Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
CHAPTER ONE
BASIC DEFINITIONS AND CIRCUIT LAWS
1.0 Introduction
This chapter deals with the basic definitions, concepts and laws that are associated with (electric) circuit. The
specific laws under consideration are: Ohm’s law, Kirchhoff’s current and voltage laws (KCL and KVL), and
the application of these laws.
1.1 Circuit Elements
An element is a basic building block of a circuit. Thus the circuit is an interconnection of the various
elements. It is a complete conducting path starting at one end of a source and ending at its other end.
Determining the currents through (or the voltages across) the elements of the circuit is termed as circuit
analysis. Generally, two types of elements are found in the electric circuits. They are either active or passive.
1.1.1 Active Elements
These elements also called sources are capable of generating electrical energy. In other words, they are acting
as sources of energy in the circuit. Examples are cells, solar modules, batteries and generators.
Electric Current
It may be defined as the time rate of net motion of an electric charge across a cross-sectional boundary. Thus,
a random motion of electrons in a metal does not constitute a current unless there is a net transfer of charge
with time.
𝑖=
𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐ℎ𝑎𝑟𝑔𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑖𝑛 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒 𝑑𝑢𝑟𝑎𝑡𝑖𝑜𝑛
𝑡𝑖𝑚𝑒 𝑑𝑢𝑟𝑎𝑡𝑖𝑜𝑛
=
𝑑𝑄
𝑑𝑡
(1.1)
Potential Difference and Electromotive Force (emf)
Potential difference (pd) between two points in an electric circuit, is that difference in their electrical state
which tends to cause flow of current between.
Emf is the force that causes an electric current to flow in an electric circuit.
1.1.2 Passive Elements
These are elements that are not capable of generating energy but can either store or dissipate energy supplied
to it. Examples are:
Resistance element (Resistor), which dissipates the energy that is supplied to it.
Symbol:
or
Resistance, R
Resistance may be defined as that property of a substance which opposes (or restricts) the flow of an electric
current (or electrons) through it.
Unit: Ω
Capacitance element, (Capacitor): It stores the energy supplied to it by an external circuit, in its electric field.
Symbol:
Capacitance, C
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
Inductance element (Inductor): It stores the energy supplied to it by an external circuit, in its magnetic field.
Symbol:
Inductance, L
In the case of the capacitance and inductance elements, the energy stored could fully be recovered.
Figure 1.1 shows an electric circuit with both active and passive elements.
Figure 1.1: An electric circuit
1.1.3 Resistivity, Conductance and Conductivity
The resistance of a wire depends upon its length, 𝑙, cross-sectional area, 𝐴, type of material, purity and
hardness of material of which it is made of and the operating temperature.
Mathematically, 𝑅 ∝
𝑙
𝐴
⟹𝑅=
𝜌𝑙
(1.2)
𝐴
Where 𝜌 (rho) is a constant and it depends on the nature of the material. It is known as the specific
resistance or resistivity of the material of the wire.
Thus, resistivity of a material is the resistance of the material of unit length having unit cross-sectional area.
Also defined as the resistance between opposite faces of a unit cube of that material.
Unit: Ω𝑚
1
Conductance is the reciprocal of resistance ( ) and is denoted by G. It is defined as the inducement offered
𝑅
by the conductor to the flow of current.
𝐺=
1
(1.3)
𝑅
Unit: S (Siemen)
Conductivity
This is the reciprocal of the resistivity. It is defined as the conductance between the two opposite faces of a
unit cube.
Substituting equation (1.2) into equation (1.3),
𝐺=
1
𝑅
=
1
𝜌𝑙
𝐴
=
1 𝐴
𝜌 𝑙
=𝜎
𝐴
𝑙
1
Where 𝜎 = 𝜌 , 𝜎 is the specific conductance or conductivity of the material
Unit: S/m (siemens/metre)
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.1.4 Nodes, Branches and Loops
10 Ω
15 Ω
30 𝑉
20 Ω
5𝐴
(a) A branch represents a single element such as voltage source or a resistor. Thus, largely, a branch
represents any two-terminal element.
