JA303 Chapter 1

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JA303 AUTOMOTIVE ELECTRICAL CHAPTER 1

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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

JA303 Chapter 1

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