ET1O1 ELECTRICAL
TECHNOLOGY
INTRODUCTION
 ET101 Electrical Technology course introduces students to the
principles of Operation of DC electrical circuit. It covers the
fundamental laws, theorems and circuit techniques. It also cover
cell and batteries, magnetic and electromagnetic circuit.
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COURSE LEARNING OUTCOMES (CLO)
Upon completion of this course, students should be able to:
1. Explain DC circuit concepts and their analyses with different method
and approach. (C2)
2. Apply the knowledge of DC circuit laws and theorems in solving
problem related to DC electrical circuit. (C3)
3. Use appropriate components and measuring equipment to perform
related DC electrical circuit laboratory exercises. (P4)
4. Demonstrate team working efficiency while doing practical work. (A3
CHAPTER 1.0
UNIT ASSOCIATED WITH BASIC
ELECTRICAL QUANTITIES
INTRODUCTION
 The system of units used in engineering
and science is the International system
of units usually abbreviated to SI units.,
and is based on the metric system.
 This was introduced in 1960 and is now
adopted by majority of countries as the
official system of measurement
1.2:
Prefixes to Signify Powers of 10.
- SI units may be made larger or smaller by
using prefixes which denote multiplication
or division by a particular amount.
- The prefixes power of ten are listed as in
the Table.
Examples of Final Examination
Questions ;
1.1/ 1000 in power of ten is
2. Convert 5mA to microamperes (µA)
3.Convert 10 kΩ to megaohms (MΩ)
4.Give the power of ten for micro
Examples ;
 Express 2.1 V in
millivolts (mV).
 Convert 4500 microvolts
(μV) to milivolts (mV).
Exercise
1)
Convert :
a) 356 mV to volts (V).
b) 500 000 Ω to megaohms (M Ω).
c) 20 000 000 picofarads (pF) into farads (F).
d) 47 000 picofarads (pF) to microfarad (μF).
e) 1800 kiloohms (kΩ) to megaohms (MΩ).
Answer ;
a) 0.356 V
b) 0.5M
c) 0.00002 F
d) 0.047 µF
e) 1.8 MΩ
Exercises
 Example 1 ;
 If a current of 5 A flows for 2 minutes , find the
quantity of electricity transferred.
 Q = I x t where I = 5 amp, t = 2x60 = 120 sec
 Q = (5)x(120) = 600 coulombs
 Example 2;
 A mass of 5000 g is accelerated at 2 m/s2 by a
force. Determine the force needed.
F=mxa
= 5 kg x 2
= 10 N
Example 3
 A mass of 1000kg is raised through a height of 10
m in 20 s. What is (a) the work done and (b) the
power developed.
 (a) W = F x d

= (m x a) x d

= (1000 kg x 9.81 m/s2) x 10

= 98100 Nm or J
 (b) P = W/t W = work done, t = time in second

=98100/20 = 4905 watt
1.3 Units and symbol of electrical
potential
a) Electrical potential and e.m.f

The units of electrical potential is the volt
(V) where one volt is one joule per coulomb.
One volt is defined as the differences in
potential between two points in a conductor
which, when carrying a current of one
ampere, dissipates a power of one watt, i.e
Volts = watts = joules/second = joules
=
ampere amperes ampere seconds
joules
coulombs
A change in electric potential
between two points in an electric
circuit is called a potential
difference.
The electromotive force (e.m.f)
provided by a source of energy
such as a battery or a generator
is measured in volts.
b) Resistance
The property of a material by which it
opposes the flow of current through it, is
called resistance.
R= ρl
A
ρ = specific resistance in ohms meter
(resistivity)
l = length of the conductor in m
A = area of cross section of the
conductor in m2
c) Electric Current )(I)
The flow of free electrons or the charge, in a
conductor is called as electric current .The unit
of current is ampere (A). One ampere of current
is said to flow at 6.24 x 1018electron pass in one
second.
Current, I = Charge (ф) = coulomb
Time (t)
seconds
I = dq
dt
d) Conductance
 Conductor : A material or element that allows
free movement of electrons and therefore allows
easy flow of electricity. Most conductors are
metals.
 The reciprocal (salingan) of resistance of
conductor is called its conductance (G). If a
conductor has resistance R, then its
conductance G is given by :

G = 1/R (Siemen)
• The SI unit of conductance is mho (i.e., ohm spelt
backward). These days, it is usual practice to use
Siemen as the unit of conductance. It is denoted
by the symbol S.
Example 1
 Find the resistance of a 100 ft length of copper
wire with a cross- sectional area of 810.1CM.
The resistivity of copper is 10.37CM –Ω/ft at
20ºC.
R= ρl
A
= (10.37CM-Ω/ft) (100 ft)
810.1 CM
= 1.280 Ω
 Example 2
 Calculate the resistance of a 2km length
of aluminium overhead power cable if the
cross sectional area of the cable is 100
mm2. Take the resistivity of aluminium to
be 0.03 x 10-6 Ωm. (answer = 0.6Ω)
Solution ;
 Given l = 2km = 2000m
 A = 100 mm2 = 100 / 1000000 = 1 x 10-4
  = 0.03 x 10-6 Ωm
 R = l / A

= (0.03 x 10-6 ) x (2000) / 1 x 10-4
 = 0.6 Ω
Go to video 1
Go to Video 2
e) Power , P
Power is defined as the rate of doing work or
transferring energy. The unit of power is the watt
(W)
One watt is one joule per second.
Power, P = W
in
t
or P = I x V
W = work done or energy transferred
joules
t = time in seconds.
I = a direct current of I amperes is
flowing in a electric circuit (A).
V = voltage across the circuits (volts)
Example 1
 An amount of energy equal to 100 J is used
in 5 s. What is the power in watts?
P = Energy = W = 100 J = 20W
Time
t
5s
f) Energy, W
Energy is work done or energy transferred
in Joules. The unit of energy is Joules
Energy, W = P x t
P = Power
t = time in seconds
Example 1;
 Determine the number of kilowatt-hours (kWh)
for each of the following energy consumption:
a) 2500 W for 2 hour.
2500W = 2.5kW
W = Pt = (2.5kW)(2h) = 5kWh
b) 100, 000 W for 5 hour.
100, 000W = 100kW
W = Pt = (100kW)(5h) = 500kWh
Take Home Exercise
 If you use 100 W of power for 10hour,
how much energy (in kilowatt-hours)
have you used?