(b) A node is a point of connection between two or more circuit elements (or branches).
(c) A loop is any closed path in a circuit.
1.1.5 Terminal Relations for Passive Elements
Resistance:
𝑣 = 𝑅𝑖
Inductance,
𝑣=𝐿
Capacitance, 𝑖 = 𝐶
𝑑𝑖
𝑑𝑡
𝑑𝑣
𝑑𝑡
1.2 Kirchhoff’s and Ohm’s Laws
1.2.1 Kirchhoff’s Current Law (KCL)
This law states that: the algebraic sum of the currents entering (or leaving) a node at any instant is zero.
i=0
(1.4)
In other words, the sum of the currents entering a node is equal to the sum of currents leaving the node.
Example 1.1
i2
i3
i1
i4
Figure 1.2: Example of KCL
Find 𝑖4 in Figure 1.2, if 𝑖1 = 4 𝐴; 𝑖2 = −5 𝐴; 𝑖3 = 1 𝐴.
Solution 1.1
Using KCL,
i=0
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
⟹ 𝑖1 − 𝑖2 – 𝑖3 + 𝑖4 = 0
or 𝑖4 = −8 𝐴
1.2.2 Kirchhoff’s Voltage Law (KVL)
It states that: the algebraic sum of the voltages around a closed path in a circuit at any instant is zero.
∑𝑉 = 0
(1.5)
NB:
In applying KVL to a closed path, the path in either direction could be traversed, beginning at any point.
Voltages could be recorded as positive values when an element is traversed in the positive direction. That is,
+
`
+
v(t)
-
-
Figure 1.3: Traversing in the negative direction
Example 1.2
If 𝑉𝑅 = 15 𝑉, and 𝑉𝐿 = 6 𝑉, what is the value of VS?
+ VR
+
+
VS
VL
-
-
Figure 1.4: Circuit for example 1.2
Solution 1.2
By KVL definition,
∑ 𝑉 = 𝑉𝑠 − 𝑉𝑅 − 𝑉𝐿 = 0
= 𝑉𝑠 − 15 − 6 = 0 (or 𝑉𝑠 = 15 + 6)
⟹ 𝑉𝑠 = 21 𝑉
1.2.3 Ohm’s Law
Ohm’s law states that, “the current, I, flowing in a circuit is directly proportional to the applied voltage, V,
provided the temperature remains constant. Thus,
𝐼∝𝑉
Therefore, 𝐼 =
𝑉
𝑅
or
𝑉 = 𝐼𝑅
or 𝑅 =
𝑉
𝐼
where R is the constant of proportionality called resistance.
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.3 Series Circuits
A series circuit is one in which the same current flows through every part of the circuit. It contains two or
more loads but only one path for current to flow from the source voltage through the loads and back to the
source.
R1
R2
R3
VT
R4
Figure 1.5: Direction of Current flow
1.3.1 Current in Series Circuit
The battery current, IT flows through the first load, R1, then through R2, R3 and then R4. Thus, if 1 A flows
through R1, then 1 A also flows through R2, R3 and R4. And of course the battery provides 1 A of current.
Thus symbolically,
𝐼𝑇 = 𝐼𝑅1 = 𝐼𝑅2 = 𝐼𝑅3 = 𝐼𝑅4 = 𝑒𝑡𝑐
(1.3)
1.3.2 Voltage in Series Circuits
The voltage in series divides up so that the sum of the individual load voltages equals the source of voltage.
V1 = IR1
V2 = IR2
V3 = IR3
VT
V4 = IR4
Figure 1.6: Voltage in series circuit
In the series circuit in figure 1.6, current flows through each resistor. The p.d. across R1 is by Ohm’s law,
𝑉1 = 𝐼𝑅1 , across R2, it is 𝑉2 = 𝐼𝑅2 etc. However, the sum of these p.d.s must be equal to the supply voltage.
