) τ C (.0 kω)(.5 μf).5 ms (a) v C i e t/c 5e.5ms/.5ms 5e 9. (b) v C i e t/c 5e 4.5ms/.5ms 5e 3.4 (c) v C i e t/c 5e 6ms/.5ms 5e m (d) v C i e t/c 5e 7.5ms/.5ms 5e 5 68 m 9. (a) v C 5 ( e t/c ) 5 ( e μs/0μs ) 5 ( e 0. ).7 (b) v C 5 ( e 5μs/0μs ) 5 ( e 0.5 ) 5.90 (c) v C 5 ( e 5μs/0μs ) 5 ( e.5 ) (a) v C i e t/c 5e 0.5ms/.5ms 5e (b) v C i e t/c 5e ms/.5ms 5e (c) v C i e t/c 5e ms/.5ms 5e SECTION 9-6 Capacitors in AC Circuits 3. (a) X C (b) X C (c) X C (d) X C πfc πfc πfc πfc 339 kω π(0 Hz)(0.047 μf) 3.5 kω π(50 Hz)(0.047 μf) 677 Ω π(5 khz)(0.047 μf) 33.9 Ω π(00 khz)(0.047 μf) 85

86 3. (a) X CT πfc 3.39 kω π( khz)(0.047 μf) (b) C T 0 μf + 5 μf 5 μf X CT 6.37 kω π(hz)(5 μf) (c) C T 0.5 μf + μf μf X CT 5.3 kω π(60 Hz)(0.5 μf) 33. X C X C.4 kω π fc π( khz)(56 nf) 970 Ω π fc π( khz)(8 nf) X CT X C + X C.4 kω kω.39 kω C T C C 56 nf 8 nf 33.3 nf C C C 33.3 nf T S C 56 nf C 33.3 nf T S C 8 nf 34. (a) For X CT 00 Ω: f khz πx C π(00 Ω)(0.047 μf) CT For X CT kω: f khz πx C π( k Ω)(0.047 μf) CT (b) For X CT 00 Ω: f 63.7 Hz πx C π(00 Ω)(5 μf) CT For X CT kω: f 6.37 khz πx C π( k Ω)(5 μf) CT 86

87 (c) For X CT 00 Ω: f 3.8 khz πx C π(00 Ω)(0.5 μf) CT For X CT kω: f 38 Hz πx C π( k Ω)(0.5 μf) CT 35. X C I rms rms 0 00 ma 0. kω 00 Ω 36. rms I rms X C X C 3.39 kω π(0 khz)( μf) rms ( ma)(3.39 kω) X C πfc 3.39 kω P true 0 W P r I rms X C ( ma) (3.39 kω) 3.39 ma SECTION 9-7 Capacitor Applications 38. The ripple voltage is reduced when the capacitance is increased. 39. X C(bypass) ideally should be 0 Ω to provide a short to ground for ac. ADANCED POBLEMS C 40. X T S C X C C T X X (μf)(8 ) μf C C X S T X S μf μf 4 87

88 4. v v i e t/c e t/c v i ln(e t/c ) ln i t C v ln i t C ln i 3 t (.0 kω)(.5 μf)ln 3.8 ms 5 4. v F ( e t/c ) v F F e t/c F e t/c F v e t/c F v F v F ln(e t/c v ) ln F t v ln C F v t C ln F 8 t ( 0 k Ω )(0.00 μ F)ln 7.6 μs Looking from the capacitor, the Thevenin resistance is TH Ω τ TH C (47 Ω)(0.00 μf) 3.4 μs v 44. t C C ln F t 0 μs 7.86 kω v 7. C C ln (000 pf)ln 0 F 88

89 45. τ ( + )C (55 kω)( μf) 55 ms τ ( + 3 )C (43 kω)( μf) 43 ms 5τ 5(43 ms) 5 ms v C 0( e 0ms/.55ms ) 3.3 See Figure 9-(a). v C 3.3e 5ms/43ms.96 See Figure 9-(b). Figure C μf, C μf, C μf, C μf C T μf, X C(tot) 68.8 kω s 0 I C 45 μa X 68.8 kω C(tot) C T μF C s C 0.0μF C s C X C 4. kω C.9 I C 95.0 μa X 4.kΩ C C μF C3 C.9.78 C μf X C kω C 3.78 I C μa X 35.4 kω C3 C4 C C m X C4.3 kω C m I C μa X.3 kω C4 C 5 6 C5 C5 X C5 53. kω C I C5 I C6 X C 4 5 C μf 505 m 303 m 0.0μF 303 m 5.7 μa 53.kΩ C6 C4 C5 505 m 303 m 0 m 89

90 47. C C3 (4 ma)x C3 (4 ma)(750 Ω) 3 f 4.5 khz πx C 3C3 π(750 Ω)(0.005 μf) X C 5 Ω πfc π(4.5 khz)(0.00 F) μ C 3 I C 5.87 ma X C 5 Ω I C I C(tot) I C + I C ma + 4 ma 9.87 ma C 5 3 C X C 03 Ω I 9.87 ma C C πfx 48. Position : C 5 C5 C C T(,5) T(,5) π(4.5 khz)(03 C Ω + C μf μf μf + + C C C μf T(3,6) C 6 C 6 3 T(3,6) Position : C C T(,5) C 5 C5 C C T(4,6) T(,5) 4 6 ) μf μf μf μf μf + C μf μf C T μf C T(3,6) μf C T(,5) μf μf μf + + C T(4,6) 0.08 μf C 0.05 μf μf 6 CT(4,6) 0.08 μf 6.54 C μf Δ increase, Δ decrease 90

91 C tot(3,5,6) μf C tot ( 3,5,6) C3 C5 C μf 0.0μF 0.05 μf C tot(,3,5,6) 0.0 μf μf μf + + C tot μf C C 0.0μF F tot C tot (, 3, 5, 6) μ C C C tot μf μF C Ctot μf Ctot(, 3, 5, 6 ) 0.063μF Ctot(3,5,6) μF C3 C C μf Ctot(3,5,6) μF C5 C.76.9 C5 0.0μF Ctot(3,5,6) μF C6 C C μf C4 C5 + C Multisim Troubleshooting Problems 50. C is leaky. 5. C is open. 5. C is shorted. 53. No fault 54. C is shorted. 9

92 CHAPTE 0 C CICUITS BASIC POBLEMS SECTION 0- Sinusoidal esponse of C Circuits. Both voltages are also sine waves with the same 8 khz frequency as the source voltage.. The current is sinusoidal. SECTION 0- Impedance and Phase Angle of Series C Circuits 3. (a) Z (b) Z + X C ( 70 Ω) + (00 Ω) 88 Ω + X C ( 680 Ω) + (000 Ω) 09 Ω 4. (a) tot 00 kω + 47 kω 47 kω C tot μf + 0.0μF 0.0 μf X C(tot) 3 kω πfc tot π(00 Hz)( μf) Z tot + X C ( tot ) ( 47 kω) + (3 kω) 74 kω θ tan X C Ω tot ) 3k tan 57.6 (I leads ) tot 47 kω (b) C tot 560 pf pf 0 pf X C(tot) 7. kω πfc π(0 khz)(0 pf) Z tot tot + X C ( tot ) ( 0 kω) + (7. kω).3 kω θ tan X Ω C ( tot ) 7.k tan 35.4 (I leads ) 0 kω 9

93 5. (a) X C Z (b) X C Z (c) X C Z (d) X C Z 6. (a) X C Z (b) X C Z (c) X C Z (d) X C Z πfc + X C 73 kω π(00 Hz)(0.00 μf) ( 56 kω) + (73 kω) 76 kω πfc + X C 45 kω π(500 Hz)(0.00 μf) ( 56 kω) + (45 kω) 55 kω 7.3 kω πfc π(.0 khz)(0.00 μf) + X C ( 56 kω) + (7.3 kω) 9.5 kω πfc + X C 8.9 kω π(.5 khz)(0.00 μf) ( 56 kω) + (3.8 kω) 63.0 kω πfc + X C 339 kω π(00 Hz)( μf) ( 56 kω) + (339 kω) 343 kω πfc + X C 67.7 kω π(500 Hz)( μf) ( 56 kω) + (67.7 kω) 87.9 kω πfc + X C 33.9 kω π( khz)( μf) ( 56 kω) + (33.9 kω) 65.4 kω πfc + X C 3.5 kω π(.5 khz)( μf) ( 56 kω) + (3.5 kω) 57.6 kω SECTION 0-3 Analysis of Series C Circuits 7. (a) I (b) I s ma Z 88 Ω s ma Z 09 Ω 93

94 8. (a) I (b) I s 50 8 μa Z 74 kω s 8 65 μa Z.3 kω 9. See Figure 0-. C tot + 0.μF 0. μf μf X C Z I tot 54 Ω π(5 khz)( μf) tot + X C ( 50 Ω) + (54 Ω) 6 Ω s.3 ma Z 6 Ω X C(0.μF) 06 Ω π(5 khz)(0.μf) X C(0. μf) 48. Ω π(5 khz)(0. μf) C I tot X C(0.μF) (.3 ma)(06 Ω).3 C I tot X C(0.μF) (.3 ma)(48. Ω) I tot tot (.3 ma)(50 Ω) 0.66 θ tan X C tot 54 Ω tan 7.0 (I tot leads s ) 50 Ω Figure 0-94

