Upcoming SlideShare
×

# Fisika

1,113 views

Published on

kumpulan rumus fisika...

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,113
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
85
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Fisika

1. 1. PHYSICS 비 오 니 다 비 나 시호 당
2. 2. We We Bundo Ee..! physic
3. 3. Electric current in a circuit can only flow in a closed circuit.
4. 4. The electric current strength flowing through a section of conductor can be determined by the equation as follow : I = electric current strength (A) Q = the amount of electric charges flowing (C) t = time (s)
5. 5. 1.5 A electric current in 240 second determined the electric charges. Known Asked Answer : I = 1.5 A t = 240 s : Q…? :Q=I.T =1.5 x 240 =360 coulomb Sample problem
6. 6. When a charge of 56 C flows past any point along a circuit in 8 seconds, the current is ??? Known : Q = 56 t=8 Asked : I…? Answer : I = 56 C / 8 S =7A Sample problem
7. 7. section of wire, then the electric current strength can also be determined by using the equation as folow. I = electric current strength (A) v = speed of electron (m/s) e=electron charges (1.6 x 10-19 c) n=amount of electron per unit of volume A=area of wire section (m2)
8. 8. Cu r r e n t d e f i n e d p o we r p e r c o n d u c t d e a s u n o r n s i t y e l e c t i t a r s e c t i i s r i c e a o f o n . = e l e c t r i c c u r r e n t s t r e n g t h (A ) J = c u r r e n t d e n s i t y (A /m 2) A= a r e a o f c o n d u c t o r s e c t i o n (m 2) CURRENT DENSITY I
9. 9. 0.5 A electric current passes through a wire (conductor) which has section area 4 mm2. Calculate current density ? Known : I= 0.5 A A= 4 mm2 = 4x10-6 m2 Asked : J ...? Answer :J = I / A = 0.5 A / 4x10-6 m2 = 0.125 x 10-6 A/m2 Sample problem
10. 10. Electric current that flows in conductor is proportional to potential different between both the edge of conductor, and influence by its resistance George Simon Ohm (1787-1854)
11. 11. HUKUM OHM Klik KKlik Kli k 0,40 0,54 0,2 0 1, 4,0 2,6 2 Klik Klik Klik From the table above, we can know if the potential different is increased then the electric current strength is also increased. The number of battery 1 2 3 V I What relationship can we get between potential different and electric current strength? Make the graphic’s relationship between potential different and electric current
12. 12. Klik Graphic of Potential different (V) to electric current strength ( I ) Data V 1,2 2,6 4,0 V(volt) 5,0 4,0 3,0 V ~ 2,0 I 0,2 0,4 0,54 V = R V= Potential different 1,0 I Klik Klik 0,1 0,2 0,3 0,4 0,5 0,6 I= Electric current strength R= Resistance ( Ω )
13. 13. KlIk Graphic of Resistance (R) Klik to electric current strength Data (I) R(Ω) Klik 50 R 10 20 30 40 I 1,0 0,5 0,3 0,25 40 If V is constant = 10 V V R V I2 = R V I3 = R V I4 = R I( A) I1 = 30 20 10 0,25 0,50 0,75 1,0 1,5 10 10 10 I2 = 20 10 I3 = 30 10 I4 = 40 I1 = 1,0 A I1 = R= I2 = 0,5 A I3 = 0,3 A I4 = 0,25 A V I
14. 14. Purpose : To investigate factors that influence wire resistance Klik 1 Klik B A Klik 1. Length of wire Klik 2. Wire resistivity 3. section area of wire IA > IB RA < RB lA < lB If the wire is longer then the resistance of wire is larger. The wire resistance is proportional to length of wire. R~ℓ
15. 15. 2 Klik A B Alluminiu m IA < IB 1. Type of wire Klik Tembaga RA > RB 2. Resistance 3. Length of wire, section area Aℓ > Cu If the resistivity of wire is larger then the resistance of wire larger Resistance of wire is proportional to resistivity. R ~
16. 16. 3 Klik A B IA < IB 1. Section area of wire 2. Resistance of wire 3. Resistivity of wire, length of wire RA > R B AA < AB If the area section of wire is larger then the resistance of wire is smaller. Resistance of wire is inversely to area section. R 1 ~ A
17. 17. Klik Factors that influence the resistance on wire is: 1. Length of wire ( l ) 2. Section area of wire( A ) 3. Resistivity of wire ( r ) R  ρ A R l = Resistance (Ω ) = Length of wire ( m ) Area section of wire( m2 ) = Resistivity of wire ( Ω m )
