In the middle of difficulty lies opportunity. في منتصف الصعوبة تكمن فرصة. “ “ www.mahmoud-magdy.com -Albert Einstein-
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • = = . • = • = = • 1 2 = 112 221 • ρe1 = ρe2 1 2 = 12 21 , 1 2 = 12 2 21 2 • 1 2 = 112 1 1 221 22 • 1 2 = 1 2 2 2 2 1 • = • = = VB I = 2 = 2 • Rt = R1 + R2 …… Rt = n R Vt = V1 + V2 + ……. • 1 = 1 1 + 1 2 +……. Rt = = 12 1+2 Vt = V1 = V2 =……. It = I1 + I2 + …….
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • = ( + ) • V = VB + I (r+R) • V = VB - I (r+R) • • ∑ I = 0 • ∑VB = IR • ϴ Фm = BASin(ϴ) • • = µ 2 • = µ 2 N = 2 = 360 • = µ = µ I = 2 = 2 • I1 d1 = I2 d2 • I11 1 = I22 L2 • ϴ F=BIL sin (ϴ) • Fg = Fm mg = BIL
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • = µ12 2 • τ = BIAN Sin (ϴ) ϴ • IAN = sin || ⃗⃗⃗⃗⃗⃗⃗⃗⃗ • • I I−I Rs = = − I • I = + I I • V−IgRg Ig Rm = V=Ig (Rm + Rg) V Rm • = + • = + = • = + ( )
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • emf = − ∆ ∆ 180˚ • emf = −2 ∆ 90˚ • emf = − ∆ 270˚ • emf = − ∆ • emf = 0 • emf = -BLV sinθ ϴ • − ∆1 ∆ emf2 = • 2 ∆2= ∆1 • emf =−L ∆I ∆t • L = 2 L L ∆I = N ∆ • emf = ABN sin = ABN 2f sin • emf = ABN = ABN 2f emf • = ABN x 2 = 4ABNf emf • = ABN x 2 = 4ABNf
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) emf • = ABN x 2 3 • = ABN x 0.707 emf emf • • emf = − ∆ ∆ = − ( 2 − 1 ) ∆ 2 90 • = 2 = 2 = • = = 2 = 2 = T • = 2 • = 2 + 1 • • = = • η = 100 = x 100 η • • = − P P - P V = V - V .R2 P I V I.R P = I.V (I)
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • = = 2 • • • • = 1 = 1 2 • • = 1 + 2 + 3 • 1 = 1 1 + 1 2 + 1 3 • Q=CV V C • • Z = √R2 + XL2 • Z = √R2 + XC 2 • Z = √R2 + (XL − XC) 2 • • tan θ = VL VR = XL R • tan θ = −VC VR = −XC R • tan θ = VL − VC VR = XL − XC R • = 1 2√ = 1 + 2 + 3 1 = 1 1 + 1 2 + 1 3
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • 1 2 = 2 () 1 () • = ℎ = ℎ = 2 = • = 2 = ℎ 2 = ℎ = • = = ℎ = ℎ = • = 2ℎ = 2 • = − = ℎ( − ) = ℎ( 1 λ − 1 ) = 1 2 2 = v V • = ℎ = ℎ m v • 2 = = −13.6 2 n • +1 − = ℎ • ∞ − = ℎ n • = 1 2 2 = ℎ = ℎ
Eng. Mahmoud Magdy Facebook: Physics society (Eng. Mahmoud Magdy) • • 1 2 = 2 2 1 2 • 1 2 = 1 2 2 2 = • 2 (n-Type) • ⸪ = + ∴ + = 2 (p-Type) • ⸪ = − ∴ − = 2 p n + − • = = − = + = − = (1 − ) • = = (1 − ) = (1 − ) • = +