95 0. (a) X C 79.6 Ω π(0 Hz)(00 μf) Z (b) I + X C ( 56 Ω) + (79.6 Ω) 97.3 Ω Ω 03 ma 56 Ω (c) s Z Ω X 79.6 Ω (d) C C s Z Ω. Z 0 s kω I 0 ma X C 589 Ω π(0 khz)(0.07 μf) + X C kω + (589 Ω) (000 Ω) ( 000 Ω ) (589 Ω) 808 Ω θ tan 589 Ω Ω. (a) X C 4.08 MΩ π( Hz)(0.039 μf) φ 90 tan X Ω C 4.08 M 90 tan kω (b) X C 40.8 kω π(00 Hz)(0.039 μf) φ 90 tan X Ω C 40.8 k 90 tan kω (c) X C 4.08 kω π( khz)(0.039 μf) φ 90 tan X Ω C 4.08 k 90 tan kω (d) X C 408 Ω π(0 khz)(0.039 μf) φ 90 tan X Ω C tan kω 95

96 3. (a) X C 5.9 MΩ πfc π( Hz)(0.μF) X C φ tan.59 MΩ tan kω (b) X C 5.9 kω πfc π(00 Hz)(0.μF) X C φ tan 5.9 kω tan kω (c) X C.59 kω πfc π( khz)(0.μf) X C φ tan.59 kω tan kω (d) X C 59 Ω πfc π(0 khz)(0.μf) X C φ tan 59 Ω tan kω SECTION 0-4 Impedance and Phase Angle of Parallel C Circuits 4. Z X C + X C (. kω)(. kω) (. kω) + (. kω).05 kω 5. B C πfc.76 ms G.33 ms 750 Ω Y Z + B C G (.33 ms) + (.76 ms) 3.07 ms Y 3.07 ms.76 ms 36 Ω θ tan 64.3 (I leads ).33 ms 6. (a) B C π(.5 khz)(0. μf).07 ms Y Z (.33 ms) + (.07 ms).47 ms Y θ tan.47 ms.07 ms.33 ms 405 Ω

97 (b) (c) (d) B C π(3.0 khz)(0. μf) 4.5 ms Y Z (.33 ms) + (4.5 ms) 4.36 ms Y θ tan 4.36 ms 4.5 ms.33 ms 30 Ω 7. B C π(5.0 khz)(0. μf) 6.9 ms Y Z (.33 ms) + (6.9mS) 7.04 ms Y θ tan 7.04 ms 6.9mS.33 ms 4 Ω 79. B C π(0 khz)(0. μf) 3.8 ms Y Z (.33 ms) + (3.8 ms) 3.9 ms Y 3.9 ms 3.9 ms 7.0 Ω θ tan ms 7. B C πfc πf(c + C ) π( khz)(0.3 μf) 4.0 ms G ms Ω Y G B ( ms) + (4.0 ms) 4.08 ms Z Y θ tan + C 45 Ω 4.08 ms 4.0 ms ms SECTION 0-5 Analysis of Parallel C Circuits 8. Z tot (68 Ω)(90 Ω) 54.3 Ω (68 Ω) + (90 Ω) C s 0 0 I tot 84 ma 54.3 Ω I I C 0 68 Ω 0 Ω 90 Ω 47 ma ma 97

98 9. X C 67.7 Ω π(50 khz)(0.47 μf) X C 45 Ω π(50 khz)(0.0 μf) I C I C X s X C s C 8 8 ma 67.7 Ω 8 45 Ω 55.3 ma I s ma 0 Ω I s ma 80 Ω ( tot) + IC( tot ( 80.8 ma) + (73.3 ma) I tot I 9 ma I θ tan C( tot ) 73.3 ma tan 65.0 I( tot) 80.8 ma 0. (a) Z X C (.0 k Ω)(. k Ω) XC.0 kω. kω 903 Ω S (b) 00 m I.0 kω (c) S 00 m I C 47.6 μ X C. kω (d) 00 m I tot S Z 903 Ω (e).0 kω θ tan tan X. kω C 5.5 (I tot leads s ). X C 6.77 kω π(500 Hz)(0.047 μf) (a) Z (4.7 k Ω)(6.77 k Ω) (4.7 k Ω ) + (6.77 k Ω) 3.86 kω (b) I s 00 m.3 μa 4.7 kω (c) s 00 m I C 4.8 μa X 6.77 kω (d) I tot s C 00 m Z 5.9 μa 3.86 kω (e) θ tan 4.7 kω kω 98

99 . tot kω, C tot 3.0 pf X C(tot) 49.8 kω π(00 khz)(3.0 pf) Z ( kω)(49.8 kω) ( kω) + (49.8 kω) 0. kω θ tan tot kω tan 3.8 X C(tot) 49.8 kω eq Z cos θ (0. kω)cos kω X C(eq) Z sin θ (0. kω)sin kω C eq 96 pf πfx C (eq) SECTION 0-6 Analysis of Series-Parallel C Circuits 3. X C 06 Ω π(5 khz)(0. μf) X C 6 Ω π(5 khz)(0.047 μf) X C3 48. Ω π(5 khz)(0. μf) The total resistance in the resistive branch is tot Ω + 80 Ω 50 Ω The combined parallel capacitance of C and C 3 is C (tot)p C + C μf + 0. μf 0.67 μf X C(tot)p 39.7 Ω π(5 khz)(0.67 μf) The impedance of tot in parallel with C (tot)p is tot X C( tot) p (50 Ω)(39.7 Ω) Z p 39.6 Ω tot + X C( tot) p (50 Ω) + (39.7 Ω) The angle associated with tot and C (tot)p in parallel is θ tan tot 50 Ω tan 85.5 X C( tot) p 39.7 Ω Converting from parallel to series: eq Z p cos θ (39.6 Ω)cos Ω X C(eq) Z p sin θ (39.6 Ω)sin Ω The total circuit impedance is Z tot eq + ( X C + X C(eq) ) (3.08 Ω) + (45.6 Ω) 45.6 Ω 99

100 The voltage across the parallel branches is Z p 39.6 Ω C C3 s Ztot Ω 330 Ω Ω 80 Ω Ω X C 06 Ω C 8.74 Ztot 45.6 Ω 4. From Problem 3, eq 3.08 Ω and X C + X C(eq) 45.6 Ω. Since 45.6 Ω > 3.08 Ω, the circuit is predominantly capacitive. 5. Using data from Problem 3: s I tot 8.4 ma Z 45.6 Ω I C I C3 X X tot C C C3 C3 I I Ω 4.4 ma ma 48. Ω Ω 6.39 ma 6. tot Ω X C 339 Ω π( khz)(0.47 μf) Z tot tot XC (89.9 ) (339 ) + Ω + Ω 35 Ω s 5 (a) I tot 4.7 ma Ztot 35 Ω (b) θ tan X C 339 Ω tan tot 89.9 Ω 47 Ω (c) s Ztot Ω 75. (I tot leads s ) (d) Ω 5 35 s.83 Ztot Ω (e) 3.83 (f) X 339 Ω C s Ztot Ω 00

101 SECTION 0-7 Power in C Circuits 7. P a true + P r ( W) + (3.5 A) P 4.03 A 8. From Problem 0: I tot 03 ma, X C 79.6 Ω P rue I tot (03 ma) (56 Ω) 0.59 W P r I tot X C (03 ma) (79.6 Ω) A 9. Using the results from Problem : eq 8.4 kω X C(eq) 8.3 kω πfc π(00 khz)(96 pf) eq X θ tan C (eq) 8.3 kω tan 3.8 eq 8.4 kω PF cos θ cos From Problem 6: I tot 4.7 ma, tot 89.9 Ω, X C 339 Ω, Z tot 35 Ω P true I tot tot (4.7 ma) (89.9 Ω) 69 mw P r I tot X C (4.7 ma) (339 Ω) 68 ma P a tot Z tot (4.7 ma) (35 Ω) 640 ma PF cos cos(75.) 0.57 I SECTION 0-8 Basic Applications 3. X Use the formula, out C. See Figure 0-. Ztot Frequency (khz) X C (kω) Z tot (kω) out ()

