JABATAN KEJURUTERAAN ELEKTRIK POLITEKNIK KOTA BHARU KM 24 KOK LANAS 16450 KETEREH KELANTAN ET 1 O 1 ELECTRICAL TECHNOLOGY
INTRODUCTION � ET 101 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.
CHAPTER 1. 0: UNIT ASSOCIATED WITH BASIC ELECTRICAL QUANTITIES
INTRODUCTION �The system of units used in enggineering 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
� The basic units in the SI system are listed with their symbols in Table 1. 1
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 ; �Express 2. 1 V in millivolts (m. V). �Convert 4500 microvolts (μV) to milivolts (m. V).
Exercise 1) Convert : a) 356 m. V to volts (V). b) 500 000 Ω to megaohms (M Ω). c) 20 000 picofarads (p. F) into farads (F). d) 47 000 picofarads (p. F) to microfarad (μF). e) 1800 kiloohms (kΩ) to megaohms (MΩ).
�Answer ; �a) 0. 356 V �b) 0. 5 M �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 = 2 x 60 = 120 sec �Q = (5)x(120) = 600 coulombs � Example 2; �A mass of 5000 g is accelerated at 2 m/s 2 by a force. Determine the force needed.
�F = m x a = 5 kg x 2 � = 10 N � �Example 3 �A mass of 1000 kg 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 s = (m x a) x s � = (1000 kg x 9. 81 m/s 2) 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 = amperes ampere seconds joules coulombs
continued �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 m 2
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 1018 electron pass in one second. Current, I = Charge (ф) = coulomb Time (t) seconds I = dq dt
d) Conductance �The reciprocal 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 • 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 3 �Find the resistance of a 100 ft length of copper wire with a cross- sectional area of 810. 1 CM. The resistivity of copper is 10. 37 CM –Ω/ft at 20ºC. R= ρl A = (10. 37 CM-Ω/ft) (100 ft) 810. 1 CM = 1. 280 Ω
�Example 1 �Calculate the resistance of a 2 km length of aluminium overhead power cable if the cross sectional area of the cable is 100 mm 2. Take the resistivity of aluminium to be 0. 03 x 10 -6 Ωm. (answer = 0. 6Ω)
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 W = work done or energy transferred in joules t t = time in seconds. or P = Ix. V I = a direct current of I amperes is flowing in a electric circuit (A). V = voltage across the circuits (volts)
Example 2 �An amount of energy equal to 100 J is used in 5 s. What is the power in watts? P = Energy = W = 100 J = 20 W Time t 5 s
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
�Determine the number of kilowatt-hours (k. Wh) for each of the following energy consumption: a) 2500 W for 2 hour. 2500 W = 2. 5 k. W W = Pt = (2. 5 k. W)(2 h) = 5 k. Wh b) 100, 000 W for 5 hour. 100, 000 W = 100 k. W W = Pt = (100 k. W)(5 h) = 500 k. Wh
Exercise �If you use 100 W of power for 10 hour, how much energy (in kilowatt-hours) have you used?