Download Calculate electrical power

Contents
Introduction
3
Energy
4
Conservation of Energy
4
Work
5
Power
7
Power in electric circuits
9
Power rating of equipment
9
Calculating electrical power in resistive circuits
12
Other power equations
14
Resistive Loads
16
Measurement of electrical energy
17
Summary
22
Answers
23
EEE042A: 5 Calculate electrical power
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EEE042A: 5 Calculate electrical power
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Introduction
Energy is the fundamental quantity that allows changes to occur. The
concept of work allows us to measure the amount of energy by mechanical
definition.
Power is the rate at which energy is delivered, or the rate at which work is
done. In an electrical circuit, voltage can do work by moving charges around
the circuit against a resistance. Electrical power in a resistive circuit is equal
to voltage times current.
After completing this topic, you should be able to:

describe the relationships between energy, work and power

calculate the power dissipated in a circuit from voltage, current and
resistance values

explain the meaning of power ratings of devices.
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3
Energy
Without energy, nothing could happen. Whenever you see movement,
sound, heat or light, there is energy. The SI unit of energy is the joule, and
the unit symbol is 'J'. The joule is a very small quantity of energy, so more
common units are the kilojoule (kJ) and the megajoule (MJ).
Energy can exist in many forms, and we can convert energy from one form
to another. Consider the following manifestations of energy:

energy in sunlight allows plants to live

chemical energy in food allows animals to live

chemical energy in fuels powers industry

electrical energy is present wherever you measure a voltage or
current.
Conservation of Energy
The Law of Conservation of Energy says that energy is 'conserved'. This
means that energy cannot be created or destroyed, but can be changed from
one form to another. Technology has many examples of conversion of
energy, including:
4

Solar cells convert sunlight into electrical energy.

The chemical energy of coal or natural gas is converted to electrical
energy at the power station.

Electrical energy is converted to heat energy in a microwave.

Electrical energy is converted to electromagnetic energy in a radio
transmitter.

Batteries convert chemical energy into electrical energy.

Light emitting diodes convert electrical energy into light energy.
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Notes:
1. The conservation law has had to be slightly modified in modern times,
because we now know that mass can be converted into energy and vice
versa. Small amounts of mass are converted to enormous amounts of energy
in nuclear power stations and nuclear weapons.
2. All energy eventually ends up as heat, which is dissipated into the
environment and cannot be recovered. So while energy cannot be
‘destroyed’, it cannot always be converted into a useful form for our use. In
addition, our machines cannot use all the energy present in an energy source,
and there will always be some wastage or loss.
Work
The concept of work gives us a very straightforward definition to determine
an amount of energy. Work is done when a force moves a body through a
distance. The work done is given by:
work  force  distance
or in symbols:
W  Fd
where:
W = work (joule)
F = force in (newton)
d = distance (metre)
Work has the same units as energy, and we can say that energy is the
capacity to ‘do work’. Work is signified by the letter 'W'.
Note that 'work' is a scientific term with a very specific meaning — it does
not mean people doing work in the everyday sense of the term. However,
when doing physical work, people are also doing some amount of work in
this technical sense.
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Example 6
A force of 600 N is required to move a body against frictional forces. The
body is moved 2 m. Determine the work done.
F
d
W
W
 600 N
 2m
?
 Fd
 600  2
 1200 J
The work done in moving the body is 1200 J.
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Power
Power is the rate of doing work, or equivalently, the rate of producing or
expending energy. The symbol for power is P, and the unit is the watt (W).
When energy is consumed at the rate of one joule per second, the power is
one watt.
Power can be illustrated by the fact that it may take several hours to dig a
trench with hand tools, but an engine-driven backhoe could do the same job
in a few minutes. We say the machine is more powerful, because it delivers
much more energy in the same period of time.
As power is the rate of doing work we can say that power is work per unit
time, or in symbols:
P
W
t
where:
P = power (watt)
W = work (joule)
t = time (second)
Don’t confuse the ‘W’ used here for watt with the ‘W’ used for ‘work’.
When dealing with small or large quantities we make use of the milliwatt,
the kilowatt, and the megawatt:

megawatt (MW) = 1 000 000 W

kilowatt (kW) = 1000 W

milliwatt (mW) = 0.0001 W
Example 7
If the energy required to raise a certain mass to the top of a building is
5000 J, determine the output power rating of an electric motor required to do
the work in:
(a) 30 s
(b) 15 s.
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Solution
W = 5000 J
(a) t 
P 
30 s
?
W
P 
T
5000