102 Figure 0-3. Use the formula, out 0. Ztot See Figure 0-3. Frequency (khz) X C (Ω) Z tot (Ω) out () Figure 0-3 0

103 33. For Figure 0-73: X C 86 Ω π(5 khz)(0.039 μf) Z ( 3.9 kω ) + (86 Ω) 3.98 kω θ tan Ω X 86 C tan kω I s 5 μa Z 3980 Ω I (5 μa)(3.9 kω) 979 m C IX C (5 μa)(86 Ω) 05 m The phasor diagram is shown is Figure 0-4(a). For Figure 0-74: X C 3.8 Ω π(5 khz)(0 μf) Z ( 0 Ω ) + (3.8 Ω) 0.5 Ω θ tan Ω X 3.8 C tan Ω I s 95.3 ma Z 0.5 Ω I (95.3 ma)(0 Ω) 953 m C IX C (95.3 ma)(3.8 Ω) 303 m The phasor diagram is shown in Figure 0-4(b). (a) For Figure 0-73 (b) For Figure 0-74 Figure X C.3 kω π(3 khz)(0.047 μf) The signal loss is the voltage drop across C. X C.3 k C Ω ( ) out A ( ) + (0 kω) + (.3 kω) in B X C 50 m 5.6 m 03

104 35. For Figure 0-73: f c.05 khz πc π(3.9 kω)(0.039 μf) For Figure 0-74: f c.59 khz πc π(0 Ω)(0 μf) 36. f c.05 khz πc π(3.9 kω)(0.039 μf) Since this is a low-pass filter, BW f c.05 khz SECTION 0-9 Troubleshooting 37. After removing C, the circuit is reduced to Thevenin s equivalent: (4.7 kω)(5 kω) th.4 kω 9.7 kω 5 kω th k 5.5 Ω Assuming no leakage in the capacitor: X C 59 Ω π(0 Hz)(0 μf) out X (4.7 k ) (59 ) C Ω 3. + X C Ω + Ω θ tan 59 Ω kω With the leakage resistance taken into account: 59 out 5.5 (.4 k ) (59) X C Ω th.83 th + X C Ω + θ tan 59 Ω kω The leaky capacitor reduces the output voltage by 0.38 and increases the phase angle by (a) The leakage resistance effectively appears in parallel with. Thevenizing from the capacitor: th leak 0 kω 0 kω kω.43 kω.67 k th leak Ω.67 k in + leak Ω 43 m 04

105 X C 3386 Ω π(0 Hz)(4.7 μf) 3386 out 43 m (.43 k ) (3386 ) X C Ω th th + X C Ω + Ω 3 m (b) X C 3386 Ω π(00 khz)(470 pf) eq ( + 3 ). kω kω.05 kω ( leak )( X C ) ( kω)(3386 Ω) X C leak X ( kω) + (3386 Ω) leak + C 7 Ω eq in.96 X C leak + eq X C leak consists of a reactive and a resistive term and cannot be added directly to eq. 3.0 kω out.96 3 k Ω 39. (a) out 0 (less than normal output) X C 3386 Ω π(0 Hz)(4.7 μf) (b) out X C 3386 Ω 0,494 in 0.3 (greater than normal output) Ω + X C (c) 0 kω out in 0 k 0.5 (greater than normal output) + Ω (d) out 0 (less than normal output) 40. (a) out 0 (less than normal) (b) 3.0 kω out in 5 3 k.5 (greater than normal output) + Ω X C 3386 (c) out Ω 5 ( ) 3933 in.7 (greater than normal output) 3 Ω + + X C (d) out 0 (less than normal output) (e) out. kω in.7 (greater than normal output) Ω + X C 05

106 ADANCED POBLEMS 4. (a) I L(A) Ω 4.8 A I L(B) 40 7 Ω 3.33 A (b) PF A cos θ 0.85; θ 3.8 PF B cos θ 0.95; θ 8.9 X C(A) (50 Ω)sin Ω X C(B) (7 Ω)sin Ω P r(a) I L(A) X C(A) (4.8 A) (6.3 Ω) 606 A P r(b) I L(B) X C(B) (3.33 A) (.48 Ω) 50 A (c) A (50 Ω)cos Ω B (7 Ω)cos Ω P true(a) I L ( A) A (4.8 A) (4.5 Ω) 979 W P true(b) I ( B) B L (3.33 A) (68.4 Ω) 758 W (d) P a(a) P true A Pr A ( ) ( ) (979 W) (606 A) 5 A P a(b) P true B Pr B ( ) ( ) (758 W) (50 A) 798 A 4. out in X + C X C + X C X (.44) C X C ( ) X C πfc C πf π(0 Hz)(00 kω) 0.08 μf 43. X C 5.9 kω π( khz)(0.0 μf) θ tan X tot C tot tan θ X C tot X C tan θ (5.9 kω)tan kω 06

107 tot + tot ( + ) tot + tot ( tot ) tot tot (47 kω)( kω tot kω).4 kω 44. X C 444 Ω π(.5 khz)(0.05 μf) X C 355 Ω π(.5 khz)(0.047 μf) 4 X C X 4 C 4 + X C (90 Ω)(355 Ω) (90 Ω) + (355 Ω) 756 Ω θ 4C tan Ω 4 90 tan 33.9 X C 355 Ω The equivalent series and X C for 4 X C : eq ( 4 X C )cos θ 4C (756 Ω)cos Ω X C(eq) ( 4 X C )sin θ 4C (756 Ω)sin Ω Z tot + X ) + ( X + tot C( tot) ( 3 eq C X C(eq) (.0 kω Ω +.0 kω + 67 Ω) + (444 Ω + 4 Ω) ( 3307 Ω ) + (4666 Ω) 579 Ω θ tan X C Ω ( tot ) 4666 tan 54.7 tot 3307 Ω s 0 I tot.75 ma Ztot 579 Ω I tot leads s by 54.7 I C I I I 3 I tot.75 ma 355 I 4 X C Ω I.75 ma tot.45 ma 4 (90 ) (355 ) + X C Ω + Ω 90 I C 4 Ω I.75 ma tot ma 4 (90) (355 ) + X C + Ω C I tot X C (.75 ma)(444 Ω) 7.43 I tot (.75 ma)(.0 Ω).75 I tot (.75 ma)(680 Ω).9 3 I tot 3 (.75 ma)(.0 Ω).75 4 C I 4 4 (.45 ma)(90 Ω).3 C lags I tot by 90. ) 07

108 Figure 0-5,, and 3 are all in phase with I tot. 4C lags I tot by See Figure θ cos (0.75) 4.4 P true P a cos θ P P a true.5 kw ka PF 0.75 P r P a sin θ ( ka)sin ka Z tot 5 A 0 Ω P true I tot tot 400 W I (5 A) x tot 6 Ω 4 Ω Ω Z + X tot tot tot tot C X C Z (0 Ω) (6 Ω) 44 Ω C x 3.3 μf π( khz)( Ω) 47. For I 0 A, A B and X C 3.39 kω π( khz)(0.047 μf). kω (. kω) + (3.39 kω) s.0 kω (.0 kω) + X C s 08

109 Cancelling the s terms and solving for X C :. kω.0 kω (. kω) + (3.39 kω) (.0 kω) + (.0 kω) + X C C (.0 kω) +.0 kω (. kω). kω X C + (3.39 kω) (.0 kω) ((. kω) + (3.39 kω) ) X (.0 kω) (. kω) X C ( ) (.0 kω) (. kω) + (3.39 kω) (. kω) C 0.03 μf π( khz)(.54 kω).54 kω 48. See Figure 0-6. in 0 peak f 00 khz 0 μ s X C 5.9 Ω π(00 khz)(0. μf) X C can be neglected because it is very small compared to kω. A kω 8 kω (8. kω +.0 kω) 4.77 kω 4.77 kω out A k 3.3 peak Ω The waveform on the scope is correct, so the circuit is OK. Figure

110 Multisim Troubleshooting Problems 49. C is leaky. 50. C is shorted. 5. No fault 5. C is open. 53. is open. 54. C is shorted. 55. Phase shift with C shorted

111 CHAPTE INDUCTOS BASIC POBLEMS SECTION - The Basic Inductor. (a) H 000 mh/h 000 mh (b) 50 μh 0.00 mh/μh 0.5 mh (c) 0 μh 0.00 mh/μh 0.0 mh (d) H 000 mh/h 0.5 mh. (a) 300 mh 0 3 μh/mh 300,000 μh (b) 0.08 H 0 6 μh/h 80,000 μh (c) 5 mh 0 3 μh/mh 5000 μh (d) mh 0 3 μh/mh 0.45 μh 3. L N N μa l Ll μa (.6 0 (30 mh)(0.05 m) 6 5 )(0 0 m ) 3450 turns 4. I dc W 0 Ω 0. A 5. W (0.H)(A) LI 50 mj 6. v induced L (rate of change of I) (00 mh)(00 ma/s) 0 m SECTION -3 Series and Parallel Inductors 7. L T 5 μh + 0 μh + 0 μh + 40 μh + 80 μh 55 μh 8. L x 50 mh 0 mh mh 8 mh