18. 18. Resistivity of a certain substance is the property of substance which influenced by the change of temperature, then Δρ= ρ₀αΔT ρt= ρ₀(1+αΔT) ρt= resistivity at temperature T ρ₀= resistivity at temperature To α = temperature coefficient of resistivity ΔT= change of temperature
19. 19. The electric resistance (R) is directly proportional with resistivity (ρ), then ΔR= R₀αΔT Rt= R₀(1+αΔT) Rt= final resistance R₀= initial resistance α = temperature coefficient of resistivity ΔT= change of temperature
20. 20. Series a R1 Vab a R2 b c Vbc Rs d Vad = Vab Vbc + Vcd + I Rs =I R1 +I R2 + R3 I + R2 d Vcd Vad Rs = R1 R3 R +3
21. 21. Parallel I R1 I= Vab 1 I a I2 R2 I3 I a b R3 Rp Vab b I1 + I2 + I3 Vab Vab Vab = + + R3 R1 R2 RP 1 1 1 1 = + + RP R1 R2 R3
22. 22. Example  Determine the resistance on the below circuit! 1 2Ω 4Ω 3Ω 5Ω Rs = R1+R2+R3+R4+R5+R6+R7 3Ω 2Ω 4Ω Rs =2+4+3+2+4+5+3 Rs =23 Ω 1 RP 2 1 6Ω 4Ω 3Ω RP 3Ω 1 RP 1 = = = 1 R1 1 6 1 6 3 RP 4Ω RP: 2 Ω 3Ω = 6 RP =2 Ω + + + 1 R2 1 3 2 6 Rs = R1+RP+R2 Rs = 4+2+3 Rs = 9 Ω
23. 23. 1.THE LENGTH OF WIRE 2. THE RESISTIVITY 3. THE AREA OF WIRE
24. 24. B A The longer wire is, higher is its electrical resistance Facrors Length of wire . A B Short Long Resistance Small Large Light Bright Dimmer R~ℓ
25. 25. A Factors B A B Reistivity Large Small Resistance Large Small Light Dimmer Bright . The higher resistivity of the wire is, higher is its electrical resistance R~
26. 26. A Factors B A B Area small Large Resistance Large Small Light Dimmer Bright R ~ 1/A The lerge area of the wire is,smaller is its electrical resistance
27. 27. L A EXAMPLE
28. 28. A wire has a length of 100 m, a diameter of 2mm, and a resistivity of 6,28 x 10-8 Ωm. Determine the resistance of the wire? Known Asked : L = 100 m r = 6,28 x 10-8 Ωm r = 2mm = 1mm = 10-3 m 2 : resistivity Answer : R = r ℓ/ A = 6,28 x 10-8 Ωm x 100 m 3,14 (10-3 m ) = 2Ω
29. 29. When the wire coil becomes hot and burn red, lamp becomes dimmer. This indicated that the current going through the lamp is reduce. wire resistance increase with temperature. So, we can conclude that
30. 30. T resistivity at temperature T O α O t = R O (1+ RO T) α= T α Rt =R O ( 1 + α α = resistivity at temperature T0 T) temperature coeficien of resistivity
31. 31. EXAMPLE A metal has resistance 50 Ω at 20 C and 76,8 Ω at 157 C. Calculate its temperature coeficient of resistivity known : R0 = 50 Ω Rt = 76,8 Ω T0 = 20 C Tt = 157 C asked : Answer : α T = 137 R = 26,8 26,8/ 50 x 137 = 3,9 x 10-3 α =
32. 32. “at any node or junction in an electrical circuit, the sum or currents flowing into that node is equal to the sum of currents flowing out of that node” “ the algebraic sum of currents in a network of conductors meeting at a point is zero” Kirchhoff's first law 11/9/2013
33. 33. Kirchhoff's first law can be expressed mathematically by the equation as follow : 11/9/2013
34. 34. Series Circuit of Resistor  The electric current which flowing circuits are the same IR1 = IR2 = IR3  The voltage source (V) is divided into V1 V 2 and V3 V = V 1 + V2 + V 3 11/9/2013
35. 35.  The series circuit can be used to increase the resistance of the circuit V = V1 + V2 I.Rs = I.R1 + I.R2 Rs = R1 + R2 for n Resistor : Rs = R1 + R2 + … + Rn 11/9/2013
36. 36.  Parallel circuit can be used to reduce the resistance of the circuit I = I1 + I2 V= V+V Rp R1 + R2 1=1 + 1 +…+ 1 Rp R1 R2 Rn 11/9/2013
37. 37. Parallel Circuit of Resistors  The Voltage source (V) in the circuit is the same V1 = V2 = V  Parallel circuit is the divider circuit I = I1 + I2 11/9/2013
38. 38. If a parallel circuit (look at the picture) with; R1 = 10 Ω, R2 = 15 Ω and R3 = 30 Ω, Find the total resistor ! Answer : Rp = 1 + 1 + 1 = 1 + 1 + 1 R1 R2 R3 10 15 30 = 6/30 =5Ω 11/9/2013
39. 39. Electromotive Force and Clamping Potential  Electromotive force is defined as the energy used to transfer the positive charge from low potential point to higher potential point per unit of charge tranferred. = electromotive force (volt)
40. 40.  Basically, every electric source, such as battery has the internal resistance (r), which simplify can be shown by the following figure. R