1 1 ليالي االمتحان الفصل األول I = = . = N = : = 1.6 × 10−19 V = = . . = R = . = = 2 . = 2 = = = . W = . = = 2 = 2 . = = = 1 1 1.5 A 7Ω 3A 1 7Ω R %50 R 2 L2 = L1 + 50 100 L1 L1 3 2 = 1L1 + 0.5 L1 = ∴ 2 = 2 3 1 ←
2 1 2 = 12 21 = 1× 2 3 1 3 2 1×1 = 4 9 ∴ 2 = 91 4 → = 2 − 1 = 9 4 1 − 4 4 1 = 5 4 1 100 5 4 1 1 x 100 = 125 % 1 2 3 4 5 1 R / N 1 = 1 1 + 1 2 + 1 3 + ⋯ = 1 + 2 + 3 … R.N 15 Ω , 12 Ω , 6 Ω 2 3A, 1A , 2A ← 12,6 12,6 R / = 6 ×12 6+12 = 4Ω 4 + 15 = 19 Ω VB = IR = 3 x 19 = 57 V R I V (V=I R)
3 1 2 I = 4 R = 30 15 30+15 =10 Ω 30,15 R = 8 24 8+24 = 6Ω 8,24 I30 = 18 x 10 = 6Ω 30 I15 = 18 – 6 = 12 A I8 = 18 x 6 = 13.5 A 8 I24 = 18 – 13.5 = 4.5 A Y X 6 + I = 13.5 I = 7.5 A Y x 18A 18A 2 A X Y 18A Y x
4 3 1 R 2 R 3 4 K2 K1 1 K2 K1 2 K2 1 K1 R` = 60 3 = 20Ω It = = 90 20 = 4.5 A I = 3 = 4.5 3 = 1.5 A K1 2 K2 Rt = 30 2 60 2 =45 Ω It = 90 45 = 2 A I = 2 2 = 1 A
5 1 5 Rt = 4R ← K1 Rt = R + R = 2R ← K2 Rt = R + R + R = 3R ← K3 ← 1 I α K2 4 – 3 – 2 1 1) V = I.R Vα I 2) V = VB 3 V = VB - I r 1 I V α 4 V = VB + I r V α I r=0 r Type equation here. r=0
6 5) V = VB – I ( r + R) 1 I Vα 6 V = I.Rv 7 # # 6 K V = VB – Ir V = VB = I (r + R3 + R4) 1 I Vα R1 R2 It Rt ∴ 5 1 2 3 4 1 1
7 1 2 7 R R A 2 = 8 4 = I1= 4R 4R 8V A 2 = 8 4 = I2= 3R 2R A 4 = 2 + 2 It= 2 = = 4 × 2 = 8 3 = = 4 × 3 = 12 2 Rt 1 It 2 + = 1+2 +1+2 − +1+2 (3 4 2 1
8 8 50V Rt 1 3 Ω , 6Ω R = 3 6 3+6 = 2Ω 8 Ω , 2Ω R = 8 + 2 = 10Ω 10 Ω , 10Ω Rt = 10 2 = 5Ω 2 = − + 1 + 2 = 50 − 14 5 + 0.6 + 0.4 = 6 3 = × 3 6 6Ω 3×2 6 = 1 3Ω = 3 − 1 = 2 PW
9 1 V = I x R Ω 10 V = 3 x 10 = 30 V 8 Ω V = 3 x 8 = 24 V Ω V = 6 x 1 = 6 V Ω V = 2 x 3 = 6 V 6 50V − × 100 = × 100 50−6×0.6 50 × 100 = 92.8% 3 PW = I2R PW = V2 /R PW = IVB = − = 50 − 6 × 0.6 = 46.4 = + = 14 + 6 × 0.4 = 16.4
10 4 1 9 K 3 2 1 3 2 2 2 1 3 4 1 (RV – ) 2 3 4 1 V2 2 V1 = VB - Ir I 3 V2 = VB - I (r+R) I V3 = IR K 1 = = 2 = 1 2 = 1.5 = 2 3 Vر2
11 1 1 = 1 3 V1 V1 V2 V2 V3 V3 3 3 2 2 1 1 2 1 VB2 VB1 VB2 10 1 2 = 10 − 2 → (1) = 28 − 2 10 → (2) 2 1 10 − 2 = 28 − 2 10 = 8 = 2 + 10 = 2 + → + 2 = 10 → (1) = 28 − 2 5 + 2 + 2 + 1 = 28 − 2 10 10 = 28 − 2 10 + 2 = 28 → (2) 2 1 = 2 2 = 8 r=2 VB1 = 28V