112 9. L T μh 50 μh 5 μh 5 μh 7.4 μh L ( mh) 0. 8 mh L + mh ( 8 mh)l + (8 mh)( mh) ( mh)l (4 mh)l 96 mh L 96 mh 4 mh 4 mh (0 H)(5 H). (a) L T H H 0 H + 5 H 00 mh (b) L T 50 mh (c) L T μh 00 μh 400 μh 57 μh. (a) (00 mh)(50 mh) (60 mh)(40 mh) L T mh + 4 mh 57.3 mh 50 mh 00 mh (b) ( mh)(6 mh) L T 4 mh 8 mh (c) ( mh)(4 mh) L T 4 mh mh 6 mh SECTION -4 Inductors in DC Circuits 3. (a) τ (c) τ L L 00 μh 00 Ω 3 H.5 MΩ μs (b) τ μs L 0 mh 4.7 kω.3 μs 4. (a) L 50 μh 5τ L 5 mh μs (b) 5τ 56 Ω Ω (c) L 00 mh 5τ μs kω.7 μs 5. τ (a) (b) L 0 mh 0 μs.0 kω v L i e t/τ 5e 0μs/0μs 5e 5.5 v L i e t/τ 5e 0μs/0μs 5e.03

113 (c) (d) (e) v L i e t/τ 5e 30μs/0μs 5e v L i e t/τ 5e 40μs/0μs 5e v L i e t/τ 5e 50μs/0μs 5e L 75 mh 6. τ 9.5 µs 8. kω I F s 0. ma 8. kω -0µs/9.5µs (a) i IF ( e ) 0.8 ma -0µs/9.5µs (b) i IF ( e ).08 ma -30µs/9.5µs (c) i IF ( e ).7 ma SECTION -5 Inductors in AC Circuits 7. The total inductance for each circuit was found in Problem. (a) X L πfl tot π(5 khz)(4.33 H) 36 kω (b) X L πfl tot π(5 khz)(50 mh).57 kω (c) X L πfl tot π(5 khz)(57 μh).79 Ω 8. The total inductance for each circuit was found in Problem. (a) X L πfl tot π(400 Hz)(57.3 mh) 44 Ω (b) X L πfl tot π(400 Hz)(4 mh) 0. Ω (c) X L πfl tot π(400 Hz)(5.33 mh) 3.4 Ω 9. L tot L + L L L + L 3 50 mh 3 + (0 mh)(40 mh) 60 mh X L(tot) πfl tot π(.5 khz)(63.3 mh) 994 Ω X L πfl tot π(.5 khz)(0 mh) 34 Ω X L3 πfl tot π(.5 khz)(40 mh) 68 Ω rms 0 I tot 0. ma 994 Ω I L X I L3 X X L ( tot ) L L X L3 + X X L + X L3 L3 I I tot tot 63.3 mh 68 Ω 0.mA ma Ω + Ω 34 Ω 0.mA ma Ω + Ω 3

114 0. (a) L tot mh 0 X L 0 Ω I 500 ma X L πfl tot X L 0 Ω f 55.6 Hz L π(57.3 mh) (b) (c) π tot L tot 4 mh, X L 0 Ω X L 0 Ω f 796 Hz L π(4 mh) π tot L tot 5.33 mh, X L 0 Ω X L 0 Ω f 597 Hz L π(5.33 mh) π tot. X L(tot) 994 Ω from Problem 9. P r I rms X L ( tot ) (0. ma) (994 Ω) 0 ma ADANCED POBLEMS. TH kω 4.7 kω kω 6.8 kω 4.57 kω L 3.3 mh τ 0.7 μs 4.57 kω TH 3. (a) v F ( i F )e t/l (for 60 μs, use 0 μs. The initial voltage is 0.) 0 + ( 0 0 )e (8. kω)(0 μs)/75 mh 3.35 (b) (c) v F ( i F )e t/l (for 70 μs, use 0 μs for calculation) 0 + ( 0 0 )e (8. kω)(0 μs)/75 mh. v F ( i F )e t/l (for 80 μs, use 30 μs for calculation) 0 + ( 0 0 )e (8. kω)(30 μs)/75 mh By KL, is equal and opposite to L (because S 0 ). Therefore, TH S S kω 0. kω TH.6 I F 569 µa 4.57 kω TH - t / τ µs/0.7µs (a) i IF ( e ) 569 µa( e ) 4.7 kω 6.8 kω 47 μa (b) i I 569 μa F 4

115 6. T ( + 3) ( + 4) 8 kω.5 kω 4.7 kω L 3.3 mh τ µs 4.7 kω - t / τ µs/0.699µs i I e 569 µa e 36 μa i ( ) 7. X L π(3 khz)(5 mh) 94. Ω X L3 π(3 khz)(3 mh) 56.5 Ω L3 I L3 X L3 (50 ma)(56.5 Ω).83 L L 7.7 I L 76. ma 94. Ω X L I L I L I L3 76. ma 50 ma 6. ma 8. Position : L T 5 mh + mh 6 mh Position : L T 5 mh + 00 μh mh 6. mh Position 3: L T 5 mh μh + 00 μh + mh 7. mh Position 4: L T 5 mh + 0 mh μh + 00 μh + mh 7. mh Multisim Troubleshooting Problems 9. L 3 is open. 30. L is shorted. 3. No fault 3. L is open. 33. L 3 is shorted. 5

116 CHAPTE L CICUITS BASIC POBLEMS SECTION - Sinusoidal esponse of L Circuits. All the frequencies are 5 khz.. I,, and L are all sinusoidal. SECTION - Impedance and Phase Angle of Series L Circuits 3. (a) Z (b) Z L (.0 k ) 500 X Ω Ω. Ω L (.5 k ) (.0 k ) X Ω Ω.8 kω 4. (a) tot 47 Ω + 0 Ω 57 Ω L tot 50 mh + 00 mh 50 mh X L(tot) πfl tot π(00 Hz)(50 mh) 94. Ω (b) tot + X L ( tot) Z ( 57 Ω) + (94. Ω) 0 Ω θ tan X L Ω ( tot ) 94. tan 58.8 tot 57 Ω (5.0 mh)(8.0 mh) L tot 3.08 mh 5.0 mh 8.0 mh X L(tot) πfl tot π(0 khz)(3.08 mh) 387 Ω Z tot + X L ( tot) ( 470 Ω) + (387 Ω) 609 Ω θ tan X L Ω ( tot ) 387 tan 39.5 tot 470 Ω 5. (a) X L πfl π(00 Hz)(0.0 H).6 Ω (b) Z + X L ( Ω) + (.6 Ω) 7.4 Ω X L πfl π(500 Hz)(0.0 H) 6.8 Ω Z + X L ( Ω) + (6.8 Ω) 64.0 Ω 6

117 (c) (d) X L πfl π( khz)(0.0 H) 6 Ω Z + X L ( Ω) + (6 Ω) 7 Ω X L πfl π( khz)(0.0 H) 5 Ω Z + X L ( Ω) + (5 Ω) 5 Ω 6. (a) Z cos θ (0 Ω)cos Ω X L Z sin θ (0 Ω)sin Ω (b) Z cos θ (500 Ω)cos Ω X L Z sin θ (500 Ω)sin Ω (c) Z cos θ (.5 kω)cos Ω X L Z sin θ (.5 kω)sin kω (d) Z cos θ (998 Ω)cos Ω X L Z sin θ (998 Ω)sin Ω SECTION -3 Analysis of Series L Circuits Ω + 0 Ω 57 Ω tot Ltot L+ L 50 mh + 00 mh 50 mh XLtot ( ) πfltot π( khz)(50 mh) 94 Ω tot 57 Ω tot ( ) 5 s 944 Ω 0.30 tot X + L( tot) 8. tot 470 Ω L tot 3.08 mh + + L L 5.0 mh 8.0 mh XLtot ( ) πfltot π(0 khz)(3.08 mh) 387 Ω tot ( ) Ltot ( ) 470 Ω tot 8 s 6.8 tot + X L (470 Ω) + (387 Ω) X ( ) 387 Ω Ltot 8 s 5.08 tot + X L( tot) (470 Ω) + (387 Ω) 9. (a) Z. kω from Problem 3. 0 I s Z. kω (b) Z.8 kω from Problem 3. I s 5.77 ma Z.8 kω 7

118 0. Using the results of Problem 4: (a) I s ma Z 0 Ω (b) I s 8 3. ma Z 609 Ω. X L π(60 Hz)(0. H) 37.7 Ω θ tan Ω X 37.7 L tan Ω. θ tan X L X L π(60 Hz)(0. H) 37.7 Ω 37.7 Ω θ tan Ω Double L: X L π(60 Hz)(0. H) 75.4 Ω 75.4 Ω θ tan Ω θ increases by 9.4 from 38.7 to The circuit phase angle was determined to be 38.7 in Problem. This is the phase angle by which the source voltage leads the current; it is the same as the angle between the resistor voltage and the source voltage. The inductor voltage leads the resistor voltage by 90. See Figure -. Figure - 4. (a) X L π(60 Hz)(00 mh) 37.7 Ω Z + X L Z L ( 50 Ω) + (37.7 Ω) 55 Ω s X Z L s 50 Ω Ω 37.7 Ω Ω 8