41. 41. If the seat of electromotive force is connected to the certain electric component(suppose resistor), then there are two important things which must be considered but before that study the figure below r A I R B S
42. 42.  If the switch (S) is not connected, then there is no electric current flowing through the circuit(I=0), therefore potential difference between A and B(VAB) is equal to the electromotive force ( ).  But is the switch (S) is connected, then there is electric current flowing through the circuit, therefore the potential difference between A and B(VAB) is not equal to the electromotive force ( ).
43. 43.  At the moment, the switch is connected (I 0), the potential difference between A and B is : called CLAMPING POTENTIAL, which can be determined by the following equation : VAB = - Ir = IR VAB = clamping potential (volt) = electromotive force (volt) I = electric current (A) r = internal resistance ( ) R = external resistance ( )
44. 44. • Free electrons can flow in a conductor because of the presence of the electromotive force or electric voltage. • Also they can flow because of the presence of potential different. • The flow is from a lower potential to the higher potential.
45. 45. Electromotive force: - +
46. 46. • Electromotive force (EMF) of an electric current source is the potential difference between the ends of the electric current source when the electric current source doesn’t conduct electric current. • Batteries are one of electric current source. • The electric voltage or potential difference between the poles of an electric current source in an electric circuit can be measured by using an instrument called voltmeter. • So, Electromotive force (EMF) is a potential different between the poles of an element before conducting electric current.
47. 47. • The potential difference measured by a using a voltmeter when electric circuit is closed shows the value of clamping voltage. Voltmeter
48. 48.  The clamping potential between the poles of voltage source is not constant, but depends on the resistance value of the circuit.  Clamping voltage is potential difference between the poles of an element while conducting electric current.  The clamping voltage can be stated by the following equation: V= ɛ - V1 V= ɛ - Ir
49. 49. In which: V = Clamping voltage (volt) ɛ = EMF V1 = Voltage in internal resistance (volt) I = Electric current flowing through the internal resistance in electric current sources (ampere) r = Internal resistance in electric current sources (volt)
50. 50. PROBLEM
51. 51. 2. Calculate the clamping voltage of an electric circuit the EMF of which is 15 volts and the voltage in an internal resistor is 4 volts! Solution: Given Asked Answer : ɛ = 15 volt Vr= 4 volt :V=? : V = ɛ - Vr = 15 volt – 4 volt = 11 volt So, the clamping voltage is 11 volt.
52. 52. Source voltage circuit Kinds of source of voltage connection  Series , , 1 , 2 3 r1 r2 r3 if several voltage source are connected by series, then the total electromotive force is : s = 1 + 2 + 3 +…+ n s = total (substitute) electromotive force in series circuit (volt)
53. 53. While the total(substitute) internal resistance of a series circuit of voltage source can be determined by the following equation : rs = internal resistance in series circuit ( ) rs = r1 + r2 + r3 + … + rn
54. 54. • The amount of electric current in the main circuit in a serial circuit of electric element or electric voltage source can be determined by the following equation: I= nɛ R + nr In which: I = electric current in a circuit R = external resistance of circuit r = internal resistance of voltage source
55. 55. PROBLEM
56. 56. 1. If ɛ = 1,5 volts, r = 0,2 Ω, and R = 3Ω calculate the current flowing in the following circuits! I + + - + ɛ, r ɛ, r - ɛ, r + - ɛ, r
57. 57. Solution: Given Asked Answer : ɛ = 1,5 volt; r = 0,2 Ω; R = 3 Ω; n = 4 : I in the serial circuit of electric element. : a. I = n ɛ = 4(1,5 volt) = 6 Volt = 1,6 A R + nr 3Ω + 4(0,2Ω) 3,8Ω So, the current flowing is 1,6 A.