30
 166.7 W
(b) t  15 s
P  ?
W
P 
T
5000

15
 333.3 W
The power required to do the work in 30 s is 166.7 W and the power
required to do the work in 15 s is 333.3 W.
If you have Hampson, read the ‘Power, work and energy’ section on page 5
through to page 12, taking note how the equations are used.
If you have Jenneson, refer to Section 2.9, ‘Electrical power and on page 35
for brief summary of key issues.
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Power in electric circuits
When electric current flows through a resistor, electrical energy is converted
to heat and leaves the circuit (it is ‘expended’ use ‘used’). For example,
current flowing through a lamp causes it to emit light and heat energy. The
energy required to light the bulb is provided by the flow of charges through
the lamp.
The rate at which energy is delivered to a circuit is the power.
Another example is the electric motor. The energy delivered to an electric
motor is converted to useful mechanical work. A motor that is efficiently
converting energy to work behaves like a resistor.
The power of an appliance is one of the most important parameters to specify.
For example, to boil a litre of water using a 750 W electric jug might take
three minutes. To boil the same amount of water using a 1500 W electric jug
would take about half that time. Likewise, an air conditioner with insufficient
power may fail to maintain a room at a comfortable temperature.
Power rating of equipment
The power rating of an electrical appliance is important because it
determines:

the capability of doing the required work,

the size of cable required for the current the appliance will draw,

the running cost of the appliance.
Information relating to the electrical characteristics of an appliance is
stamped on a manufacturer’s nameplate which is attached to the appliance.
The information always includes the:

voltage to which the equipment can be safely connected, both the type
(ac/dc) and the value.

maximum power the appliance will consume.
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Figure 1: A typical electrical nameplate
The nameplate in Figure 1 belongs to an electrical stapling machine.
The information on it shows that:

it is suitable for operation on 200–240 V alternating current with
a frequency of 50–60 hertz,

its maximum power consumption is 500 W.
Electric light globes are a common example. A globe rated at 240 V, 100 W
is brighter and costs more to operate than a 240 V, 40 W lamp.
Input and output power
Lamp wattages refer to the input power. However, this is not always the
case. The power rating of electrical equipment may specify either:

the input power drawn from the supply, or

the output power.
The input power of an appliance is usually quoted when it is not practical to
quote the output power. Common examples of such equipment are a bread
toaster, a hot-water service, a welding machine or a lamp. In these examples
electrical energy is being converted to heat, and this represents the ‘final’
state of the energy – the energy will not be converted to another form. It
therefore makes sense to simply quote the power used by the device.
For industrial equipment such as transformers or motors, the output power is
quoted, because this figure is required to make use of the device For
example, it is important to know that mechanical output power that is
provided by a motor in order to select a motor for a particular job. Note that
where the output power is quoted, the electrical power that is actually used
will be greater, because of the losses within the device.
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Activity 1
1
What is the unit of electric power?
_____________________________________________________________________
2
What can we learn from the power rating of an electrical appliance?
_____________________________________________________________________
_____________________________________________________________________
3
What information is usually given on an appliance nameplate?
_____________________________________________________________________
_____________________________________________________________________
4
If work of 100 J is used to move a body a certain distance in 0.25 s, what power would
be required?
_____________________________________________________________________
_____________________________________________________________________
5
What is the relationship between a watt and a milliwatt?
_____________________________________________________________________
6
Define power.
_____________________________________________________________________
_____________________________________________________________________
7
What work is done if a force of 20 N moves a body a distance of 5 m?
_____________________________________________________________________
_____________________________________________________________________
Check your answers with those given at the end of the section.
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Calculating electrical power in resistive
circuits
In a resistive circuit, electrical power is determined from the equation:
P  VI
where:
V = voltage (volt)
I = current (ampere)
P = power (watt)
To make sense of this equation, we can compare the electrical situation to a
water pump that is lifting water uphill. The height through which the water
is lifted is similar to the circuit voltage. The flow rate of water is similar to
the current. But the power expended depends upon both the amount of
liquid pumped, and the height.
Example 1
An electric circuit comprises a voltage source of 20 V and a resistance of
5 . Determine the power dissipated by the resistor.
First find I:
V
R
20

5
4A
I
Draw the complete diagram with all the relevant information.
Now:
P  VI
 20  4
 80 W
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Example 2
If, in the circuit of Example 1, the voltage is doubled determine the power.
Again first find I:
V
R
40