119 (b) (c) (d) X L π(00 Hz)(00 mh) 6 Ω Z + X L Z L ( 50 Ω) + (6 Ω) 96 Ω s X Z L s 50 Ω Ω 6 Ω Ω X L π(500 Hz)(00 mh) 34 Ω Z + X L Z L ( 50 Ω) + (34 Ω) 348 Ω s X Z L s 50 Ω Ω 34 Ω Ω X L π( khz)(00 mh) 68 Ω Z + X L Z L ( 50 Ω) + (68 Ω) 646 Ω s X Z L s Ω Ω Ω Ω 5. (a) X L π( Hz)(0 H) 6.8 Ω φ tan X L 6.83 Ω tan kω (b) X L π(00 Hz)(0 H) 6.8 kω φ tan Ω 6.8 k tan 39 kω (c) X L π( khz)(0 H) 6.8 kω φ tan Ω 6.8 k tan 39 kω (d) X L π(0 khz)(0 H) 68 kω φ tan Ω 68 k tan 39 kω 6. (a) φ 90 tan X Ω L tan kω (b) φ 90 tan X Ω L 6.8 k 90 tan kω 9

120 (c) φ 90 tan X Ω L 6.8 k 90 tan kω (d) φ 90 tan X Ω L 68 k 90 tan kω SECTION -4 Impedance and Phase Angle of Parallel L Circuits 7. X L π( khz)(800 μh) 0 Ω Y tot Z Ω Y tot 0.3 S + 0 Ω 7.69 Ω 0.3 S 8. (a) X L π(.5 khz)(800 μh) 7.54 Ω Y Z Y + X L Ω 6.37 Ω 0.57 S Ω 0.57 S (b) X L π(3 khz)(800 μh) 5. Ω Y Z Y + X L Ω 9.43 Ω 0.06 S + 5. Ω 0.06 S (c) (d) X L π(5 khz)(800 μh) 5. Ω Y Z Y + X L Ω 0.9 Ω 0.09 S X L π(0 khz)(800 μh) 50.3 Ω Y Z Y + X L Ω.6 Ω S Ω Ω 0.09 S S 0

121 9. X L πfl X f L Ω πl π(800 μh).39 khz SECTION -5 Analysis of Parallel L Circuits 0 0. I 4.55 ma. kω 0 I L.86 ma 3.5 kω I (4.55 ma) + (.86 ma) 5.37 ma tot. (a) X L π( khz)(5 mh) 34 Ω X L (560 Ω)(34 Ω) Z X (560 Ω) + (34 Ω) (b) I (c) I L (d) I tot + L s ma 560 Ω X s L Ω 59 ma s ma Z 74 Ω (e) θ tan 560 Ω tan 60.7 X L 34 Ω 74 Ω ( ) XL (.5 k Ω)(5.0 k Ω). Z tot 4.59 kω (.54 k ) X Ω L.5 k θ tan Ω 5.0 kω 66.5 eq Z tot cos θ (4.59 kω)cos kω X L(eq) Z tot sin θ (4.59 kω)sin kω SECTION -6 Analysis of Series-Parallel L Circuits 3. X L π(00 khz)(.0 mh) 68 Ω X L (500 Ω)(68 Ω) Z p X (500 Ω) + (68 Ω) + L θ tan X L 500 Ω tan Ω 579 Ω

122 eq Z p cos θ (579 Ω)cos Ω X L(eq) Z p sin θ (579 Ω)sin Ω Z tot ( ) + X (444 Ω) + (534 Ω 694 Ω + eq L(eq) ) 5 I tot 36 ma 694 Ω I tot (36 ma)(0 Ω) 7.9 L I tot Z p (36 ma)(579 Ω) The circuit is predominantly inductive because X L(eq) > tot. 5. Using the results of Problem 3: I tot 36 ma L 0.8 I L 33. ma X 68 Ω I L ma 500 Ω SECTION -7 Power in L Circuits 6. P a true + P r ( 00 mw) + (340 ma) P 354 ma 7. X L π(60 Hz)(0. H) 37.7 Ω Z I tot P true P r + X L ( 47 Ω) + (37.7 Ω) 60.3 Ω s 6.6 ma Z 60.3 Ω I tot I tot X L (6.6 ma) (47 Ω) 3.0 mw (6.6 ma) (37.7 Ω) 0.4 ma 8. θ tan. kω tan X L 3.5 kω PF cos θ cos Using the results of Problems 3 and 5: PF cos θ cos P true s I cos θ (5 )(36 ma)(0.386) 347 mw P r I L X L (33. ma) (68 Ω) 69 ma P a s I tot (5 )(36 ma) 900 ma

123 SECTION -8 Basic Applications 30. Use the formula, out in. Z tot See Figure -. Frequency (khz) X L (kω) Z tot (kω) out () Figure - 3. X Use the formula, out L in. Z tot See Figure -3. Frequency (khz) X L (kω) Z tot (kω) out (m) Figure For Figure -55: See Figure -4(a). X L π(8 khz)(0 H) 503 kω Z X ( 39 kω) + (503 kω) 505 kω + L 3

124 θ tan Ω X 503 k L tan kω 39 kω in 505 k 77. m Z Ω L X Z L in 503 kω 505 k 996 m Ω For Figure -56: See Figure -4(b). X L π(8 khz)(0 H) 503 kω Z + X L ( 39 kω) + (503 kω) 505 kω θ tan Ω X 503 k L tan kω Z L in X Z L in 39 kω 50 m 505 k 3.86 m Ω 503 kω 50 m 505 k 49.8 m Ω Figure -4 SECTION -9 Troubleshooting 33. L 8 3 L (a) out 0 (b) out 0 (c) out 0 (d) out 0 4

125 ADANCED POBLEMS 35. See Figure -5(a). th Ω + 56 Ω Ω 48.8 Ω Ω th Ω See Figure -5(b): L tot 5 mh + 50 mh 50 mh X L(tot) πfl tot π(400 Hz)(5 mh) 6.8 Ω X L( tot) 6.8 L Ω th th + X L( tot) (48.8 Ω) + (6.8 Ω) Figure Since X L(tot) > th, the circuit is predominantly inductive. 37. From Problem 35, X L(tot) 6.8 Ω Z tot X Ltot ( ) 56 Ω + Ω (33 Ω ) + (6.8 Ω) 7.8 Ω s 5 I tot 343 ma Z 7.8 Ω tot 38. (a) X L πfl π(80 khz)(8.0 mh) 4.0 kω Z tot ( ) XL. kω (.0 kω kω (5.6 kω) + (4.0 kω) ) 875 Ω s 8 (b) I tot 0.6 ma Z 875 Ω tot (c) tot ( + 3) + 4. kω 4.3 kω kω 6.54 kω θ tan X L 4.0 kω tan 3.6 tot 6.54 kω 5

126 (d) See Figure -6(a). th kω +.0 kω 3.3 kω 6.37 kω kω th s kω X 4.0 k L L Ω th 3.8 th + X L (6.37 k Ω ) + (4.0 k Ω) 7.37 (e) See Figure -6(b) and (c): th 3.0 kω 3.3 kω 767 Ω kω th s k Ω The voltage across the 4 -L combination is the same as the voltage across 3. X L th ( th + 4 ) + X L (5.6 k ) (4.0 k ) 6.89 k Ω + Ω Ω (6.37 k ) (4.0 k ) 7.53 kω Ω + Ω Figure -6 6

127 39. (a) Z X L X L + X L (68 Ω)(00 Ω) (68 Ω) + (00 Ω) 56. Ω Ω 68 θ tan tan X 34. L X L 00 Ω Converting the parallel combination of and X L to an equivalent series form: eq Z X cosθ (56. Ω)cos Ω X L(eq) Z B I Z L X L sinθ (56. Ω)sin Ω X L X L + eq) + X L(eq) (47 Ω Ω) + (3.6 ) ( Ω 98.7 Ω Z s B ma 98.7 Ω (b) 68 I L Ω I 405 ma 8 ma (68 ) (00 ) + X L Ω + Ω (c) X L L3 X L + X L3 75 Ω + 45 Ω 0 Ω s 40 I L X L L3 0 Ω 333 ma (d) 00 I X L Ω I 405 ma 335 ma (68 ) (00 ) + X L Ω + Ω kω kω 0.7 kω 3 ( ) 4.7 kω 0.7 kω 3.7 kω + 3 ( ) 5.6 kω kω 8.87 kω tot ( + 3 ( )) 3.3 kω 8.87 kω.4 kω X L π(0 kω)(50 mh) 3.4 kω θ tan Ω X 3.4 k L tan kω out lags in.4 k (.4 k ) (3.4 k ) tot Ω in 609 m + tot X L Ω + Ω 3 ( 4 + ) k 3 Ω 609 m 3 ( 4 5) 3.7 k 5.6 k 5 m + + Ω + Ω kω out m 43 m kω+ 6.8 kω out 43 m Attenuation 0.43 in 4. L tot (( L 4 + L5 ) L3 + L ) L (.0 mh.0 mh).0 mh.0 mh.0 mh (.0 mh.0 mh).0 mh.0 mh.0 mh.0 mh 7