That is
Or
𝑉𝑇 = 𝑉1 + 𝑉2 + 𝑉3 + 𝑉4 + … ….
(1.4)
𝑉𝑇 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + 𝐼𝑅4
(1.5)
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.3.3 Resistance in Series Circuit
Suppose that the circuit in figure 1.6 is replaced by the equivalent circuit of figure 1.7 containing the single
resistor, 𝑅𝑇 , the value of the resistor being selected so that the current flowing in both circuits is the same; that
is, the resistance of the circuit in figure 1.7 is equivalent to that in figure 1.6.
RT
I
VT
Figure 1.7: Equivalent resistance of a series circuit
In the case of figure 1.7, the circuit equation is 𝑉𝑇 = 𝐼𝑅𝑇
(1.6)
Where, 𝑅𝑇 , is the equivalent resistance of the series circuit.
From equations 1.5 and 1.6
𝐼𝑅𝑇 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + 𝐼𝑅4
Or 𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 + ….
(1.7)
Thus the total resistance in a series circuit is equal to the sum of the individual resistances around the series
circuit. That is, the total resistance RT can also be determined by Ohm’s law if the total voltage, VT and total
current IT, are known.
For example
A
R1 = 5 Ω
2A
R2 = 15 Ω
100 V
R3 = 20 Ω
R4 = 10 Ω
Figure 1.8: For calculating total resistance
Using Ohm’s law
𝑅𝑇 =
𝑉𝑇
𝐼𝑇
=
100
2
= 50 Ω
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
Or
𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4
= 5 + 15 + 20 + 10 = 50 Ω
1.3.4 Voltage Divider Equation
To find the voltage across only one of the resistors in a series circuit, the voltage divider equation can be used.
This is generally given by
𝑉𝑁 =
𝑅𝑁
𝑅𝑇
𝑉𝑇 , where RN is any one of the resistors in the series circuit.
In the example above
𝑉1 =
𝑅1
𝑅𝑇
𝑉𝑇 =
5
50
× 100 = 10 𝑉;
15
𝑉2 = 50 × 100 = 30𝑉;
20
10
𝑉3 = 50 × 100 = 40𝑉;
𝑉4 = 50 × 100 = 20𝑉
1.3.5 Summary
In summary, for a series circuit:
i.
Same current flows through all parts of the circuit
ii.
Applied voltage is equal to the sum of voltage drops across the different parts of the circuit
iii.
Different resistors have their individual voltage drops
iv.
Voltage drop across individual resistor is directly proportional to its resistance, current being the
same in each resistor
v.
Voltage drops are additive
vi.
Resistances are additive
vii.
Powers are additive
Exercise 1
R1
25 Ω
R2 = 40 Ω
50 V
R4 = 15 Ω
R3 = 20 Ω
Figure 1.9: For exercise
Find: 𝑅𝑇 , 𝐼𝑇 , 𝑉1 , 𝑉2 , 𝑉3 , 𝑎𝑛𝑑 𝑉4
[𝑹𝑻 = 𝟏𝟎𝟎 Ω; 𝑰𝑻 = 𝟎. 𝟓 𝐀; 𝑽𝟏 = 𝟏𝟐. 𝟓 𝑽; 𝑽𝟐 = 𝟐𝟎 𝑽; 𝑽𝟑 = 𝟏𝟎 𝑽; 𝑽𝟒 = 𝟕. 𝟓𝑽]
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.4 Parallel Circuits
These are multiple-load circuits which have more than one path for current. Each different current path is
called a branch.
R1
R2
R3
Figure 1.10: Parallel Circuit
The current in figure 1.10 splits up among three branches. Each branch has its own load, and each branch is
independent of all other branches. Examples of parallel circuits can be observed in figure 1.11 (a) to (e).
