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CHAPTER 1
INTRODUCTION TO
Prepared By
Sheilani binti Shaari
Mahizan binti Mustapha
Department of Electrical Engineering/PKB
INTRODUCTION TO
ELECTRIC CIRCUIT
Upon completion of this course, students
Course Learning Outcome (CLO)
Department of Electrical Engineering PKB
Upon completion of this course, students
should be able to:
Explain clearly basic electrical quantities,
types of electrical circuits, electrical power,
electrical energy and solve related
problems.
INTRODUCTION
ELECTRICAL QUANTITIES
ELECTRIC CIRCUIT
OHM’S LAW
CONTENT
Department of Electrical Engineering PKB
OHM’S LAW
ELECTRICAL POWER
ELECTRIC ENERGY
General Objective (GO)
Student should be able to:
1. Understand basic electrical quantities
LEARNING OUTCOMES
Department of Electrical Engineering PKB
quantities
INTRODUCTION
Electric is an energy which cannot
seen but can be felt and be used by
human on today and future.
Department of Electrical Engineering PKB
Electric energy can be created impact from action:
a) Friction
b) Heat
c) Electromagnetic field
Electric energy can be change to another some energy like:
a) Heat energy – Iron
INTRODUCTION
Department of Electrical Engineering PKB
a) Heat energy – Iron
b) Sound energy – Radio
c) Kinetic Energy – Motor
d) Light energy – Lamp
2 types of electric:
i) Static electricity
no electron movement in certain
INTRODUCTION
Department of Electrical Engineering PKB
no electron movement in certain direction.
ii) Dynamic electricity
electron movement in certain direction.
Item Symbol Unit
Electromotif Force E Volt (V)
Electrical Charge Q Coulomb (C)
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
Electrical Charge Q Coulomb (C)
Current I Ampiar (A)
Voltage V Volt (V)
Resistance R Ohm (Ω)
Resistivity ρ Rho (Ωm)
Electromotive Force (Emf)The force or electric pressure that causes the flow of electrons or current in the circuit
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
current in the circuit
Battery
Electrical Charges
The Positive and negative charges
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
charges
+ve -ve
Current
The movement of the electrical charge cause by free electron movement from
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
electron movement from positive to negative terminals.
R
V
-ve+ve
Voltage
The Potential difference between 2 points.
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
between 2 points.
+ve -ve
V
Resistance
The property of material which is opposition to current flow
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
RI
R
I
Resistance
4 factor that influence the value of resistance l
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
1. Length of conductor,
2. Surface area , A
3. Resistivity, ρ
4. Conductor Temperature, T
Resistance1. Length of conductor,
The length of conductor is proportional to the resistance. The longer the length of the wire, the higher the resistance value.
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
higher the resistance value.
2. Surface area , A
Area is inverse proportional to the resistance. As the resistance increase the cross section area of a conductor will decreases.
l∝R
AR
1∝
Resistance3. Resistivity
Resistivity is proportional to the resistances. Higher the resistance, higher the resistivity.
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
4. Conductor Temperature, T
The conductor temperature is proportional to the resistance. As the conductor temperature increase the value of resistance also increase.
ρ∝R
TR ∝
Resistivity, ρThe characteristic of conductive material to opposition or decrease the current flow
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
the current flow
R=ρℓ/A
R=ρℓ/A
Where;
A = ∏(r)2
A = ∏(d/2)2
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
Where;
A= cross section area (m)
ρ= Resistivity (Ωm)
ℓ = Length (m)
R= Resistance
1. Calculate the resistance of a
1.5m length of alumuinium wire. Given diameter wire 10mm and
R?Example
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
Given diameter wire 10mm and resistivity of alumuinium is 0.025µΩm
ℓ
dρ
Item Spesification Material
conductor Material that current is easily flow
Copper
iron
Semi Material that has Silicon
ELECTRICAL QUANTITIES
Department of Electrical Engineering PKB
Semi conductor
Material that has conductance value between conductor and insulator
Silicon
germinium
Insulator Material that does not allowed current flow in normal condition
Rubber, glass, air
INTRODUCTION
ELECTRICAL QUANTITIES
ELECTRIC CIRCUIT
OHM’S LAW
CONTENT
Department of Electrical Engineering PKB
OHM’S LAW
ELECTRICAL POWER
ELECTRIC ENERGY
General Objective (GO)
Student should be able to :
1. Explain types of electrical circuit
LEARNING OUTCOMES
Department of Electrical Engineering PKB
2. Understand Ohms Law
3. Understand Electrical Power
4. Understand Electrical Energy
Electric circuit is a combination or cable to bring current from voltage sources to electrical components (load)
ELECTRIC CIRCUIT
Department of Electrical Engineering PKB
(load)
It consists of two type;
i) Complete electric circuit
ii) Non Complete electric circuit
Complete electric circuit
Also called basic/
simple circuit
it is closed end
I
V
R
ELECTRIC CIRCUIT
Department of Electrical Engineering PKB
it is closed end connection that can make current through completely.