128 X L(tot) πfl 6.8 Ω X L( tot ) L(tot) + X L( tot ) 6.8 (00 ) (6.8 ) Ω 0.53 Ω + Ω 6.8 Ω L Ω 6.8 Ω out Ω out 0.33 Attenuation 0.33 in in 4. Ω A.5 k.5 kω A See Figure -7. When the switch is thrown from position to position, the inductance will attempt to keep A through, thus a.5 k spike is created across for a short time. This design neglects the arcing of the switch, assuming instantaneous closure from position to position. The value of L is arbitrary since no time constant requirements are imposed. Figure See Figure -8. The correct output voltage is calculated as follows: X L πfl π(0 khz)(50 mh) 34 Ω 3.9 kω kω 0.7 kω 4.7 kω 0.7 kω 3.7 kω 5.6 kω kω 8.87 kω 3.3 kω 8.87 kω.4 kω.4 k A (.4 k ) (3.4 k ) Ω Ω + Ω 8

129 3.7 kω B k 5.6 k 0.5 Ω + Ω 6.8 kω out k 3.9 k 0.43 Ω + Ω The measured output is approximately 0.3 peak, which is incorrect. After trial and error, we find that if the 4.7 kω is open we get: tot 3.3 kω (5.6 kω kω kω).74 kω.74 k A (.74 k ) (3.4 k ) Ω Ω + Ω 6.8 kω out k 3.9 k 6.8 k 0.74 Ω + Ω + Ω This is relatively close to the measured value. Component tolerances could give us the scope reading. Figure -8 Multisim Troubleshooting Problems 44. is shorted. 45. L is open. 46. L is shorted. 47. is open. 48. No fault 49. L is shorted. 9

130 CHAPTE 3 LC CICUITS AND ESONANCE BASIC POBLEMS SECTION 3- Impedance and Phase Angle of Series LC Circuits. X C πfc 677 Ω π(5 khz)(0.047 μf) X L πfl π(5 khz)(5 mh) 57 Ω Z + ( X L ) C X ( 0 Ω ) + (677 Ω 57 Ω) (0 Ω) + (50 Ω) 50 Ω θ tan X C X L 50 Ω tan 88.9 ( s lagging I) 0 Ω X tot X C X L 50 Ω Capacitive. Z + X L X C ( ) (4.7 kω) + (8.0 kω 3.5 kω) 6.5 kω 3. Doubling f doubles X L and halves X C, thus increasing the net reactance and, therefore, the impedance increases. SECTION 3- Analysis of Series LC Circuits 4. Z tot I tot Z + X L X C ( ) (4.7 kω) + (4.5 kω) 6.5 kω s tot 4 64 μa 6.5 kω I tot (64 μa)(4.7 kω).89 L I tot X L (64 μa)(8.0 kω) 4.9 C I tot X C (64 μa)(3.5 kω).5 5. θ tan Ω X 4.5 k tot tan kω The voltage values were determined in Problem 4. lags s by 43.8 because it is in phase with I. L and C are each 90 away from and 80 out of phase with each other. See Figure

131 Figure 3-6. tot 0 Ω 390 Ω 4 Ω L tot L + L 0.5 mh +.0 mh.5 mh C tot C + C 0.0 μf pf 0.08 μf X L(tot) 36 Ω X C(tot) 540 Ω Z tot tot + X L( tot) X C( tot) ( ) (4 Ω) + (304 Ω) 335 Ω s (a) I tot 35.8 ma Ztot 335 Ω (b) P true I tot tot (35.8 ma) (4 Ω) 8 mw (c) P r I tot X tot (35.8 ma) (304 Ω) 390 ma (d) P a P ) + ( P 430 ma ( true r ) SECTION 3-3 Series esonance 7. Because X C < X L, f r is less than the frequency indicated. 8. X C X L at resonance. s 9. f r π LC π 734 khz (mh)(47 pf) X L πf r L π(734 khz)( mh) 4.6 kω X C X L 4.6 kω Z tot 0 Ω s I 54.5 ma Z 0 Ω tot 0. C L 00 at resonance s 0 Z 00 Ω I 50 ma X L X C max I L max ma kω 3

132 . f r 454 khz π LC π (8 μh)(.5 nf) X L π fl π(454 khz)(8 μh) 34 Ω X 34 Q L Ω 6 39 Ω f 454 khz BW r 75.7 khz Q 6 BW 75.7 khz f c f r 454 khz 46 khz BW 75.7 khz f c f r khz + 49 khz 3.0. I max s 77 ma 39 Ω I half-power 0.707I max 0.707(77 ma) 54 ma SECTION 3-4 Series esonant Filters 3. (a) f r π LC (b) f r π LC 4.5 khz π ( mh)(0.0μf) 4.0 khz π ( mh)(0.0 μf) These are bandpass filters. 4. (a) tot 0 Ω + 75 Ω 85 Ω f r π LC π ( mh)(0.0μf) (b) X L π(4.5 khz)( mh).09 kω X L.09 kω Q 3 85 Ω BW tot f 4.5 khz r. khz Q 3 tot 0 Ω + Ω 3 Ω f r π LC π ( mh)(0.0 μf) X L π(4.0 khz)( mh) 30 Ω X L 30 Ω Q Ω BW tot f 4.0 khz r.54 khz Q khz 4.0 khz 3

133 5. (a) f r π LC π (00 μh)(0.00 μf) X L π(339 khz)(00 μh) 3 Ω X Ω Q L Ω BW f 339 khz r 39 khz Q khz (b) f r π LC π (5 mh)(0.047 μf) X L π(0.4 khz)(5 mh) 37 Ω X Ω Q L Ω BW f 0.4 khz r.6 khz Q khz SECTION 3-5 Parallel LC Circuits 6. X L πfl π( khz)(5 mh).3 kω X C 603 Ω πfc π( khz)(0.0 μf) Y Z tot (0.0 S) Y + X C X L 4 00 Ω Ω.3 k + ( S S) 0.03 ms 0.03 ms 99.7 Ω 7. X L π( khz)(5 mh).3 kω X C 603 Ω π( khz)(0.0 μf) Since X C < X L, the parallel circuit is predominantly capacitive. The smaller reactance in a parallel circuit dominates the circuit response because it has the largest current. Ω 8. I tot Z s tot ma I 99.7 Ω s 5 I L 4.4 ma I C X L.3 kω L C s 5 s ma 00 Ω X s C Ω 8.9 ma 33

134 9. X π fl π ( )( ) B L L X C B C 0 khz 0 mh 68 Ω.59 ms X 68 Ω L.06 k π fc π 0 khz 5 nf Ω ( )( ) 94 µs X.06 kω C W 80 Ω X L 68 Ω Q Ω ( ) ( ) (eq) Q + 80 Ω kω G p p(eq) W 00 µs 5.0 kω p(eq) The equivalent circuit is shown in Figure 3-. Figure 3- tot p tot Y G (eq) + B 00 µs + (.59 ms 94 µs) 678 µs Z tot.47 kω 678 µs SECTION 3-6 Parallel esonance 0. Z r is infinitely large.. f r π W L LC C (0 Ω) (47 pf) 50 mh 04 khz π (50 mh)(47 pf) X L πf r L π(03.8 khz)(50 mh) 3.6 kω X L 3.6 kω Q Ω W Z r W (Q + ) 0 Ω(630 + ) 53. MΩ. From Problem : Z r 53. MΩ and f r 04 khz s 6.3 I tot 9 na Z 53. MΩ r I C I L (0 Ω) kω) 93 μa 34

135 SECTION 3-7 Parallel esonant Filters 3. Q BW X L kω 80 5 Ω f 5 khz r 6.5 Hz Q BW f c f c 800 Hz 400 Hz 400 Hz 5. P (0.5)P r (0.5)(.75 W).38 W 6. Q f 8 khz r 0 BW 800 Hz X L(res) Q W 0(0 Ω) 00 Ω X L 00 Ω L.99 mh πf r π(8 khz) X C X L at resonance C π f 0.99 μf π(8 khz)(00 Ω) r X C 7. Since BW Q f r, the bandwidth is halved when Q is doubled. So, when Q is increased from 50 to 00, BW decreases from 400 Hz to 00 Hz. 8. f r π LC L 35. mh 4π f C 4(3.4) (60) (00 μf) r 9. See Figure 3-3. Figure

136 ADANCED POBLEMS 30. (a) 0 log out 0 log 0 db in (b) 0 log out 3 0 log 4.4 db in 5 (c) 0 log out log 3 db in 0 (d) 0 log out 5 0 log 4 db in 5 3. X L(tot) 3.33 kω + + X L X L 5 kω 0 kω X L( tot ) (0 kω)(3.33 kω) Z p X L(tot) X (0 kω) + (3.33 kω) + L( tot ) 3.6 kω θ p tan 0 kω tan 7.6 X L ( tot) 3.33 kω Converting the parallel combination of, X L, and X L to an equivalent series circuit: eq Z p cos θ p (3.6 kω)cos Ω X L(eq) Z p sin θ (3.6 kω)sin kω Z (tot) I tot I I C + eq) + ( X L (eq) + X C ) + (497 Ω) + (000 ) ( Ω 4740 Ω Z s tot Ω. ma I tot (. ma)(3.3 kω) 6.96 C I tot X C (. ma)(.0 kω). L-L- I tot Z p (. ma)(3.6 kω) 6.67 L 6.67 I L.33 ma X 5 kω I L I X L L L kω 667 μa kω 667 μa 36

