58- محمود مجدي - ليالي الامتحان

42 -1 -2 emf emf -3 emf emf emf emf 2 emf 1 emf emf 3 emf 2 emf 1 emf emf -4 Pw = V.I -5 6 3 6

43 3– -3 -1 = = PW PWP = PWS VP IP = VSIS , 2 , I2 S R 20 -2 = X 100 = X 100 = X 100 3

44 I = =I 2 R PW = PW - PW X 100 200 PWS = 500W 200 120 = () X 100 = 500 120x10 √2×√2 X100 = 500×100 600 = 83.3% 4 -1 2 3 4 emf 120V 10A I , V t

45 4 ليالي االمتحان الفصل الرابع 1 2 3 4 5 6 1 2 3 4 5 : 1 2 PW α I 2 %10 %10 %10

46 R -1 I = ISin V = VSin = 0 = V ∝ F ∝ F ----------------------------------------------------------------------------------------- L -1 = Sin = Δ Δ XL = 2 f L (XL ) { L N 2 L -2 α F 3 ∴ 0.1 H : I F XL I = V XL = 2f 2 f L = L , ,

47 4 I : = 0.2 = 2 = 18000 → ∴ f = 50 Hz = 2 50 0.2 = 62.83 I = Veff XL = 120x 1 √2 62.83 = 1.3505A ----------------------------------------------------------------------------------------- C C = Q V V = VSin I = CΔV Δt XC = 1 2= 1 ω. C 1 2 3 V = 120 sin(18000t) = ∝ 1 ∝ 1 I = V XC = 2f 1 2 f C = 4 2 f 2 I ∝ 2 ,

48 I1 I2 Ω4 I3 B A 5V r = 1Ω + - + - 1 3 2 2 2 : = 5 = 100 = = 5 100 = 500 V10F = = 500 10 = 50V ∴ V = VB - V10F = 100 – 50 = 50 V 5 = . = 50 5 = 250 – 3 : 1 + 2 − 3 = 0 51 + 103 = 15 52 + 103 = 5 500

49 4 I1 = 1.4 A , I2 = 0.6 A , I3 = 0.8 A ∑ = ∑ = 0.8 10 – = 8 = . = 2 8 = 16 ---------------------------------------------------------------------------------------- R - L 0 < ϴ < 90 , V = √ 2 + 2 = √2 + 2 = ,tan = = R – L 1 V 2 × V = I × ZL = √2 + 2 = 50 = 2 × 50 × 1 = 100Ω Tan = = 100 300 = = 18.43 = 2 2 180 50 = 18.4 = 1

50 > > I XL=XC 3A 24V 4.8A (48V 50Hz ) R = V I = 24 3 = 8Ω Z = = 48 4.8 = 10Ω ∵ Z = √R2 + XL2 ∴ = √ 2 − 2 = √102 − 8 2 = 6Ω = 2 = 6 2×50 H = 0.019 ---------------------------------------------------------------------------------------- R – C −90° < < 0° V = √VR2 + VC 2 Z = √R2 + XC 2 I = , tan = −XC R = − ---------------------------------------------------------------------------------------- R – L – C 90º VR VL 90º VR VC 180º (VC) VL Z = √R2 + [X − XC ] 2 Vt = √VR2 + [V − VC ] 2 I = V Z tan = X−XC R = V−VC R = − Z= XL - Xc ,VR

51 4 X , , X, 1 2 ( − ) = √2 + 2 ( − ) = √2 + 2 3 ( − ) = − A 90 I V O < ϴ < 90 I V 90 I V ϴ I V XC > XL XL > XC RLC R-R L-L C-C

52 = = 10Ω = = 50Ω = = 200 50 = 4A = √2 + ( ) 2 = 4 × √(50) 2 + 102 = 203.96 = 2 = 2= 10 250 = 0.03 90 2 1 LC 3 VL I I VC VT I VC ˂ VL I VT VC ˃ VL

53 4 2 1 - - - - - 1 F = 1 2√LC L = 1 4 2F 2C C = 1 4 2F 2L 2 1 2 = √ 22 11 9 400 m C = (9 × 10 , = −12F) L = ? , = 400m F = C = 3 × 108 400 = 750 × 103HZ L = 1 4 2 F 2C = 1 4 2 × (750 × 103) 2 × (9 × 10−12) L = 5.003 × 10−3H = 5 mH

54 50 Hz F1 = 50 , C2 = 2C1 , L , F = ? 1 2 = √ 2 1 = √ 21 1 = √2 = 50 2 2 = 50 √2 = 25√2HZ Rt = 100 + 200 = 300Ω → 0.4 , 0.2 → = 0.6 H 0.6 , 0.3 → = 0.2 H 0.6 , 0.6 → = 0.3 H 0.3 , 0.2 → = 0.5 H = 0.5 → 10 , 10 → = 5 5 , 5 → = 10F F = 10 = . = 1000 × 0.5 = 500Ω = 1 ω.C = 100Ω = √R2 + [− ] 2 = √(300) 2 + [500 − 100] 2 = 500Ω = = 200 500 = 0.4A tan = − = 500−100 300 = 4 3 → Shift⁡tan = 53. 130 = V=