The circuit must have voltage supply (V), Electric current (I) and resistance (R).
Non Complete electric circuit
It is a circuit without one of three components either voltage sources or load resistance
ELECTRIC CIRCUIT
Department of Electrical Engineering PKB
or load resistance
The current flow will never happen with perfect in non complete circuit.
2 type of Non Complete electric circuit;
i) Open circuit
ELECTRIC CIRCUIT
Department of Electrical Engineering PKB
Open circuit
ii) Short circuit
Non Complete electric circuit
i. Open circuit
The load in that circuit will open
No current flow occur
ELECTRIC CIRCUIT
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No current flow occur
Resistance value is higher (∞)
Load (R)Pull outV
Non Complete electric circuit
ii.Short circuit
Connection at the load will short with a conductor which no resistance.
ELECTRIC CIRCUIT
Department of Electrical Engineering PKB
a conductor which no resistance.
The current which go through is high.
Fuse will burntV
R
Short withcable
Definition
The current flow through the electrical circuit is
OHM’S LAW
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electrical circuit is directly proportional to the potential difference across the circuit and inversely proportional to the resistance of the circuit.
l∝RI V
OHM’S LAW
The electric current (I) is :
directly proportional to the voltage
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AR
1∝
l∝RI V
IR
inversely proportional to the resistance.
OHM’S LAW
IRV =
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V = IR
Voltage (V) = Current (A) x Resistance ( Ω)
IRV =I = V
R
R = V
I
Linear Resistance
With Constant Value of Resistance:
Increasing Voltage will increasing
OHM’S LAW
Department of Electrical Engineering PKB
Increasing Voltage will increasing the current
l∝RV IR CONSTANT
Linear Resistance
Constant Resistance and temperatureV
OHM’S LAW
Department of Electrical Engineering PKB
temperatureV
I
R constant
Graph V vs I with constant R
Non Linear Resistance
Changing Value of R
Increasing Voltage will increasing
OHM’S LAW
Department of Electrical Engineering PKB
Increasing Voltage will increasing the current to a constant level
l∝RV IR CHANGING
Non Linear Resistance
Changing value of R
V
OHM’S LAW
Department of Electrical Engineering PKB
V
I
R changing
Graph V vs I with changing value of R
Constant level
increasing
Calculate the current flow in the circuit ifthe resistance is 10Ω and the supply
I?
R
Example
OHM’S LAW
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the resistance is 10Ω and the supplyvoltage is 15V.
Then, calculate the current value, if thevalue of resistance is change to 10 kΩ.
R
V
I?
R
ELECTRIC POWER
is a job that can be done in one time unit
IVP =
Department of Electrical Engineering PKB
Power (P) = Current (A) x Voltage (V)
Unit : Watt (W)
IVP =
Power (P) = Current 2 (A) x Resistance (R)
RIP2
=
ELECTRIC POWER
Department of Electrical Engineering PKB
R
VP
2
=
Power (P) = Voltage 2 (v) / Resistance (R)
WATT METER
used to measure the value of the power that has been used
ELECTRIC POWER
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W
Internal Connection
Meter Symbol
WATT METERThere are two coils in it:
The CURRENT COIL is connected in series with the circuit.
ELECTRIC POWER
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VS
LoadVOLTAGE COIL connected in parallel
with the circuit.
ELECTRIC ENERGY
Electrical energy is the product of power and time.
T = Pt
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T = PtEnergy (T) = Power (P) x Time (hour)
Unit : Kilowatt hour (kWh) or Joule (watt second)
T = VItEnergy (T) = Voltage (V) x Current (I) x Time (hour)
T = I2Rt
ELECTRIC ENERGY
Department of Electrical Engineering PKB
T = I2RtEnergy (T) = Current2 (I) x Resistance (R) x Time (hour)
T = (V2/R)tEnergy (T) = [Voltage2 (V) / Resistance (R) ] Time (hour)
is the energy absorbed to supply load of 1 KW for a period of 1 hour.
Work (w) = Power (P) x Time (h)
WORK
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Work (w) = Power (P) x Time (h)
Watt
is the power used when 1A current flow between two points with a 1 volt potential
Unit for work or energy is Joule.