5
8A
I
P  VI
 40  8
 320 W
Note that this answer is four times that of Example 1. That is, when the
applied voltage is doubled, the power is increased four times (power is
proportional to the voltage squared).
Example 3
Determine the power in the circuit of Example 1 if the supply voltage of
remains at 20 V but we increase the resistance from to 10 .
Again first find I:
V
R
20

10
2A
I
P  VI
 20  2
 40 W
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Note that when compared to Example 1, the power is halved. Therefore,
we can say when the resistance is doubled the power is halved. In other
words, power is inversely proportional to resistance.
Other power equations
We know from Ohm’s law that:
V
R
and
V  IR
I
If we substitute for I or V in the power equation:
P  VI
V 

V
R
V2
R
P  VI
 IR  I
 I 2R
We now have three options when calculating power in resistive circuits. We
can use:
P  VI
V2
R
P  I 2R
P
If we are given:

voltage and current we use equation (a).

current and resistance we use equation (b).

voltage and resistance we use equation (c).
All of the equations will give exactly the same results. It is purely a matter
of available information which determines the one you use.
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Example 4
Using the original values from Example 1, determine power using all
three equations.
Given:
V = 20 V
R=5
I=4A
(a)
P 

VI
20  4

80 W
(b) P =

=
(c)
P =
=
=
I2R
42  5
80 W
V2
R
20  20
5
80 W
Example 5
A 300  resistor takes a current of 2 A. Determine the power dissipated by
the resistor.
P  I 2R
 22  300
 1200 W
 1.2 kW
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Example 6
Consider example 5 but with the current increased to 4 A. What power will
be dissipated in the resistor?
P  I 2R
 42  300
 4800 W
 4.8 kW
Note: Doubling the current through the same resistance increases the power
four times.
Example 7
A 3  resistor is connected to a 12 V DC power supply as shown below.
Determine the power dissipated by the resistor.
V2
R
122

3
 48 W
P
Resistive Loads
The above equations can be used to determine the power consumed in a
resistive circuit. So for example, it can be used for incandescent lamps,
electric radiators or water heaters, which all use resistive elements.
Note however that the circuit elements need not be resistors to behave like
resistive loads! Any device that simply removes energy from the circuit will
look like a resistor to the circuit. For example, electric motors when working
a maximum efficiency behave almost like a resistive load.
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Measurement of electrical energy
The utility that sells you electrical energy must bill you for the amount of
energy used. Electrical energy is measured by adding up the power usage
over time.
Since power is equal to energy used per unit time, it follows that energy is
equal to power multiplied by time:
energy
T
energy  P  T
P
Thus one joule of energy is expended when a device with a power of one
watt operates for one second. (We could also call this amount of energy one
'watt-second'.)
In practice, electrical energy is measured by a device called a watt-hour
meter or electrodynamometer. The watt-hour meter in your home's main
switchboard measures your energy usage by measuring the total power at
every instant, and continuously accumulating this over time.
Domestic electrical energy usage is therefore quoted as kilowatt-hours
(kWh). Kilowatt-hours are the so-called 'units' on your electricity bill. The
kilowatt-hour is not an SI unit, so we need to be able to convert to joules for
comparison purposes. As there are 3600 seconds in one hour, we have:
1 kilowatt-hour = 3600 kilowatt-second = 3600 kilojoule = 3.6 megajoule.
If you have Hampson, read the ‘Electrical power, work and energy’ section
on pages 10-12, checking to see how the equations are used.
If you have Jenneson, refer to Section 2.9, ‘Electrical power and energy’ on
page 35 for brief summary of key issues.
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Activity 2
1
When a 50  resistor is connected to a 10 V supply, 200 mA of current flows.
Determine the power consumed using each of the three power equations.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
2
What current is taken by a 100 W incandescent lamp when connected to a 240 V
supply?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
3
What is the resistance of a toaster element if it consumes 900 W from a 240 V supply?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
4
A 240 V electric jug takes 8 A from the supply. What is its power rating?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
5
Across what voltage must a 3.3 k resistor be placed for it to consume 5 W of power?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
Check your answers with those given at the end of the section.
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Check your progress
1
A generator is to be lifted vertically to a position nine metres above the ground by
means of a crane. If the work done by the crane is 45 kJ, determine the force required.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
2
An electric motor of mass 500 kg is lifted through a height of 10 m. Determine the
work done on the motor.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
3
Neatly draw a circuit diagram that shows a 100  resistor connected to a 9 V battery.
Include the following in the circuit:
(a) a switch controlling current to the resistor
(b) an ammeter measuring the current through the resistor
(c) a voltmeter measuring the load voltage
(d) calculate the current.
4
Using Ohm’s law, fill in the blanks in the table below.
Voltage (V) volts
Current (I) amperes
30 V
12 V
3 k
40 mA
600 
4.8 mA
2.2 
330 A
20 V
10 V
12 M
10 A
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Resistance (R) Ohm
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5
Calculate the resistance of the lamp in the following circuit.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
6
A 24  heating element requires a current of 10 A to produce its specified heat output.
Calculate the required supply voltage for this heater.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
7
Calculate the resistance of a resistor that takes 100 mA when connected to a 10 V
battery.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
8
Calculate the voltage across a 4.7 k resistor that has 3.5 A passing through it.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
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Calculate the power consumed by an electrical appliance that is drawing 5 A from a
240 V supply.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
10 A motor consumes 12 kW when connected to 240 V. Calculate the motor current.
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
11 A resistance of 30  is connected to an 80 V DC power supply. Determine the:
(a) circuit current
__________________________________________________________________
__________________________________________________________________
(b) power dissipated by the resistor.
__________________________________________________________________
__________________________________________________________________
12 An electric toaster element has a voltage rating of 240 V and a power rating of 550 W.
Determine the:
(a) circuit current when the rated voltage is applied to the element
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
(b) resistance of the element.
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
Check your answers with those given at the end of the section.
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Summary