(b)
(a)
(c)
(e)
(d)
Figure 1.11: Different types of Parallel Circuits
1.4.1 Voltage in Parallel Circuits
All the voltages in a parallel circuit are the same. That is, the source voltage appears across each branch of a
parallel circuit.
⟹ 𝑉𝑇 = 𝑉1 = 𝑉2 = 𝑉3 = ⋯
(1.8)
1.4.2 Current in Parallel Circuits
IT
I2
A
I1
VT
R1
R2
Figure 1.12: Currents in Parallel Circuit
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
The relationship of the currents in a parallel circuit is as follows:
𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3 + ⋯
That is, the total current is equal to the sum of the individual branch currents.
(1.9)
The total current 𝐼𝑇 splits into two smaller currents at junction A (I1 and I2). If there are more junctions, then
the current will split further into smaller currents. The movement and direction of these currents are guided by
the Kirchhoff’s Current Law.
1.4.3 Resistance in Parallel Circuits
The total resistance of a parallel circuit is always less than the lowest branch resistance.
Thus with reference to figure 1.12,
From equation (1.9)
𝐼𝑇 = 𝐼1 + 𝐼2
But by Ohm’s Law
𝐼=
𝑉
⟹
𝑅
𝑉𝑇
𝑅𝑇
𝑉1
=
𝑅1
+
But from equation (1.8)
1
Thus
𝑅𝑇
Or 𝑅𝑇 =
=
1
𝑅1
+
𝑉2
𝑅2
𝑉𝑇 = 𝑉1 = 𝑉2
1
⟹
𝑉𝑇
𝑅𝑇
=
𝑉𝑇
𝑅1
+
𝑉𝑇
𝑅2
(1.10)
𝑅2
𝑅1 ×𝑅2
𝑅1 +𝑅2
Eqn. (1.10) is often referred to as the reciprocal formula because the reciprocals of the branch resistances are
added, and then the reciprocal of this sum is taken to obtain the total (equivalent) resistance.
Example
What is the total resistance of three resistors: 4 Ω, 6 Ω and 12 Ω connected in parallel?
1
Total resistance, 𝑅𝑇 = 1 1 1 = 2Ω
+ +
4 6 12
4×6×12
Or 𝑅𝑇 =
=2Ω
4×6+4×12+6×12
The total resistance is 2 Ω, which is less than the lowest resistance (4 Ω).
1.4.4 Resistors with the Same Value
When all the resistors in a parallel circuit have the same value, just divide the value of a resistor by the
number of resistors. That is,
1
𝑅𝑇
=
𝑁
𝑅
𝑅
or 𝑅𝑇 = , where N is the number of resistors.
𝑁
E.g. two 24 Ω resistors in parallel have a total resistance of 𝑅𝑇 = 24⁄2 = 12Ω
Now, if five 25 Ω resistors are arranged in parallel then the total resistance, RT will be 𝑅𝑇 = 25⁄5 = 5Ω
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Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.4.5 Current – Divider Rule
When there is the need to find the current through only one of two parallel resistors, then the current divider
𝐼1 =
formula could be used. Thus
This is derived from 𝐼1 =
⟹ 𝐼1 =
𝑅𝑇 𝐼𝑇
𝑅1
=
𝑅1 𝑅2
𝐼
𝑅1 +𝑅2 𝑇
𝑅1
𝑉1
𝑅1
𝑅2
𝐼
𝑅1 +𝑅2 𝑇
𝐼2 =
𝑅1
𝐼
𝑅1 +𝑅2 𝑇
but 𝑉1 = 𝑉𝑇 = 𝑅𝑇 𝐼𝑇
𝑅2
=𝑅
1 +𝑅2
I2
IT
I1
𝐼𝑇
𝑅1
𝑅2
Example: Find 𝐼1 and 𝐼2 if 𝐼𝑇 = 14𝐴.
I2
IT
Solution
I1
𝐼1 =
25 Ω
45 Ω
45
45+25
× 14 = 9𝐴
𝐼2 =
25
45+205
× 14 = 5𝐴
Figure 1.12: Circuit for example
1.4.6 Summary
In summary, for a parallel circuit:
i.
ii.
iii.
iv.
v.
vi.