12 3 11 1 3 – 2 1 3 2 – 3 – 4 12 I , R , VB a 20 V b 36 V c 10 V d 36-20=16V ← b,a = 16 4 = 4 20 − 10 = 10V ← = 10 5 = 2 ← ∑ I = 0 4 − 2 − = 0 → = 2 = = 2 × 3 = 6 ← 20 − = 6 V = 14 V I 4A 2A VB r = 1 I a 5V r = 1Ω
13 1 a ∑I = 0 → 1.4 – 0.6 – I = 0 I = 0.8 A → (1) ∑VB = ∑IR 1 VB = 1.4 x 1 + 1.4 x 4 + 0.8 x 10 VB = 15 V → (2) ∑VB = ∑IR 2 5 = 0.8 x 10 – 0.6 x 1 – 0.6 R R = 4 Ω → (3) = + = 5 + 1 × 0.6 = 5.6 ------------------------------------------------------------------------------------------ , , 13 a 3Ω = 2 + 4 = 6 ∑VB =∑ IR 0 = 4 x 6 + 6 x 3 – 3I2 I2 = 14 A I3 =14 + 6 20 A I1 = 14 + 4 18A 14 VB I 20V VXY : I = 2 + 3 = 5 A = 3 × 24 + 5 × 15 − = 72 + 75 − 20 = 127 a VB
14 ليالي االمتحان الفصل الثاني Τ m ϴ BIAN sin (90 –ϴ) BA sin ϴ ϴ τ = ΦmIN 1 = 1 Md Md = IAN 2 Md 2 1 10 0.3 T 4 A 1 |Md| = IAN = 4 x 2 x 1 x10 = 80 J/T 2 ∴ 1 2 = ϴ 0 = 90 X X X X X 1m X X X X X 2m
15 2 1 1 2 3 1 2 3 B= =n L= 2π N r B= μIN 2r L= 2π rN N= ϴ 360 N1 N2 = r2 r1 B= μI 2лd ❖ • • • • • BT = √B 1 2 + B2 2 d = r • n • → • B=μ I n B B= μIn I 1 R α R → I → B
16 1 2 A 3 d d = 2 + 0.5 = 2.5 cm 4 dA = dB BA = BB 5 3A A 40cm e/S 20 10 A 5 x 10 cm IB = N.e = 5 x1020 x 1.6 x10-19 = 80A BT = BB – BA = 4 πx10−7x80 2 πx50x10−2 − 4 πx10−7x3 2 πx10x10−2 =2.6 x 10-5T A A B A B 10cm 40cm 3A X 1cm 2cm d
17 2 6 BT = √B1 2 + B2 2 7 D B A E ∴ d E 2I 2d E I 2d A 2I A d A I A A I d 2I 2d A ∴ 8 10V 11cm 1.5 x 10-5 Ω.m 1 Ω 0.1 cm L= 2πrN = 7 2π ×11 10−2= 4.84 m R=ρ R= 1.5 x10−5x4.84 π x (0.1×10−2) 2 = 23.1 Ω I= + = 10 23.1+1 = 0.415 A B = μIN 2r = 4 πx10−7x 0.415 x 7 2 x11x10−2 = 1.7x10-5 T x E D A B 2I I X X X X (-) (+) (-) (+)
18 2 R R 2I I X 3I 2R 9 B1 = μ x 1 2 I x 1 2 2 r = 1 8 μ I r B2 = μ x 1 2 I x 1 2 2 r = 1 8 μ I r BT= zero B1 = μ x I x 1 2 2 r = 1 4 μ I r B2 = μ x I x 1 2 2x2 r = 1 8 μ I r μ I r 1 8 BT= B1 = μ x I x 1 2 2 r = 1 4 μ I r B2 = μ x I x 1 2 2x2 r = 1 8 μ I r μ I r 3 8 BT= 10 ∴ I 11 3 4 B = µnI B 1 4 R2 = ¼ R1 I2 = 4 I1 B2 = 4 B1 12 C BT = B1 – B2 BT =B1+B2 BT = √B BT = B2-B1 1 2 + B2 2 r r r r I r r I r r I I r 0.5I 0.5I
19 2 : BT = B1+B2 13 1Kg Kg 1 (1.1 1 0.9) 0.9 2 Kg 3 (0.8 1 0.9) 0.8 F=BIL sinϴ * F=μ12 2πd F=IL 3 = 1 1A 1m F = BIL sinϴ =0.8N -2 L = 40 cm → :. F = BIL=2 x 1x 40 x 10 1kg 40cm I L=1m N S 2T
