INTRODUCTION TO ELECTRIC CIRCUIT

DET10013 - ELECTRICAL TECHNOLOGY

Chapter 1:

Introduction to
Electric Circuit

COURSE LEARNING OUTCOME

1. Apply the concept and principles of the related
electrical circuit theorems and law to solve DC
electrical circuit using various method and
approach ( C3 , PLO 1 )

2. Construct DC circuit and measure related
electrical parameters using appropriate electrical
equipments ( P4 , PLO 5 )

3. Demonstrate ability to work in team to complete
assigned tasks within the stipulated time frame (
A3 , PLO 9 )

TOPIC TITLE

CHAPTER TITLE

1.0 INTRODUCTION TO ELECTRIC
CIRCUIT
2.0
DC EQUIVALENT CIRCUIT AND
3.0 NETWORK THEOREMS
4.0
5.0 CAPACITORS AND CAPACITANCE

INDUCTORS AND INDUCTANCE

MAGNETIC CIRCUIT,
ELECTROMAGNETISM AND
ELECTROMAGNETIC INDUCTION

LEARNING OUTCOME (2 Hours)

1.1 Remember standard symbols for electrical components.
1.1.1 Identify common symbols in electrical circuit diagrams.
1.2 Apply the general features of cells and batteries.
1.2.1 Show the differences between cells and batteries.
1.2.2 Show the effects of different cell connections:

a. series
b. parallel
c. series-parallel
1.2.3 Calculate the total voltage of series sources
a. with the same polarities.
b. with opposite polarities.

Standard Symbol for Electrical Components

conductor /wire switch ground/earth

Cell (dc supply) Battery (dc supply) AC supply

resistor inductor capacitor
G V A

galvanometer voltmeter ammeter

Cell

• A single unit of a primary or secondary battery
that converts chemical energy into electric
energy.

Battery

• A battery is a series of two or more connected
cells, which changes chemical energy into
electrical energy.

Relationship of Cells & Batteries

• A Battery is a combination of cells
• Cell combination could be in

SERIES, PARALLEL & SERIES-
PARALLEL
• Practically, a cell is also notified
as a battery.

Series Connection Cells

Series Connection Cells

Example 1.1
Calculate total e.m.f. of the circuit below

Total e.m.f., ET = E1 + E2 + E3 + E4
= 2.0 + 2.0 + 2.0 + 2.0
= 8V

Parallel Connection Cells

Parallel Connection Cells

Example 1.2
Calculate total e.m.f. of the circuit below

Total e.m.f., ET = E1 = E2 = E3 = 2.0V

Series-Parallel Connection Cells

Series-Parallel Connection Cells

Example 1.3
Calculate total e.m.f. of the circuit below

Total e.m.f. for series cells, ESeries = E1 + E2 + E3 + E4
= 2.0 + 2.0 + 2.0 + 2.0
= 8V

Total e.m.f., ET = ESeries = 8V

Series Connection with same Polarities

Example 1.4
Calculate total e.m.f. of the circuit below

Total e.m.f., ET = E1 + E2
=8+6
= 14V

Series Connection with opposite Polarities

Example 1.5
Calculate total e.m.f. of the circuit below

Total e.m.f., ET = E1 + E2
=8-6
= 2V

SELF-EXERCISE

QUESTION: Calculate total e.m.f. of each cells connection as
follow.

i) 5V 11V 4V
BA

ii) 44V A AnAsNwSeWr:E2R0V
B 44V ANSWER
44V
Answer: 44V

SELF-EXERCISE

QUESTION: Calculate total e.m.f. of each cells connection as
follow.

iii) 4V 3V

B 2V 5V A

AnsAwNeSrW: 7ERV

6V 1V

iv) 4V
4V 4V A

B

AnsAwNeSrW: 1E2R0V

30 cells

SELF-EXERCISE

QUESTION: Calculate total e.m.f. of each cells connection as
follow.

v) 14V A 20 cells
B 14V

14V AnAsNwSeWr:E1R4V

5V 5V 5V ANSWER
Answer: 50V
vi) 5V 5V A
B 5V 5V

5V 5V

10 cells

LEARNING OUTCOME (1 Hour)

1.3 Remember electric current and quantity of electricity.
1.3.1 State the definition of electric current.
1.3.2 State the unit of charge.
1.3.3 Calculate charge or quantity of electricity Q from Q=It.