58. 58.  Parallel , 1 r1 , 2 r2 , 3 r3 if several voltage source are connected by parallel, then then electromotive force of the circuit is : p = 1 = 2 = 3 =…= n
59. 59. while the total(substitute) internal resistance of a parallel circuit of voltage source can be determined by the following equation : rp = internal resistance in parallel circuit ( )
60. 60.  The total voltage in the parallel circuit of electric element is as follow: V = ɛ = ɛ = ɛ3 = ɛ 1 2 The amount of electric current in a circuit in the parallel connection of voltage sources can be determined by the following equation: I= ɛ R+1 r n
61. 61. In which: I = e = n = R = r = current flowing in a circuit EMF voltage sources the number of voltage source external resistance circuit internal resistance
62. 62. PROBLEM
63. 63. 1. If ɛ = 1,5 volts, r = 0,2 Ω, and R = 3Ω calculate the current flowing in the following circuits! ɛ, r I ɛ, r ɛ, r ɛ, r
64. 64. MENGUNJUNGI RUMAH FRANKENSTEIN YANG PENUH BAHAYA © Surya Institute, 2007
65. 65. Tebak, mengapa tindakan mereka ini berbahaya? © Surya Institute, 2007
66. 66. © Surya Institute, 2007
67. 67. © Surya Institute, 2007
68. 68. If Kirchhoff’s 1 law relates with electric current in the branched electric circuit, then Kirchhoff’s 2 law relates with voltage and electromotive force in closed elecric circuit. Kirchhoff’s law states that Kirchhoff’s ii law in the closed electric circuit, the algebraic sum
69. 69. Mathematically, the Kirchhoff’s ii law can be expressed by following equation :  ΣVtertutup =0  ΣE +Σ(I.R) = 0
70. 70. In using the loop theorem to solve problems in the closed electic circuit, we have to consider the following several things : a.Pick a loop for each closed circuit in a certain direction ( the direction of the loop is free) b.If the direction of the loop is the same as the direction of electric current then the
71. 71. Sample problem If R1= 4ῼ, R2 = 2ῼ, R3= 6ῼ, r1=r2=0, ɛ1= 8 v and ɛ2= 18 , determine the electric current in each branch! Solution: determine the direction of electric current assumption in each branch and also determine the direction of the loop 
72. 72. Solve the problem in each loop using the loop theorem ( Kirchhoff’s ii law ) Loop 1 ΣE +Σ(I.R) = 0 -ɛ+ I₁R₁+ I₃R₃ = 0 ɛ = I₁R₁ + I₃R₃ 8 =4 I₁+6I₃... (1) Loop 2 Σɛ + ΣIR =0 -ɛ+ I₁R₁+ I₃R₃ = 0 ɛ₂ = 0 18 = 2 I₂ + 6 I₃... (2) 
73. 73. Because I₃ = I₁ + I₂ (remember the Kirchhoff’s 1 law), then 18 = 2 ( I₃ -I₁ ) + 6 I₃ 18 = 2 I₃ - 2 I₁ + 6 I₃ 18 = -2 I₁ + 8 I₃…(3) Solve the equation (1) and (3) 8 = 4 I₁ + 6 I₃ 1 18= -2 I₁ + 8 I₃ 2 8 = 4 I₁ + 6 I₃ 36 = -4 I₁ + 16 I₃ 44= 0 + 22 I₃ I ₃ = 44/22 A = 2A
74. 74. Therefore 8 = 4 I₁ + 6(2) -4 I₁ = 4 I₁ =-1 A Because I₃ = 2A, and I₁ = -1 A, then I₂ = I₃ – I₁ = 2A - (-1A) = 3A Thus, I₁ = -1A, I₂ = 3A and I₃ = 2A
75. 75. To measure electric quantities are used certain electric meters which have the measuring limit to the value of the electric quantities measured. Electric meters are instruments for measuring and indicating magnitudes of electric quantities values, such as current, charge, potential and power along with the electrical characteristic of circuit such as resistance capacitance and inductance.
