ja303 chapter 1
TRANSCRIPT
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INTRODUCTION TO ELECTRIC CIRCUIT
1.1DEFINITION OF ELECTRIC QUANTITIES
a) Electromotive Force (EMF)
Force or electric pressure causes the flow of electric charge. Source
of electric energy is battery and power plant.
Symbol : E
Unit : Volt (V)
b) Electric charge
Consist of positive and negative charge. Charge quantity is
Coulomb.Symbol : Q
Unit : Coulomb(C)
c) Current
The movement of electric charge causes by the movement of free
electrons. Current moves from positive to negative terminal
Symbol : I
Unit : Ampere (A)
d) Voltage
The potential difference between two points in an electric circuit.
Symbol : V
Unit : Volt (V)
e) Resistance
The property opposes the movement of current.
Symbol : R
Unit : Ohm ()
1.1.1 Factors that affect the resistance of conductor material:
a) Material (Resistivity)
It is a conductor where it opposes or reduces flow of current
through it.
Symbol : (rho)
Unit : Ohm meter (m)
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Example 1.1
Calculate resistance of aluminium cable which has length of 1.5 km. Given that the diameter for the
wire is 10 mm and resistivity is 0.025 .m.
Solution:
Given, mxd 31010 , mx 3105.1 , mx 610025.0
Equation,A
R
,
where262
32 1054.78)
2
1010()
2( mx
xdA
477.0
1054.78
)105.1)(10025.0(6
36
x
xxR
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1.2TYPES OF ELECTRICAL CIRCUIT
Electric circuit is an arrangement of conductor or cable for current flow from
voltage source into electric components. There are 4 types of electric circuit which
are simple circuit, complex circuit, open circuit and short circuit.
a) Simple Circuit
It is also known as basic circuit as shown in figure below. It is close circuit
which allows flow of current perfectly from source to other components and
back to the source. The circuit consists of voltage supply (V), electric current
(I) and resistance (R).
Figure 1 Simple Circuit
b) Complex CircuitA complex set of electronic components and their interconnections that are
etched or imprinted onto a tiny slice of semiconducting material.
Figure 1.2 Complex circuit
R
I
V
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c) Open Circuit
The electric circuit is no source of burden results to no current flow through
the circuit.
Figure 1.3 Open circuit
d) Short Circuit
The connection is short by a conductor with no burden and has no resistance
value. The current flow is big. Commonly, when short circuit occurs, the fuse
will burn.
Figure 1.4 Short circuit
V
V
Short by a cable
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Measurement of Voltage, Current and Resistance.
a) Voltmeter
Voltmeter is used to measure the potential different or voltage (V) in a circuit. Placeone pole of voltmeter at one end of resistor and another pole at another end as shown
below.
Figure 1.5 Voltmeter
b) Ammeter
Ammeter is used to measure current (I) in an electric circuit. The ammeter must be
connected in circuit in order to get the reading from ammeter as shown in diagram
below.
Figure 1.6 Ammeter
V
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c) Ohmmeter
Ohmmeter is used to measure the resistance in a circuit. Place ohmmeter according to
the diagram below.
Figure 1.7 Ohmmeter
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1.3 OHMS LAW
Ohms Law states that current in complete circuit is directly proportional to pressure or
voltage but inversely proportional to resistance. If the resistance value is remained but the
voltage increase, the current also will increase. Equation for Ohms Law is :
Where, I = Current (A)
V = Voltage (V)R = Resistance( )
Linear and Non Linear ResistanceFrom the exaperiment, relationship between current and voltage is explain in the graph below
where resistance and temperature is keep constant.
V (volt)
R (pemalar)
I (Ampere)
Figure 1.8: V vs I Graph (R constant)
From experiment with vary of resistance, graph is shown below
V
I
Figure 1.9: V vs I Graph (R not constant)
IRV
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Example 1.2
Calculate current when resistance is 10 dan voltage supply is 15V. Then, calculate current when
resistance change to 10 k.
Penyelesaian :
Given , V= 15V
i). For R = 10,
Ohms Law, V= IR
A
R
VI 5.1
10
15
ii). For R = 10k ,
mAAxxR
VI 5.1105.1
1010
15 33
Practice makes perfect.