1.4 Remember the main effects of electric current.
1.4.1 Identify the three main effects of electric current, giving
practical examples of each.

1.5 Apply resistance and resistivity
1.5.1 Explain that electrical resistance depends on four
factors.Calculate that resistance, =


where ρ is the resistivity.

Electric current, I

• Current: - motion of charge

- depends on the rate of flow of charge

- electric fluid

- unit of current is ampere (A)

• Equation: dq = changing of charge

I = dt = changing of time
I = current (ampere)
Q = charge (coulomb)
• For steady state condition:
t = time (second)
I = ( ℎ ) , thus Q = It
( )

Electric current, I

Example 1.6
If a current of 5 A flows for 2 minutes, find the
charge transferred.

Q = It = 5 x 2 x 60 = 600 C

Main Effect of Electric Circuit

1. Heat Effect - Example: soldering iron, water
heater, fuse, bulb, cookers, electric fires,
furnaces, kettles, iron

2. Magnetic Effect - Example: bells, relays, motors,
generators, transformers, telephones, lifting
magnets, car ignition

3. Chemical Effect - Example: cell and battery,
electroplating

Resistance & resistivity

• Resistance – property of a component which
restricts the flow of electric current.

• The value of resistance depends upon 4
factors:
1. Length, l
2. Cross-sectional area, A
3. resistivity, ρ
4. temparature

Resistance & resistivity

• Equation:

R = ρl [Unit = Ω]



R = resistance [Ω]
l = Length [m]
A = Cross-sectional area [m2]

ρ = resistivity [Ω.m]

• Resistivity is difference for different material

Resistance & resistivity

Example 1.7
Calculate resistance of a 5m long conductor if it
has cross sectional area 10 2 and resistivity
0.3 10−5 Ω.m

0.3 x 10−5 x 5
Resistance, R= A = 10 x 10−6

= 1.5Ω

Resistor (R)

• A device that is manufactured to have specific
resistance.

• Used to limit current flow and reduce voltage
applied to other components.

• Basic unit is ohm (Ω)

Resistor (R)

• Different examples of resistors

SELF-EXERCISE

i) In what time would a current of 1 A transfer a
charge of 30 C?

AnsAwNSeWr:ER30s

ii) What would be the resistivity of 2m length
conductor wire if the resistance value is 500Ω
and the cross sectional area 0.5 2

AnsweArN:SW12ER5µΩm

LEARNING OUTCOME (1 Hour)

1.6 Understand Ohm’s Law.
1.6.1 Explain Ohm’s Law.
1.6.2 Outline the procedure adopted when using Ohm’s Law

1.7 Apply Ohm’s Law in circuit.
1.7.1 Construct circuit to explain Ohm’s Law.
1.7.2 Use Ohm’s Law to find current, voltage and resistance in
a circuit.

.

Ohm’s Law

• Ohm’s Law states that the current (I) through a
conductor between two points is directly
proportional to the potential difference or
voltage (V) across the two points, and inversely
proportional to the resistance (R) between them.

I =


Ohm’s Law Triangle

V = IR

V I =


IR R =


Simple Circuit

E = E.M.F. (Electromotive force)

- Generates from voltage
source

- Example: cells / batteries

From Ohm’s Law:

Current =


I =


Simple Circuit

V drop = Voltage drop
---------------------------
- appears when current, I flows

through resistor,R.
- Inverse polarity from E

From Ohm’s Law:

Voltage = Current x Resistance

V drop = IR

Simple Circuit

I R • A complete circuit should consist of at
least 1 electricity source (battery) and 1
+ Load load (resistor)

E • A battery possess e.m.f. that produces
DC current.
-
• Current will only produce when the
Source source (battery) is connected to the load
(resistor) in close loop connection.

Simple Circuit

I + • A complete circuit should consist of at
least 1 electricity source (battery) and 1
+ R Vd load (resistor)

E • A battery possess e.m.f. that produces
DC current.
-
- • Current will only produce when the
Source Load source (battery) is connected to the load
(resistor) in close loop connection.