20 2 F1 = F2 L 3 A, B , C 1 C B A ∴ A 2 C B A B 3 C B A C ∴ 1 1 2 3 2 3 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X A B C
21 2 4 5 1 2 3 6 1 2 3 • 1 2 1 80 0.08 m , 0.12 m 3μA 0.5 T 0 60 1 60 3 20 Deg/µA 2 0 60 τ = BIANsinϴ = 0.5 x 3 x10-6 x0.12 x 0.08 x 80 sin 90 τ = 1.152 x10-6 N.M • 1 2 3
22 3 0 3 0 1 2 3 4 ❖ 1 2 I ϴ B , A, N 1 3 3 2
23 2 ❖ • 1 2 3 Rs • < 1 • > 1 , • VB • 1 2 − RS = = + Rm= − V= ( + ) = = + RS = RX= Rm= المقلوب -1 المقلوب -1 المقلوب -1
24 • 1 , 2 3 < 1 • > 1 , • VB • 1 2 • 1 8 Ω 10Ω 30mA 1.5V : = = 30 × 10−3 × 10 = 0.3 (8Ω) = 1.5 − 0.3 = 1.2 =0.15A 1.2 8 It = × 10 Is = It- Ig = 0.15 - 30 −3 = 0.12 = 2.5 Ω RS =
25 2 2 0.3Ω : = 1 4 → = 4 = − = 4− = 3 0.3 = 3 → = 0.9Ω = 1 10 → = 10 = − = 10− = 9 = 0.9 9 = 0.1 Ω 1 4 3 = 1 − 4 ← 4 ← RS = O → 0.3 = 3 → = 0.9Ω RS = 0.9 9 = 0.1Ω 3 30 Ω 1/3 : = 1 6 → = 6 Rm = − = 6− = 5 = 5 30 = 5 Rg Rg = 6Ω = 1 3 → = 3 Rm = − = 3− = 2 = 2 = 2 × 6 = 12Ω 1 6 5 = 1 − 6 ← 6 ← Rm = O x Rg 30 = 5 x Rg Rg = 6 Ω Rm = 2 x 6 = 12 Ω
26 = 1 5 + = 1 5 5R = R + Rx Rx = 4R = 4× 500 = 2000 Ω 1 2 1 = 1 − 2 ← 2 ← Rx = O x R 500 = ( 2 – 1) R R = 500 Ω Rx = ( 5 – 1) x 500 Rx = 4 x 500 = 2000Ω 4 500Ω : = 1 2 = = 1 2 I = + + = 1 2 2R = R + Rx R = Rx = 500 Ω 5 VB : Ig = 500 μA 9 K Ω 1 4 3 = 1 − 4 ← 4 ← Rx = O x R 9 = (4 -1) x R R = 3 KΩ VB = Ig x R VB = 500 x10-6 x 3x103 = 1.5V 500 Ig
3 emf emf 1 emf 2 emf = - N ∆Ф ∆ t ∆Ф (Ф= BA sin ϴ) Ф 1 6 3 0.5 12 emf = IR = ∆t × R = ∆t × R = N∆∅ ∆ t QR = N∆ Q = N∆ R = 12 3 12 0.5 = 6 C N = 12 R = 0.5 Ω ∆Ф = Ф(2) − Ф(1) = 6 − 3 = 3 ∆∅ = ∆ = ( − ). Ф ∆∅ = . ∆ = ( − ).
28 1 2 3 N N N I I I (2) (1) S N N S S N
29 3 A = 25 7cm 100 . . . . . . . . . . . . 1 2 emf =N∆∅ ∆ t N2 < N1 I2 < I1 ………… ∴ 1 500 T/s emf . . . . . . 2
30 I =VB – emf R emf =N∆∅ ∆ t = ∆ ∆t . = 1 x 500 x × (7 x 10-2 ) 2 = 7.7V I =100 – 7.7 25 = 3.03288A emf 2 = − sin Ө L ϴ 20 m T -4 0.8 V 4 x10 L = 20 m , B = 4 x 10-4 T , emf = 0.8 V emf = B L V sin ϴ V = Emf BL sin ϴ = 0.8 410−420 sin 90 = 100 / V = ? km / hr 18 5 V km/hr = 100x =360 km / hr 50 cm 2T 1m L 5 m/s = 50 emf = BLV sin ϴ = 2 x 50 x 10-2 x 5 x sin ( ) = = 2 50 = 1 ← B ← V