55 4 > V1 = IXL = 0.4 × (ωL) = 0.4(1000 × 0.3) = 120V V2 = IZ = 0.4 × √(200) 2 + (100) 2 = 89.44V V3 = IZ = 0.4 × √1002 + (ωL) 2 = 0.4 = √1002 + (1000 × 0.2) 2 = 89.44V -1 F = 1 2√LC -2 = 1 4 22 ⋅ C -3 = 1 42F 2⋅ PW = 2, = 2 , = Z R R C R L 60 V

56 -1 VX VY = VZ -2 RLC 60 40 20 -3 CCL – CCR – LLC – LLR – RRC – RRL RRL or RRC 1 √602 + 802 = 100 ≠ 20 CCL – LLC 2 Vt = 80-60 = 20 V = VL ,VC XL = XC = 8Ω = = 8Ω V = . = 10 × 8 = 80 = = 220 10 = 22Ω = 22 − 16 = 6Ω VL = I√RL 2 + XL2 = 10 × √6 2 + 8 2 = 100V #

57 5 1الفصل الخامس - 2 3 2 1 4 3 1 1 2 1 310 2 8 3

58 ❖ 300K %20 %80 6000K %10 %40 %50 300K ) %100 1 1 2 = 2 1 39º 1 ( ) 2 = 2 1 9.66 2 = 39 + 273 300 2 = 9.288 ∴ ∴ ∴ = ℎ. … = × ∴ 1 λ

59 512342 ← , ← 1 ← 2❖

60 ❖ ❖ ➢ = 1 2 2 = ➢ = √ 2 ❖ ▪ 2 1 4 3 2 1 • • • ≥

61 5 • ≥ E < < ≥ ≥ = − = − 1 1 2 2 = ℎ. − ℎ. = ℎ. − ℎ. 3 eV 3000A = 3108 = 9.110−31 = 1.610−19 ℎ = 6.62510−34 1 2 2 = ℎ. − = √ ℎ. − 3 0.5 = √ (6.62510−34)(3108) 300010−10 − 31.610−19 0.59.110−31 = 6.33 × 105\ E

62 . λ − . − → = 1.610−19 2 → = 6.410−19 ( ) =? = + ℎ. = ( + 1.610−19) 2ℎ. = ( + 6.410−19) … … ℎ. 2ℎ. 2 +6.410−19 +1.610−19 2 2 + 3.210−19 = + 6.410−19 = 3.210−19 B A 2 = = ℎ → ℎ 3 C B , A B, A A KE A B C K E B A

63 5 C , B , A C 4 λ. 1 ≥ e 2 . 3 X … A B C

64 ❖ ✓ ∆ = ∆ 1 − 2 = 1 2 ∆ 2 ℎ1 − ℎ2 = 0.5 2 = √ ℎ1 − ℎ2 0.5 = √ 6.62510−3441018 − 6.62510−3431017 0.59.110−31 = 73.4106\ + = + ✓

65 5 = ℎ 2 = . = ℎ = ℎ F = 2 ℎ ∅ = 2 ℎ ∅ = 2 = × ∅ = ℎ∅ = ℎ ∅ ∅ = λ = ℎ – X – λ = h m. V = h me . V = h m. C ❖ …… – – – – e e e e e e

66 X-ray …… …… …… …… < = ℎ → = √ 2 3 º \ = ℎ = 6.62510−34 9.110−3118105 = 410−10 = 4 º

67 5 %…… 25% 1 = 1 2 1 2 2 = 1 2 2 2 2:1 1 2 2 2 : 1 2 1 2 1 2 = 1 2 2 2 = 16 25 = 2 = 1 + 0.25 2 = 1.25 2 = 5 4 1 m2 = 5 4 1 2 = 5 4 1 2 = 25 16 1 = 2 − 1 = 25 16 1 − 1 = 9 16 1 = 9 161 1 100 = 56,25% 3 12µA 0.9mw = () × 100 = N ∅ × 100 = . ∅ ∴ ∅ = N = , = × × 100 × = 12 × 10−6 × 3 × 1.610−19 × 100 1.610−190.910−3 = 4%

68 -الفصل السادس- .1 • → 2 = = 2 = ℎ π = = 2 2 = 1 → = 2 → ( ) .2 .1 .2 .3 .4 .5 • •

6 69 ∆ = E − 1 ∆ = (ℎ. ) = ( ℎ. ) ∞ − = 0 − −13.6 2 2 ∆ = (ℎ. ) = ( ℎ. ) = +1 − ∆ = ℎ. = 2 − 1 → (1) ∆ = ℎ. = 3 − 2 → (2) −13.6 4 − −13.6 1 −13.6 9 − −13.6 4 (− 1 4 ) − (−1) (− 1 9 ) − (− 1 4 ) = 27 5 → = 5 27 = 2 − 1 3 − 2 = −13.6 2 3 – 1 2 3