Exercise 2
Same voltage acts across all branches of the circuit
Different resistors (or branches) have their individual currents
Total circuit current is equal to the sum of individual currents through the various resistors
Branch currents are additive
Powers are additive
The reciprocal of the equivalent or combined resistance is equal to the sum of the reciprocals of
the resistances of the individual branches.
I1
I
I2
18 Ω
32 Ω
Figure 1.13: Circuit for example
In figure 1.13, if 𝐼 = 20 𝐴, Calculate:
(i)
RT (ii) VT
(iii) I1 using Ohm’s Law
[11.52 Ω; 230.4 V; 7.2 A; 7.2 A]
10
(iv) I1 using the current divider rule
Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.5 Series – Parallel Circuits
1.5.1 Network Reduction
R1
R1
R2
R2
R1
R3
.R
R3
3
(a)
R1
R3
R1 + R2
R2
𝑅3 + 𝑅4
(R1 + R2)
R4
(b
) Reduction
Figure 1.14: Network
Example
i)
ii)
R2 + R3
For figure 1.14 (a), if 𝑅1 = 10 Ω; 𝑅2 = 15 Ω; 𝑅3 = 24 Ω, find 𝑅𝑇
For figure 1.14 (b), find 𝑅𝑇, if 𝑅1 = 12 Ω; 𝑅2 = 8 Ω; 𝑅3 = 10 Ω; 𝑎𝑛𝑑 𝑅4 = 20 Ω
Exercise
R0
R3
R1
R6
R4
R2
R5
Figure 1.15: Network Reduction Exercise
Find the total resistance, RT
if 𝑅0 = 24 Ω; 𝑅1 = 60 Ω; 𝑅2 = 100 Ω; 𝑅3 = 140 Ω; 𝑅4 = 200 Ω; 𝑅5 = 150 Ω and 𝑅6 = 50 Ω
11
R3 + R4
Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.6 Series – Parallel Problems Using KCL and KVL
R1
I2
R3
100 V
R2
40V
6A
I1
2A
R4
30V
Figure 1.16: Series – Parallel Example
Example
From figure 1.16, determine:
(𝑖) 𝐼2
(ii) 𝐼1
(ii) 𝑉𝑅1
(iv) 𝑉𝑅3
Solution
(i)
From figure, 𝐼2 = 2𝐴
(ii)
Using KCL, 𝐼2 + 𝐼1 = 6𝐴
⟹ 𝐼1 = 6 − 2 = 4𝐴
(iii)
Let the voltages across 𝑅1 , 𝑅2 , 𝑅3 𝑎𝑛𝑑 𝑅4 𝑏𝑒 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 𝑉1 , 𝑉2 , 𝑉3 𝑎𝑛𝑑 𝑉4
Then 𝑉1 + 𝑉2 = 100𝑉
𝑜𝑟 𝑉1 + 40 = 100𝑉
⟹ 𝑉1 = 60𝑉
(iv)
𝑉2 is parallel to 𝑉3 𝑎𝑛𝑑 𝑉4
⟹ 𝑉2 = 𝑉3 + 𝑉4 or 𝑉3 + 30 = 40𝑉
⟹ 𝑉3 = 10𝑉
Exercise
R1
0.8A
0.3A
R3
R2
100 Ω
100V
R4
30 V
Figure 1.17: Circuit for Exercise
Using figure 1.17, determine:
(a) 𝐼2
(b) 𝑉2
(c) 𝑅3
(d) 𝑉1
(e) 𝑅1
12
(f) 𝑅4
Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.7 Electric Power and Energy
Work is said to be done by or against a force, when its point of application moves in or opposite to the
direction of the force and is measured by the product of the force and the displacement of the point of
application in the direction of force
Simply, work is done whenever a force causes motion. For example, voltage is an electric force and it forces
current to flow in a circuit. This constitutes work done. The rate at which this work is done is called Electric
Power. The basic unit of power is Watt and it is equal to the voltage across a circuit multiplied by the current
through it. Thus, the basic equation for power (P) is
𝑃 = 𝑉𝐼 𝑤𝑎𝑡𝑡𝑠
This is when V is in Volts and I is in Amperes
NB: Watt is a small unit, hence, kilowatt is more widely used, as well as Megawatts.