31 3 B L emf V emf α L A, B -1 A B -2 B A -3 B A B A B A -4 A B B A A B 1m 1T A 10m/s = 0 I =2emf R = 2BLV 2 = 2x1x1x10 2 = 10A X X X X X X X X X X X X X X X X X X X X X X X X X X X X V L B X X X X V A B X X X X A 1Ω 1Ω X X X X X X X X X X X X X X
32 1 3 2 1 6 5 4 5 3 1 -1 emf 6 4 2 -2 emf 1 2 (2) = − ∆1 ∆ ( M) V M N D μ : 2 emf emf
33 emf = - L ∆ ∆ t I 3 L = L μAN2 L emf emf emf 1 ∆t α emf < emf 100 2 40 cm 25cm = = ∆X = X2 – X1 emf = - L ∆I ∆t = 2 ∆ ∆ = 4πx10−7 x 40 x10−4 x (100) 2 25 x10−2 × 2−(−2) (8 – 4) x10−3 emf = 0.201 V emf emf I -2A 4 8 +2A I t(ms)
34 30% 70% Imax I = 0 ∆I ∆t = −N∆∅ ∆t emf = - L ∆I1 ∆t = −N1 ∆∅ (1) ∆t emf 1 = - L ∆I1 ∆t = −N2 ∆∅ (2) ∆t emf 2 =− L= N1 ∆∅ (1) ∆I = N2 ∆∅ (2) ∆I 60 40 6 x 10-4 4A 20% : N1 = 40 ∆ Ф2 = 80 100 x 6 x10-4 = 4.8 x10-4 wb ∆I1 = 4 – 0 = 4A , ∆ Ф1= 6 x 10-4 wb , N2 = 60 L= N1 ∆∅ (1) ∆I = 40x6x10−4 4 = 6x10-3H = N2 ∆∅ (2) ∆I = 60x4.8x10−4 4 = 7.2 x10-3H I = 0% I I =VB – emf R I = 0 0 = VB – emf emf = VB emf = 100% VB I = 0% emf = 100%VB I = 30% emf = 70%VB I = 90% emf = 10%VB I = 100% emf = 0 VB % 100 = emf I
35 3 20% 500A/s 0.2H 80% = ∆ ∆ = 0.2500 = 100 = 20% → = 80% = 0.2 ∆ ∆ = 500 / 100 = 0.8 x VB = 100 0.8 = 125 = ∆ ∆ ∆ ∆ = = ? ? 0.2 I = 80% Imax → emf = 20% VB emf = 0.2 x 125 = 25 V ∆ ∆ = = 25 0.2 = 125 A/s ❖ -1 -2 -3 ❖ 2 1 4 3 -1 -2
36 -3 -4 ∆ ∆ 3 3x 3 -3x 3 ϴ ϴ B Ф emf = − ∆∅ ∆ 3x -3x 3 3 -3 cos sin Ө → cos Ө
37 3 1 emf emf = sinӨ → = 2 = ( ) → = 1 = ϴ ϴ x (360ᵒ ) (2 π Ft) π = 180ᵒ 90 ∅ t emf t
38 X 360ᵒ= 60ᵒ ϴ = 90ᵒ − 60ᵒ = 30ᵒ 50Hz 3 ϴ 2πFt = 2πF(3 x 10-3 ) = 2 x 180 x 50 x 3 x 10−3 = 54ᵒ ᵒ = 90ᵒ − 60ᵒ = 30ᵒ emfmax emf= ABN 1 - emf = emfmax Sinϴ 2- emf = emfmax × 2 3- emfeff = emfmax Sin 45 = √2 = 0.707 emf emf ¼ emf ¼ = emf ½ = 4 ABN F emf ½ emf 100V : 1 2f 1 2f max 2f 2f +1 Max 1 2 t X X max X max X
39 3 1 0 300 300 t 300 90 60 x X = 60 t 30 = 2t 2 0ᵒ 300 30 t 45 90 45 x X = 45 t 30 = 1.5t 3 0 → 30ᵒ → 30ᵒ → 30ᵒ → 45ᵒ → 15ᵒ → X = 15 t 30 = 1 2 t -1 -2 180 I(A) t(s) t(s) I(A)
40 ½½ 1 t(s) I(A) t(s) I(A) t(s) I(A) t I t I t I t I
41 3 2 – 3 B A B 10Ω 100 W = Pw.t W = emfeff 2 t R ∅ = = 410−2 = 1 = 1 810−3 = 125 = = 410−21002125 = 3141.59 = sin 45 = 3141.59 sin 45 = 2221.44 t = T = () 2 . = (2221.44) 2810−3 10 = 3947.84 A 4 8 ∅ B (\) 410−2 A 4 8 B = − ∆∅ ∆