1 Joule = 1 watt second
WORK
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1 Joule = 1 watt secondWork (J) = Power (P) x Time (s)
Notes: Convert hour to seconds
1 H = 1 x 60 minutes x 60 sec
= 3600 s
Example
A toaster used 5A current and 240 V
I = 5A
ELECTRIC ENERGY
Department of Electrical Engineering PKB
A toaster used 5A current and 240 V supply for 15 minutes. Calculate,
i. Power (P) usedii. Energy (T) which absorbed the
heat in kJ
v = 240vt = 15 min
INTRODUCTION
RESISTOR CIRCUIT ANALYSIS
KIRCHOFF’S LAW
CONTENT
Department of Electrical Engineering PKB
General Objective (GO)
Student should be able to:
1. Understand the characteristics of series and parallel circuits.
LEARNING OUTCOMES
Department of Electrical Engineering PKB
series and parallel circuits.
2. Explain Kirchhoff’s Law in complex electrical circuits.
Electrical circuit
Is a complete connection to the load by using a conductors
INTRODUCTION
Department of Electrical Engineering PKB
The load or resistor in the circuit can be connected in 3 different ways:
INTRODUCTION
Department of Electrical Engineering PKB
1. Series Circuit
2. Parallel Circuit
3. Combination of Series and Parallel Circuits
refer to the connection of the resistor in the circuit.
connected from end to end.
SERIES CIRCUIT
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VT
R1 R2 R3
V2 V3V1
Total Resistance RT
Is the sum of all the resistor that exist in the circuit
SERIES CIRCUIT
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RT = R1+R2+R3
Total Current ITThe current through each resistor is equal to the total current .
I1
SERIES CIRCUIT
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IT=I1=I2=I3
ITI2
I3
Total voltage
is the sum of all voltage drop (voltage down) on each of the resistance
SERIES CIRCUIT
Department of Electrical Engineering PKB
VT=V1+V2+V3V1
V2
V3
Voltage drop
is the reduction of the voltage supply in every resistance
V2 = I2 R2
SERIES CIRCUIT
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V1
V2
V3
V1 = I1 R1
V2 = I2 R2
V3 = I3R3
voltage divider law
is used to determine voltage value which across every resistance in series circuit
R
SERIES CIRCUIT
Department of Electrical Engineering PKB
VT
R1
R2
TVRR
RV )(
21
1
1+
=
TVRR
RV )(
21
2
2+
=
V1
V2
voltage divider law
V1R1
TVRRR
RV )(
321
11
++=
SERIES CIRCUIT
Department of Electrical Engineering PKB
VT
V2R2
R3 V3
TVRRR
RV )(
321
22
++=
TVRRR
RV )(
321
3
3++
=
Example
Based on the circuit below. Calculate:
a. Total resistance, RT
b. Total Current, IT
SERIES CIRCUIT
Department of Electrical Engineering PKB
b. Total Current, ITc. Voltage drop across each resistor by
using:
i. Ohm’s Law
ii. Voltage Divider Law
VT = 120 v
R1 = 15Ω
R2 = 10Ω
Solution
a. Total resistance RT = R1 + R2
RT = 15 + 10
RT = 25 Ω
R1 = 15Ω
SERIES CIRCUIT
Department of Electrical Engineering PKB
RT = 25 Ω
b. Total Current IT = VT / RT
IT = 120/25
IT = 4.8 A
Notes: Series Circuit : IT = I1 = I2
VT = 120 vR2 = 10Ω
Solution
c. Voltage drop across each resistor by using Ohm’s Law
V1 = I1 R1
= 4.8 x 15
SERIES CIRCUIT
Department of Electrical Engineering PKB
Notes: Series Circuit : V1 + V2 = VT
VT = 120 v
R1 = 15Ω
R2 = 10Ω
V1
V2
1 1 1
= 4.8 x 15= 72 v
V2 = I2 R2
= 4.8 x 10= 48 v
Solution
c. Voltage drop across each resistor by using Voltage Divider Law
R = 15Ω R=
SERIES CIRCUIT
Department of Electrical Engineering PKB
VT = 120 v
R1 = 15Ω
R2 = 10Ω
V1
V2
= 72 v
= 48 v
TVRR
RV )(
21
11
+=
TVRR
RV )(
21
22
+=
120)1015
10(2
+=V
120)1015
15(1
+=V
Notes:
The answer for V1 and V2 is same for both Laws
connection of resistors which is against between each other
PARALLEL CIRCUIT
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VTR1 R2
Total resistance
PARALLEL CIRCUIT
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VTR1 R2
R3
321
1111
RRRRT
++=
Total voltage
voltage across each parallel resistor is equal to the voltage source
PARALLEL CIRCUIT
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VT V2 V3
321 VVVVT ===
V1 V2 V3
Total current
the summation of all current flow in each branch is equal to the current source.