Energy is a fundamental quantity that drives change of any kind. Energy
comes in many forms, and one form can often be converted to another.

Work is done when a force moves its point of application. The unit
of work is the joule. Work and energy use the same unit, which is the
joule.

Power is the rate of doing work. If work is to be done more quickly,
more power is required. The unit of power is the watt.

Power rating determines work capacity, cable size requirements and
running costs.

Nameplates usually nominate voltage and power ratings.

Power rating can indicate input or output depending on the type of
equipment.

The SI derived unit of power is the watt.

Commonly used multiple and sub-multiple units used for nominating
power are:
1 MW = 1 000 000 W = 106 W
1 kW = 1000 W = 103 W
1 mW = 0.001 W = 10-3 W

Power may be calculated from any of the following equations. Use the
most convenient.
P  VI
P  I 2R
P
22
V2
R

Power is proportional to the current squared. For example twice the
current gives four times the power, if the resistance is the same.

Power is inversely proportional to the resistance. For example, half the
resistance gives twice the power, if the voltage is the same.
EEE042A: 5 Calculate electrical power
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Answers
Activity 1
1
The unit of electric power is the watt.
2
Power rating gives us an indication of:
3

capability to do a required amount of work

size of cables necessary to wire device
 cost of running device.
Nameplate particulars always include:

voltage rating

type of voltage: ‘AC’ or ‘DC’

power rating.
4
W
 100 J
t
 0.25 s
P
 ?
P

W
t
100

0.25
 400 W
The power required to move the body in 0.25 s is 400 W.
5
There are 1000 milliwatt in a watt, or conversely a milliwatt is one
thousandth of a watt.
6
Power is the rate of doing work.
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7
F
d
W
W
 20 N
 5m
?
 Fd
 20  5
 100 J
So the work done in moving the body is 100 J.
Activity 2
1
P  VI
 10  0.2
2W
P  I 2R
 0.2 2  50
2W
V2
R
10 2

50
2W
P
2
P  VI
P
I
V
100

240
 0.417 A
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V2
R
V2
R
P
240  250

900
 64 
P
4
P  VI
 240  8
 1920 W
5
P

PR 
V2
R
V2
V

PR
V

5  3300
V

16 500
V

128.5 V
Check your progress
1
5000 N
2
49 000 J
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P  VI
 10  0.2
2W
P  I 2R
4
 0.2 2  50
2W
Voltage (V) volts
Current (I) amperes
Resistance (R) Ohm
30 V
0.01 A
3 k
24 V
40 mA
600 
0.0106 V
4.8 mA
2.2 
12 V
330 A
36 364 
20 V
1.67× 10–6.A
12 M 
10 V
10 A
1
2
V
R
10 2

50
2W
P
5
1200 
6
240 V
7
100 
8
16 450 V
9
1200 W
10 50 A
11 (a) 2.67 A
(b) 21.3 W
12 (a) 2.3 A
(b) 105 
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