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1. PHYSICS
비 오 니 다 비 나 시호 당

2. We
We
Bundo Ee..!
physic

5. Electric current in a circuit can
only flow in a closed circuit.

6. 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)

7. 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

8. 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

9. 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)

10. 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

11. 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

12. 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)

13. 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

14. 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 )

15. 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

16. 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

17. 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 ~

18. 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

19. 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 )

20. 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

21. 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

22. 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

23. 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

24. 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

25. 1.THE LENGTH OF WIRE
2. THE RESISTIVITY
3. THE AREA OF WIRE

26. 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

27. 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~

28. 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

29. L
A
EXAMPLE

31. 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

33. 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

34. 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

35. 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 =

37. “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

38. Kirchhoff's first law can be expressed
mathematically by the equation as follow :
11/9/2013

39. 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

40. 
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

41. 
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

42. 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

43. 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

44. 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)

45. 
Basically, every electric source, such as
battery has the internal resistance (r),
which simplify can be shown by the
following figure.
R

46. 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

47. 
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 ( ).

48. 
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 ( )

49. • 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.

50. Electromotive force:
-
+

51. • 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.

52. • The potential difference measured by a using a voltmeter
when electric circuit is closed shows the value of clamping
voltage.
Voltmeter

53. 
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

54. 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)

55. PROBLEM

57. 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.

58. 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)

59. 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

60. • 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

61. PROBLEM

62. 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

63. 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.

64. 
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

65. 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 ( )

66.  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

67. 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

68. PROBLEM

69. 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

70. MENGUNJUNGI RUMAH FRANKENSTEIN
YANG PENUH BAHAYA
© Surya Institute, 2007

71. Tebak, mengapa tindakan mereka ini berbahaya?
© Surya Institute, 2007

72. © Surya Institute, 2007

73. © Surya Institute, 2007

74. 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

75. Mathematically, the
Kirchhoff’s ii law can be
expressed by following
equation :
 ΣVtertutup
=0
 ΣE +Σ(I.R) = 0

76. 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

78. 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


79. 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)


80. 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

81. 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

82. 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.

83. Measurement of Electric Quantities



Measurement of Current
Measurement of Voltage
Measurement of Resistance

84. 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

86. 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

87. 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:

88. 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

89. 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

91. 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

92. • 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.

93. 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

94. 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

96. 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= ?
:

97. 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.

98. 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.