I I
PARALLEL CIRCUIT
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VT
I1 I2 I3IT
321 IIII T ++=
Current divider law
is used to determine the value of the current in each branch
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
Ij
R1 R2V1
V2
I1I2
VT
TIRR
RI )(
21
2
1+
=
jIRR
RI )(
21
12
+=
Example
Based on the circuit below. Calculate:
1. Total resistance, RT
2. Voltage drop across each resistor
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
2. Voltage drop across each resistor
3. Current I1 and I2 by using:
Ohm’s Law
Current Divider Law
Vj = 240v
IT
I1 I2
R1 = 2Ω R2 = 4Ω
Solution
1. Total resistance
1/RT = 1/R1 + 1/R2
1/RT = 1/2 + ¼
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
1/RT = 1/2 + ¼
1/RT = ¾
RT = 4/3
RT = 1.33 Ω
VT = 240v
IT
I1 I2
R1 = 2Ω R2 = 4Ω
Solution
2. Voltage drop across each resistor
IT
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
VT= 240v
IT
R1 = 2Ω R2 = 4ΩV1 V2
Parallel circuit:
V1 = V2 = VT = 240 v
Solution
3. Current by using Ohm’s Law
I1 = V1 /R1
= 240/2
IT = I1 + I2= 120 + 60
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
Vj = 240v
IT
I1I2
R1 = 2Ω R2 = 4Ω
1 1 1
= 240/2= 120 A
I2 = V2 /R2
= 240/4= 60 A
= 120 + 60= 180 A
Solution
3. Current by using Current Divider Law
IR
I )( 2=
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
VT= 240v
IT
R1 = 2Ω R2 = 4Ω
I1
= 120 A
TIRR
I )(21
2
1+
=
180)42
4(1
+=I
Solution
3. Current by using Current Divider Law
IR
I )( 1=
PARALLEL CIRCUIT
Department of Electrical Engineering PKB
VT= 240v
IT
R1 = 2Ω R2 = 4Ω
I2
= 60 A
TIRR
RI )(
21
12
+=
180)42
2(2
+=I
Notes:
The answer for I1 and I2 is same for both Laws
Most of the electric circuit are the combination of series and parallel circuits.
COMBINATION CIRCUIT
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R1 R2
R3
IT
I1
I2I3
+ -
VT
to solve problems in calculating the value of total resistance, total voltage and total current for combinations circuit have to used both formulas in series and parallel circuit respectively.
COMBINATION CIRCUIT
Department of Electrical Engineering PKB
parallel circuit respectively.
Example
Based on the circuit below. Calculate:
1. Total resistance, RT
2. Current IT, I1 and I2
COMBINATION CIRCUIT
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2. Current IT, I1 and I23. Voltage drop across
each resistor
R1 =10Ω R2 =20Ω
R3 15Ω
IT
I1
I2I3
+ -
VT =120v
used to solve complex electrical circuit which have 2 or more supply
There are 2 types of Kirchoff’s law;
KIRCHOFF’S LAW
Department of Electrical Engineering PKB
There are 2 types of Kirchoff’s law;
a) Kirchoff’s current law
b) Kirchoff’s voltage law
Kirchoff’s current law(first order of Kirchoff’s law)
sum of the current into a node is equal to the sum of current out of the node
algebraic sum of all the current entering and leaving a
KIRCHOFF’S LAW
Department of Electrical Engineering PKB
algebraic sum of all the current entering and leaving a node is equal.
I1321 III +=I3
I2
NODE
Kirchoff’s voltage law(second order of kirchoff’s law)
sum of the voltage drop and voltage source around the closed path is equal to zero
sum of voltage drop must equal to voltage source
KIRCHOFF’S LAW
Department of Electrical Engineering PKB
sum of voltage drop must equal to voltage source
321 VVVVs ++=Vs
V1
V2
V3
Example
By using Kirchoff’s Law, calculate the current at each branch
KIRCHOFF’S LAW
IKirchoff’s Current Law
Department of Electrical Engineering PKB
5v10v
R1 = 1Ω R2 = 6Ω R3 = 2Ω
I1
I3
I2
Kirchoff’s Current Law
Kirchoff’s Voltage Law
C1
Kirchoff’s Voltage Law
C2