137 3. For ab 0, a b in both magnitude and phase angle. X L 6 Ω; X L 5 Ω 6 a L (80 ) (6 ) Ω 9.38 Ω + Ω It is not possible for ab to be 0 because the LC branch has no resistance; thus, the voltage from a to b can only have a phase angle of 0, 90, or 90 (the branch will be either resonant, purely inductive, or purely capacitive depending on the value of X C ). Therefore, it is not possible for a to equal b in both magnitude and phase angle, which are necessary conditions. 33. X C 4 Ω π(3 khz)(0. μf) X L π(3 khz)( mh) 6 Ω X L π(3 khz)(8 mh) 5 Ω s I -L X (80 Ω) + (6 Ω I C-L X C + L ) s 33 ma X 4 Ω 5 Ω L 4.5 ma θ L tan X Ω L 6 tan Ω The resistive component of current in the left branch is: I I L cos θ L (4.5 ma)cos ma The reactive component of current in the left branch is: I X I L sin θ L (4.5 ma)sin ma In the right branch X C >X L, so I C L is totally reactive and is 80 out of phase with I X in the left branch. I tot I + IC L I X ( ) (5.8 ma) + (33 ma 3.5 ma) 04 ma 34. For parallel resonance: f r W C L π L C X L π(.6 khz)(5 mh) 408 Ω Q p p Z r W ( + ) 4 Ω(0 + ) Q 4.6 kω.6 khz π(5 mh)(0.5 μf) X L W 408 Ω 0 4 Ω X L π(.6 khz)(0 mh) 63 Ω Since Z r is much greater than, W, or X L and is resistive, the output voltage is: out s 0 For series resonance: f r 4. khz π L C π (0 mh)(0.5 F) and X C X L π(4. khz)(0 mh) 58 Ω X L π(4. khz)(5 mh) 644 Ω μ 37

138 Since X C < X L, the parallel portion of the circuit is capacitive. XC Z r ( W + XL) W+ XL 7 Ω W + ( XL XC) Assuming that Z r is almost totally reactive: Z r 7 Ω out 0 s.96 + Z r (860 Ω ) + (7 Ω) 35. f r BW Q (500 Hz)(40) 0 khz.5 X C 5 Ω 0 ma C π f μf π(0 khz)(5 Ω) Q W f r W r X C X L 40 X Q L π X 40 W C L LC L 0.05X L 0.05(πf r L) ( 0.05(πf L) ) C C W r L L 0.05 fr LC f r 4π LC 4π LC 4π LC In the above derivation, the term (0.05(π)) 0.05 f 4π r LC 0.05 f LC r LC(4π ) f r L f r 989 μh C(4π ) 36. efer to Figure 3-4. f r Choose C 0.00 μf π LC f r 4π LC (a) f r 500 khz: L 4π f r C 4π (500 khz) (0.00 μf) 0 μh (b) f r 000 khz: L 4π f r C 4π (000 khz) (0.00 μf) 5.3 μh (c) f r 500 khz: L 3 4π C 4π (500 khz) (0.00 μf).3 μh f r 38

139 (d) f r 000 khz: L 4 4π f r C 4π (000 khz) (0.00 μf) 6.3 μh Figure See Figure 3-5. The winding resistance is neglected because it contributes negligibly to the outcome of the calculations. f r π LC f r 4π LC C 4π f r L For f r 8 MHz, 9 MHz, 0 MHz, and MHz: C 40 pf 4π (8 MHz) (0 μh) C 3 pf 4π (9 MHz) (0 μh) C 3 5 pf 4π (0 MHz) (0 μh) C 4 pf 4π (MHz) (0 μh) Figure

140 Multisim Troubleshooting Problems 38. L is open. 39. No fault 40. C is open. 4. L is shorted. 4. C is shorted. 43. L is shorted. 40

141 CHAPTE 4 TANSFOMES BASIC POBLEMS SECTION 4- Mutual Inductance. L M k L L 0.75 ( μh)(4 H).5 μh μ. L M k L L k L μh M 0.5 L L (8 μh)( H) μ SECTION 4- The Basic Transformer 3. n N N sec pri (a) N n N (b) N n N sec pri sec pri (a) In phase (b) Out of phase (c) Out of phase SECTION 4-3 Step-Up and Step-Down Transformers 6. n N N sec pri sec.5 pri.5(0 ) N sec N pri (50 turns) 500 turns 4

142 8. n N sec sec 0 N pri pri pri n sec See Figure 4-(a). N sec (a) sec n pri pri N pri See Figure 4-(b). N sec (b) sec n pri pri N pri 0(0 ) 00 rms (50 ) 00 rms Figure 4-0. sec pri n N N N N sec pri sec pri sec (0.)(00 ) 40. sec sec n 0. pri N N pri pri sec (6 ) n (a) L Nsec n pri 0 pri 6 N pri 0 (b) L 0 (The transformer does not couple constant dc voltage) (c) N sec L npri pri 4(0 ) 40 N pri 4. No, the voltages would not change. 4

143 5. (a) L (0.) sec (0.)(00 ) 0 (b) pri 0 L 0( ) Meter would indicate 0. SECTION 4-4 Loading the Secondary 7. I I sec pri N pri N n I sec n sec I pri 3 00 ma ma 8. sec pri I I pri sec (a) sec n pri Nsec 30 pri 5 N pri sec 5 (b) I sec 50 ma 300 Ω L (c) I pri sec 5 I 50 ma sec 5 ma pri 30 (d) P L I sec L (50 ma) (300 Ω) 0.75 W SECTION 4-5 eflected Load 9. pri L 680 Ω 680 Ω n n Ω pri n L 50 (8 Ω) 0 kω. I pri sec pri ma 0 kω 43

144 . pri 300 Ω; L.0 kω n L pri L.0 kω n Ω pri SECTION 4-6 Impedance Matching 3. pri n n n pri L L pri L 6 Ω 4 Ω 4 n n 0.5 from Problem 3. pri speaker n 5 I pri.56 A 6 Ω Ω 6 Ω I sec I pri n (.56 A) 3. A P speaker I sec speaker (3. A) (4 Ω) 38.9 W 5. n 0 n L pri n L n pri 0 (50 Ω) 00(50 Ω) 5 kω 6. pri L L (0.0) L n 0 For L l kω, pri (0.0)( kω) 0 Ω 0 Ω (pri) Ω P (.67 ) 0 Ω 0.78 W L pri 44

145 For L kω, pri (0.0)( kω) 0 Ω 0 Ω (pri) Ω P (.86 ) 0 Ω W For L 3 kω, pri (0.0)(3 kω) 30 Ω 30 Ω (pri) Ω (3.75 ) P 30 Ω W For L 4 kω, pri 40 Ω 40 Ω (pri) Ω (4.44 ) P 40 Ω W For L 5 kω, pri 50 Ω 50 Ω (pri) Ω P (5 ) 50 Ω W For L 6 kω, pri 60 Ω 60 Ω (pri) Ω (5.45 ) P 60 Ω W For L 7 kω, pri 70 Ω 70 Ω (pri) Ω (5.83 ) P 70 Ω W For L 8 kω, pri 80 Ω 80 Ω (pri) Ω (6.5 ) P 80 Ω W 45

146 For L 9 kω, pri 90 Ω 90 Ω (pri) Ω P (6.43 ) 90 Ω W For L 0 kω, pri 00 Ω 00 Ω (pri) Ω (6.67 ) P 00 Ω W See Figure 4-. Figure 4- SECTION 4-7 Transformer atings and Characteristics 7. P L P pri P lost 00 W 5.5 W 94.5 W 8. P 94.5 W % efficiency out 00% 00% 94.5 % Pin 00 W 9. Coefficient of coupling

147 Pa ka 30. (a) I L(max).67 A sec 600 sec 600 (b) L(min) 359 Ω I L(max).67 A sec (c) X C 359 Ω IL C max πfx π(60 Hz)(359 Ω) C 7.4 μf 3. ka rating (.5 k)(0 A) 5 ka SECTION 4-8 Tapped and Multiple-Winding Transformers For secondary : sec 4 n pri For secondary : sec 6 n 0.5 pri For secondary 3: sec 3 n 0.5 pri 34. (a) See Figure 4-3. (b) turns: sec turns: sec turns: sec turns: sec

148 Figure For both primaries: 00 Top secondary: n Next secondary: n Third secondary: n Bottom secondary: n 000 SECTION 4-9 Troubleshooting 36. Open primary winding. eplace transformer. 37. If the primary shorts, excessive current is drawn which potentially can burn out the source and/or the transformer unless the primary is fused. 38. Some, but not all, of the secondary windings are shorted or the primary voltage is lower than expected. ADANCED POBLEMS 39. (a) N sec 400 turns turns 700 turns L n pri Nsec pri 35 N pri 00 L 35 I L Ω.9 A L 300 L L 5 I L.5 A 0 Ω L 48