• When current flows across resistor, R,
voltage drop, Vd will be produced
across R

Simple Circuit

I + • A complete circuit should consist of at
least 1 electricity source (battery) and 1
+ R Vd load (resistor)

E • A battery possess e.m.f. that produces
DC current.
-
- • Current will only produce when the
Source Load source (battery) is connected to the load
(resistor) in close loop connection.

• When current flows across resistor, R,
voltage drop, Vd will be produced
across R

Simple Circuit (Example)

Example 1.8
QUESTION: By referring to the circuit below,
calculate:
i) Current, I
ii) Voltage drop across resistor 10Ω, Vdrop

I

+ +

15V 10Ω Vdrop

--

Simple Circuit (Example)

I

+ +

15V 10Ω Vdrop

--

15

i) Current, I = = 10 = 1.5A

ii) Voltage drop, = IR = 1.5 x 10 = 15V

SELF-EXERCISE

A 100 V battery is connected across a resistor and
causes a current of 5 mA to flow. Determine the
resistance of the resistor. If the voltage is now
reduced to 25 V, what will be the new value of the
current flowing?

RA=NSW20ERkΩ

I = A1N.S2W5EmR A

LEARNING OUTCOME (2 Hours)

1.8 Understand series, parallel and series-parallel connections.
1.8.1 Identify a series circuit.
1.8.2 Explain the flow of current and voltage division in
the series circuit.
1.8.3 Identify a parallel circuit.
1.8.4 Explain the voltage drop and the current division in
the parallel circuit.
1.8.5 Explain the equivalent resistance in series and parallel
circuits.
1.8.6 Identify a combination of series and parallel circuit.
1.8.7 Explain the total resistance for the combination of
series and parallel circuit.

LEARNING OUTCOME (2 Hours)

1.9 Apply series, parallel and series-parallel connections to dc circuit.
1.9.1 Construct a series connection circuit
1.9.2 Calculate the flow of current and voltage division in the series
circuit.
1.9.3 Construct a parallel circuit.
1.9.4 Calculate the voltage drop and the current division in the
parallel circuit.
1.9.5 Construct a series-parallel connection circuit.
1.9.6 Calculate the equivalent resistance in series and parallel
circuits.
1.9.7 Calculate the total resistance for the combination of series and
parallel circuit.
1.9.8 Use of voltage divider in series circuit and use of current divider
in parallel circuit.
1.9.9 Solve problems related to series, parallel and combination of
series and parallel circuits.

Series Circuit

• Is formed when any number of devices are
connected end-to-end so that there is only
one path for current to flow.

Series Circuit Characteristics

Series Circuit Characteristics

1. Resistances are additive
RT = R1 + R2 + R3

2. The current flows throughout the circuit is same.
I = IR1 = IR2 = IR3

3. Different resistors have their individual voltage drop
VR1 ≠ VR2 ≠ VR3

4. Total e.m.f equals to the sum of voltage drops across each
resistor

E = VR1 + VR2 + VR3

Equivalent resistance in series

RT = R1 + R2 + R3 Equ.
1

• Applicable to any means of resistors.
• Standard equation of series connection

resistors.

Equivalent resistance in series
(resistors with same value)

RT = r x n Equ. r = resistance value
2 n = amount of resistors

• Applicable for any means of resistors with same
value.

Voltage Divider Rule

VR1 = R1+ R1 R3 xE
R2 +

Series Circuit (Example)

Example 1.9
By referring to the circuit above, calculate:
i) Total resistance of the circuit,
ii) Current, I
iii) Voltage drop across resistor 6Ω, 2

4Ω 6Ω

15V 8Ω

Series Circuit (Example)

4Ω 6Ω i) Rtotal = 4 + 6 + 8
= 18Ω

15V 8Ω

ii) I = = 15 = 0.833A
18

iii) VR2 = IR2 = 0.833 x 6 = 5V use VDR

or

VR2 = R1+ R2 R3 xE = 6 x 15 = 5V
R2 + 4+ 6 + 8