Do many exercise to keep
you master the topic
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1.4 ELECTRIC POWER
Electric power is a work done in a time. Equation below shows the relation between
powe, current and voltage.
Symbol: PUnit : Watt (W)
From Ohms law, IRV danR
VI ,
RIP2
R
VP
2
IVP
Where P = Power (W),
I = Current (A)
R = Resistance () danV = Voltage (V)
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THE USE OF WATT METER
Watt meter is used to measure power consumption. There has two coils inside it
where voltage coil is connected in parallel and current coil is conneted in series for an
electric circuit. Watt meter symbol and connectiod are shown below.
Figure 1.10 : Watt meter symbol
Gelung arus
Gelung voltan
Beban (R)
Figure 1.11 : Watt meter connection
W
VS
Info
Electric power converts to horse power:1 horse power = 746 watt
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Example 1.3
A toaster has current of 5A and home voltage supply is 240V. Calculate the power consumptionneeded for the toaster.
Given I = 5 A , V = 240V
Solution
WIVP 1200)240)(5(
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1.5A SERIES CIRCUIT
It is called as series cicuit because of the resistor connection in the circuit. Series
connection is a connection of resistor is in line from end to end as shown in diagram
below.
IT
Figure 1.12 Series Circuit Connection
From the arrangement of the circuit above, equations are derived with involvement of
total resistance, current and total volatge.
Total Resistance, RTTotal resistance is the sum of all resistors in the series circuit (1.5a).
(1.5a)
Total Current, ITCurrent through in every resistor is same as given in equation (1.5b).
(1.5b)
nT RRRRR .....321
nT IIIII ......221
nj IIIII ......221
R1 R3
Rn
R2
+ V1 - + V2 - + V3 -
+
Vn
-
VT
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Total Voltage, VTTotal voltage is the sum of all voltage reduction on every resistor as described in
equation (1.5c).
(1.5c)
While the reduction in every resistor can be calculated using Ohms Law as we
mentioned before. Equation (1.5d) below shows the method to calculate the voltage
reduction..
nTn
T
T
T
RIV
RIV
RIVRIV
33
22
11
Voltage Divider Rule
We can use volatge divider rule in order to calculate the value of individual voltage
through every resistor in series circuit as described in equation(1.5e) and (1.5f).
i). For circuit has 3 resistors :
TV
RRR
RV )(
321
11
TV
RRR
RV )(
321
22
(1.5e)
TV
RRR
RV )(
321
33
ii). For circuit has 2 resistors:
TV
RR
RV )(
21
11
TV
RR
RV )(
21
22
nT VVVVV ......321
(1.5d)
(1.5f)
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Exercise
From the diagram, determine ;
i). Total resistance, RT
ii). Circuit current, IT
iii). Voltage reduction in every resistor.
Solution
i). Total resistance, RT
RT = R1 + R2 = (15 + 10) = 25
ii). Circuit current, IT
IT =T
R
V=
25
120= 4.8 A
iii) Voltage reduction in every resistor
VR1 = ITR1 = (4.8)(15) = 72 V
VR2 = ITR2 = (4.8)(10) = 48 V
R1=15
R2=10
V=120V
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1.5 B PARALLEL CIRCUIT
Parallel connection is shown in figure below.
Figure 1.13 Parallel Circuit
Total Voltage, VT
Voltage across every junction is the same value of voltage supply as given in
equation(1.5g);
(2.7)
Total Current, IT
Total current is value of current supply or the summation of junction currents (1.5h);
nT VVVVV ......221
nj IIIII ......221
nT IIIII .....321 (1.5h)
VT
IT
I1 I2 I3
R1 R2 R3V1 V2 V3
(1.5g)
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Total Resistance, RT
In parallel circuit, resistance is calculated using equation (1.5i);
For a circuit has 3 resistors;
321
1111
RRRRT
atau
313221
321
RRRRRR
RRRR
T
Current Divider Rule
We can use Current Divider Rule for determine current value in every junction in
circuit (1.5j).
i). For circuit has 2 resistors
Figure 1.14
TI
RR
RI )(
21
21
TI
RR
RI )(
21
12
(1.5i)
(1.5j)
VT
IT
I1 I2
R1 V1 R2 V2