149 (b) Z pri Z pri + + N (.94)( Ω) (6)(0 pri N pri L L N700 N S S S 35.3 Ω 60 Ω S 8.9 Ω Ω) 40. (a) pri 400 Nsec sec 0 n 0.05 N pri pri 400 (b) I sec 5 ka sec 0 (c) I pri ni sec (0.05)(4.7 A).09 A 4. (a) The lower 00 Ω resistor is shorted out by the meter ground; so, the full secondary voltage measured by the meter is: meter sec Nsec n pri N pri 6 (b) The common point between the 00 Ω resistors is ground. Both resistors are still in the secondary with one-half of the secondary voltage across each. meter sec Position : L 560 Ω + 0 Ω +.0 kω 780 Ω n L 780 Ω 0 Ω 3.3 pri N sec N sec + N sec + N sec3 nn pri turns Position : L 0 Ω +.0 kω 0 Ω L 0 Ω n.0 0 Ω pri N sec N sec + N sec3 nn pri turns Position 3: L.0 kω L n pri 000 Ω 0 Ω 0 N sec N sec3 nn pri turns 49

150 43. pri kω 3 ma pri n n L n pri L pri L 8 Ω kω This is 70.7 primary turns for each secondary turn n A I pri(max) 83.3 ma ma Isec(max) 794 ma L(min) 5.9 Ω 794 ma 45. n I pri(max) (0.0833)( A) 83.3 ma A fuse rated at 0. A should be used. Multisim Troubleshooting Problems 46. Partial short in transformer. 47. Secondary is open. 48. No fault. 49. Primary is open. 50

151 CHAPTE 5 TIME ESPONSE OF EACTIE CICUITS BASIC POBLEMS SECTION 5- The C Integrator. τ C (. kω)(0.047 μf) 03 μs. (a) 5C 5(47 Ω)(47 μf).0 ms (b) 5C 5(3300 kω)(0.05 μf) 48 μs (c) 5C 5( kω)(00 pf) μs (d) 5C 5(4.7 MΩ)(0 pf) 35 μs 3. τ 6 ms, C 0. μf τ C τ 6 ms 7.3 kω C 0. µf Use a standard 7 kω resistor. 4. t (min) 5τ 5(6 ms) 30 ms W SECTION 5- esponse of C Integrators to a Single Pulse 5. v C 0.63(0 ).6 6. (a) v 0.86(0 ) 7. (b) v 0.95(0 ) 9 (c) v 0.98(0 ) 9.6 (d) v 0.99(0 ) 9.8 (considered 0 ) 7. See Figure 5-. Figure 5-5

152 8. τ C (.0 kω)( μf) ms v out 0.63(8 ) 5.06 The time to reach steady-state with repetitive pulses is 5 ms. See Figure 5- for output wave shape. Figure 5- SECTION 5-3 esponse of C Integrators to epetitive Pulses 9. τ (4.7 kω)(0 μf) 47 ms 5τ 35 ms See Figure 5-3. Figure T 00 μs t W 0.5(00 μs) 5 μs f 0 khz st pulse: 0.63( ) 63 m Between st and nd pulses: 0.05(63 m) 3.6 m nd pulse: 0.63( 3.6 m) m 644 m Between nd and 3rd pulses: 0.05(644 m) 3. m 3rd pulse: 0.63( 3. m) + 3. m 644 m 5

153 See Figure 5-4. Figure 5-4. The steady-state output equals the average value of the input which is 5 with a small ripple. SECTION 5-4 esponse of C Differentiators to a Single Pulse. See Figure 5-5. Figure τ (.0 kω)( μf) ms See Figure 5-6. Steady-state is reached in 5 ms. Figure

154 SECTION 5-5 esponse of C Differentiators to epetitive Pulses 4. τ (.0 kω)( μf) ms See Figure 5-7. Figure The output voltage is approximately the same wave shape as the input voltage but with an average value of 0. SECTION 5-6 esponse of L Integrators to Pulse Inputs 6. 0 mh τ 0 kω μs 5τ 5 μs out(max) 0.63(8 ) 5.06 See Figure 5-8. Figure mh 7. τ 50 μs.0 kω 5τ 50 μs out(max) See Figure 5-9. Figure

155 SECTION 5-7 esponse of L Differentiators to Pulse Inputs 8. (a) τ 00 μ H 45.4 ns. kω (b) At end of pulse, v out (0 )e.. See Figure ns 37 ns 8.89 Figure τ 00 μ H 45.4 ns. kω 5τ 7 ns See Figure 5-. Figure SECTION 5-8 Applications 0. τ C ( kω)(0.00 μf) μs v B F ( e t/c ) 0 ( e 440μs/μs ) 0.0. The output of the integrator is ideally a dc level which equals the average value of the input signal, 6 in this case. SECTION 5-9 Troubleshooting. τ C (3.3 kω)(0. μf) 76 μs 5τ 5C 5(76 μs) 3.63 ms (b) Since the output looks like the input, the capacitor must be open or the resistor shorted because there is no charging time. (c) The zero output could be caused by an open resistor or a shorted capacitor. 3. τ 76 μs; 5τ 5(76 μs) 3.63 ms (b) No fault. (c) Open capacitor or shorted resistor. 55

156 ADANCED POBLEMS 4. (a) Looking from the source and capacitor, (. kω)(.0 kω +.0 kω) tot.05 kω 4. kω τ tot C (.05 kω)(560 pf) 588 ns µs (b) See Figure 5-..6µs/0.588µs v 0 e 0 m Figure 5-5. (a) Looking from the capacitor, the Thevenin resistance is 5 kω. τ (5 kω)(4.7 μf) 3.5 ms; 5τ 5(3.5 ms) 8 ms (b) See Figure 5-3. Figure L tot 8 μh + 4 μh μh (0 kω)(4.7 kω) tot 5.95 kω 4.7 kω Ltot μh τ.0 ns tot 5.95 kω This circuit is an integrator. 56

157 7. v F ( e t/τ ).5 5( e t/τ ).5 5 5e t/τ 5e t/τ 5.5 e t/τ ln e t/τ ln 0.5 t τ t τ s s 8. The scope display is correct. See Figure 5-4. Figure 5-4 Multisim Troubleshooting Problems 9. Capacitor is open. 30. is open. 3. No fault 3. L is open. 57

158 CHAPTE 6 DIODES AND APPLICATIONS SECTION 6- Introduction to Semiconductors. Two types of semiconductor materials are silicon and germanium.. Semiconductors have 4 valence electrons. 3. In a silicon crystal, a single atom forms 4 covalent bonds. 4. When heat is added to silicon, the number of free electrons increases. 5. Current in silicon is produced at the conduction band and the valence band levels. 6. Doping is the process of adding trivalent or pentavalent elements to an intrinsic semiconductor in order to increase the effective number of free electrons or holes, respectively. 7. Antimony is an n-type impurity. Boron is a p-type impurity. 8. A hole is the absence of an electron in the valence band of an atom. 9. ecombination is the process in which an electron that has crossed the pn junction falls into a hole in the p-region, creating a negative ion. SECTION 6- The Diode 0. The electric field across a pn junction is created by the diffusion of free electrons from the n-type material across the barrier and their recombination with holes in the p-type material. This results in a net negative charge on the p side of the junction and a net positive charge on the n side of the junction, forming an electric field.. A diode cannot be used as a voltage source using the barrier potential because the potential opposes any further charge movement and is an equilibrium condition, not an energy source.. Forward biasing of a pn junction is accomplished by connecting the positive terminal of a battery to the p-type material. 3. A series resistor is necessary to limit the diode current when a diode is forward-biased to prevent overheating. 58

159 SECTION 6-3 Diode Characteristics 4. To generate the forward bias portion of the diode characteristic curve, use the set-up shown in Figure 6-. Figure 6-5. The barrier potential would decrease from 0.7 to 0.6 is there were an increase in junction temperature. 6. (a) The diode is reverse-biased because the anode is at 5 and the cathode is at 8. (b) The diode is forward-biased because the anode is at ground and the cathode is at 00. (c) The diode is forward-biased by the positive voltage produced by the voltage divider. (d) The diode is forward-biased because its cathode is more negative than the anode due to the 0 source. 7. (a) (reversed biased) (b) F 0.7 (c) F 0.7 (d) F (a) The diode should be forward-biased with F 0.7. The 5 measurement indicates an open diode. (b) The diode should be forward-biased with F 0.7. The 5 measurement indicates an open diode. (c) The diode should be reverse-biased and the measured voltage should be 0. The.5 reading indicates that the diode is shorted. (d) The diode is reverse-biased. The 0 reading across the resistor indicates there is no current. From this, it cannot be determined whether the diode is functioning properly or is open. 9. A S 5 B A C S D S 8 59