Example 1
What is the maximum power in kilowatts that can be carried by a 220-V power line that has a 15 A fuse?
Solution
The purpose of a fuse is to limit the current in a circuit to a safe level. Now, the maximum power here is
𝑃 = 𝐼𝑉 = 15 × 220
= 3.3 𝑘𝑊
Example 2
An electric heater rated at 1000 𝑊 is connected to a 120-V power line. How much current does it draw?
Solution
𝐼=
𝑃
𝑉
=
1000
120
= 8.33 𝐴
Example 3
The horsepower (hp) is a unit of power equal to 746 W. A generator driven by a diesel engine that develops
15 hp delivers 40 A at 270 V. What is the efficiency of the generator? [Efficiency of a device is the ratio
between the energy or power it delivers and the energy or power it takes in, so that
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑂𝑢𝑡𝑝𝑢𝑡
𝐼𝑛𝑝𝑢𝑡
Solution
1ℎ𝑝 = 746 𝑊 and 𝑃 = 𝐼𝑉
⟹ 𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 = 15ℎ𝑝 × 746𝑊/ℎ𝑝
= 11190 𝑊
𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 = 40 𝐴 × 270𝑉
= 10800 𝑊
Thus efficiency of the generator is
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟
𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟
=
10800
11190
= 0.908
13
or 90.8%
Applied Electricity
Chapter One: Basic Definitions and Circuit Laws
1.7.1 Other Power Relations
If the conductor or device through which a current passes has resistance R, the power consumed may be
𝑉
expressed in terms of I and R or in terms of V and R by substituting 𝑉 = 𝐼𝑅 or 𝐼 = 𝑅 in the formula for
power, 𝑃 = 𝐼𝑉.
= 𝐼(𝐼𝑅) = 𝐼 2 𝑅
Thus
𝑉
Or 𝑃 = 𝐼𝑉 = (𝑅) 𝑉 =
𝑉2
𝑅
1.7.2 Energy
This is defined as the ability to do work. This is spent when work is done because energy has to be spent to
maintain a force when that force acts through a distance.
In electricity, the total electrical energy spent is the rate at which the work done is multiplied by the length of
time during which work is done. In other words, electrical energy, E is equal to the electrical power (P)
multiplied by time (t).
𝐸 = 𝑃𝑡 = 𝑉𝐼𝑡
or 𝐸 = 𝐼 2 𝑅𝑡
If I is in 𝐴, V in volts and t in hours, then E will be in watt hours.
If t is in seconds, E will be in watt seconds or Joules. (1 Joule = 1 watt second)
NB: Electrical energy is purchased and sold in units of kilowatt hours.
1 𝑘𝑊ℎ = 1000 𝑊ℎ = 3.6 × 106 𝐽
Example 4
An electric iron has a rating of 1.5 𝑘𝑊 at 240 𝑉 and works for 5 hours a day, 24 days in a month.
Calculate:
(i)
The current taken
(ii)
The resistance of its heating element
(iii)
The cost of electrical energy consumed in one month, if the cost of 1 kWh is GH¢1.50.
Solution
1.5×1000
= 6.25 A
(i)
Current taken =
(ii)
Resistance of its heating element, R =
(iii)
Electrical energy used in one month, E is given by
240
𝑉
𝐼
=
240
6.25
𝐸 = 𝑃𝑡 = 1.5 × 5 × 24 = 180 𝑘𝑊ℎ
Thus total cost = 180 × 1.50
= GH¢270.00
14
= 38.4 Ω