70 1 2 3 4 1 2 3 X-Ray 4 10−8 10−13 ……… º 1 2 3 3 2 4 1 2

6 71 = 1 2 2 = . = √ 2 = 1 ℎ. . ∴ 2 ℎ. ∆ 28 → 30 … … 28 → 30 →

72 % (1) = ℎ. . = 6.62510−343108 1.610−1950103 = 24.8410−12 (2) = . = 701032010−3 = 1400 (3) −= 2 100 1400 = 28 (4) = 98 100 1400 = 1372 = . = . = 1010−330 1.610−19 = 1.8751018

73 7 - الفصل السابع- 10 s −8 – – – -

74 -1 ● ● 2 ● -2 --- --- --- -3 -1 -2 -3 –

75 7 0.6 10 ☺ %99.5 %98 2 1 0 3 2 1 0 3 0 2 0 3 3 3 2 1 10 s −3 1 2 1 10 1 0 E2:E1 E2:E1 E1 E2

76 2 E1 E1 (1) (2) UV (3) (1) 3 2 1 (3) (2) 3 -1 -2 3

77 7 x 2 2 = فرق الطور 2 × 2 = 180 1 2 3 4 5

78 1الفصل الثامن- 2 3 -1 −273℃ = 0 -2 – N-type e – P-type . = 1 2 3

79 8 p-type n-type = + p= − = 2 + = 2 − -1 -2

80 …… → () −10 > −20 -10V

81 8 2V = = 2 R = . = 20010−3 2 = 0.1 = 2 − 0.5 = 1.5 = = 1.5 0.1 = 15 : E B C – 1 V N P N P N P

82 2 – 1 = 2 on = 0 = 0 off ∴ = + 1 1 ← ← ← 2 ← ← ←

83 8 1 1 = ℎ = ℎ = 6.62510−15 6.62510−34 = 1019 = 1019 > 1015 ∴ 2 2 2 1 3 2 3 4 3 2 2 0.2KΩ 0.5KΩ 5 / 0.1% 100 = − → = 1.7 − 0.5103 = = 100 = 2 = 0.1 100 ∅ = 0.1 100 1017 = 1014\ = = 10141.610−19 = 1.610−5 = 1001.610−5 = 1.610−3 = 1.7 − 1.610−30.5103 = 0.9 ∴ = (0.9) 2 0.2 × 103 = 4.05 × 10−3 = 1015 1.7V = 6.62510−15

84 6 ← 1 ← 2 ← 3 0,1 ← 19 ← 2 10011 4 → = 19 19 2 9 1 9 2 4 1 4 2 2 0 2 2 1 0 1 2 0 1 1 0 0 1 1 2 4 2 3 2 2 2 1 2 0 16+0+0+2+1 = 19 X 10011 log Bin

85 8 10011 1 ( 10011) ← Mode 4 = 10011 ← = 19 *************************** 1 1 AND 0 0 OR 1←0 0←1 Not OR AND ← , AND A log Bin 2 Dec A C B

86 Y X 7 9 – 11 ……… 1 0 1 1 1 × 2 1 × 2 1 0 × 2 2 1 × 2 3 8 0 2 1 11 X Y out 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 0 1 1 1 X Y

87 1 1 Rاالمتحان األول - A1 A2 2 2 …6 3 2A 1.5A 3 A 6 A R VA = 12 V 4 2 Ω 3 Ω 4 Ω 6 Ω 240v 120w , 180v 5 75w 60w 133w 213.3w R 2A R R R 1A R R 0.5A R 1A

88 K 6 A 3 2 1 0.5 7 … I,2I D C B A 8 10−5T … P 3.5x− 5.5x− 6.5x− 4.5x− V2 9 3 10 6 3 9 I 10 A 11 A2 1 4× − T 2× − T − T A C D B 2I I . . . . . . . . . . . . 30cm 10cm.P )1( )2( 30A 15A

89 1 I 12 > > > > > > > > > = = = 20m 13 3x10−4 0.6v … 100m/s 360m/s 360km/h 100km/h 14 8 4cm B 6 3cm A 15 r I (4) I r (1) S S r r r (2) I r r r (3) I

90 16 2 Ω emf 17 emf 141.4 127.32 20 18 2 1 6v BA 19 A B r ,3r 20 50J 25watt 25J 50watt 30J 15watt 15J 30watt : 2:1 3 1:2 6 1:1 3 1:1 6 emf 200v 40 (ms)

91 1 21 XC1 XC2 = 4 C2 C1 22 6 A 11.8A 8.5 1.67A 23 6H … 264Ω 10.5Hz 105Hz 35Hz 70Hz 60° RLC 24 R xL−xc − √ -√3 √ √3 25 12A 26 12A 6√ A 12A 6√ A 27 Z R.L.C √ 3R …… 28 F2 = F rad/s