76. 76. Measurement of Electric Quantities    Measurement of Current Measurement of Voltage Measurement of Resistance
77. 77. Measurement of Current To measure the electric current strength in circuit is used an instrument which is called ammeter. Basically ammeter consist of:  Galvanometer  One or more resistors which are called shunt resistor
78. 78. To obtain accurate measurements of current, then the resistance of ammeter is made much smaller than the circuit resistance. While for upgrade the ability of measurement of an ammeter , then a shunt resistor much be set parallel with to galvanometer , therefore the surplus of electric current will flow pass through the shunt resistor
79. 79. If the current of complete scale is ammeter is expressed with I which has value n times greater than the current of complete scale in galvanometer( Ig ) , then the multiple of the maximum measuring limit of the ammeter can be determined by the equation as follow:
80. 80. The ammeter circuit in the figure above shows that Rg and Rsh are connected bu parallel, therefore • • • • • Ig : Ish =1/Rg : 1/Rsh I = Ig + Ish nIg = Ig + Ish Ish = ( n – 1 ).Ig Because the parallel circuits are current divider circuit, the • Ig = Rsh /(Rg + Rsh) • Ig = Rsh /(Rg + Rsh) nIg • Rg + Rsh = n Rsh • Rg = (n-1)Rsh Rsh = Rg/(n-1) Where : • Ish = current in the shunt resistor • Rsh = shunt resistance • Rg = galvanometer resistance
81. 81. In it`s using the measure electric current in a circuit, ammeter must be set in series to the circuit, therefore in this case the substitude resistance of Rsh and Rg is the internal resistance of ammeter, which the value is : RA =(Rg x Rsh):(Rg + Rsh) Where : RA = internal resistance of ammeter
82. 82. A galvanometer which its resistance is 30 Ω ohm will experience the complete scale deflection at 500 mA, calculate the resistance of shunt resistor so that can be used to measure the current of 3A ! Known : R = 30 Ω Ig = 500 mA = 0,5 A
83. 83. • The instrument used to measure the potential difference or voltage is voltmeter. Voltmeter is arranged by a galvanometer and one or more resistors which are connected in series. • To obtain the accurate measurement of voltage, then the resistance of a voltmeter is made much greater than the circuit resistance. Therefore to upgrade the ability of measurement of a certain voltmeter, then must be set a series resistor which are connected in series with the galvanometer, it will cause the surplus of voltage will given to the series resistor.
84. 84. If the voltage of complete scale in voltmeter is expressed with V which has values of times greater than the voltage of complete scale in galvanometer (Vg ) can be determined by the equation as follow : n = V/Vg • or • • • • • V = n.Vg Where : n = multiple of the maximum measuring limit V = voltage of complete scale in voltmeter Vg = voltage of the complete scale in galvanometer The voltmeter circuit in the vigure above, shows that Rs and Rg is connected in series, therefore : • Vs : Vg = Rs.Rg
85. 85. Because the series circuits are the voltage divider circuit, then : Vg = Rg/(Rs + Rg) . V Vg = Rg/(Rs + Rg) .nVg Rs + Rg = nRg Rs = (n-1)Rg Rg = Rs/(n-1) • Where : Vs = voltage in the series resistor Rs = resistance of series resistor Rg = galvanometer resistance In it`s using to measure the voltage in the circuits, voltmeter must set in parallel to the circuits, therefore in this case the substitude resistance of Rs and Rg is the internal resistance of voltmeter, which the value is : RV = Rs + Rg • Where : RV = internal resistance of voltmeter
86. 86. A galvanometer has the resistance of 50 Ω and has the maximun deflection if passed through a current of 0.01 A. For measuring voltage up to 100 V, calculate the resitance of the series resistor which must be set! Known : Rg= 50 Ω Ig = 0.01 A Asked Answer :Rs= ? :
87. 87. To measure electric resistance we can be use an instrument which is called ohmmeter. Basically ohmmeter is made by ammeter circuit, where the electric current measured by the ammeter and the emf ( ) known can be used to determine the value of a certain resistance (Rx) by a certain calibration.
88. 88. Usually the function of voltmeter, ammeter and ohmmeter is combined in an instrument called multimeter. Voltmeter, ammeter and ohmmeter principle explained above is the analogue of voltmeter, ammeter, and ohmmeter principle. However, now beside those there are digital voltmeter, ammeter, and ohmmeter which can show the measurement value of voltage, current and resistence in form of numbers.