Energy & Power in Appliances (Oxford AQA IGCSE Combined Science Double Award)
Revision Note
Written by: Dan Mitchell-Garnett
Reviewed by: Caroline Carroll
Power & Energy in Appliances
The power of an appliance is defined as:
The rate at which energy is transferred by an appliance
Power can be calculated using:
Where:
P = power measured in watts (W)
The watt is equivalent to joules per second (J / s)
E = energy transferred measured in joules (J)
t = time measured in seconds (s)
This equation can also be written as
Where:
W = work done measured in joules (J), which is equivalent to energy transferred
Examiner Tips and Tricks
The equation will be given in both forms on your equation sheet for your exam.
Time is an important consideration when it comes to power
Two cars transfer the same amount of energy, or do the same amount of work to accelerate over a distance
If one car has more power, it will transfer that energy, or do that work, in a shorter amount of time
Two cars with different amounts of power
Two electric motors:
lift the same weight
by the same height
but one motor lifts it faster than the other
The motor that lifts the weight faster has more power
Two motors with different amounts of power
Power ratings are given to appliances to show the amount of energy transferred per unit time
Common power ratings are shown in the table below:
Power Rating Table
Appliance | Power Rating |
Torch | 1 W |
Light bulb | 10 - 100 W |
Electric cooker | 10 000 W or 10 kW (1 kW = 1000 W) |
Railway engine | 1 000 000 W or 1 MW (megawatt) |
Saturn V space rocket | 100 MW |
Large power station | 10 000 MW |
Global demand for power | 10 000 000 MW |
Star (similar in size to the Sun) | 100 000 000 000 000 000 000 MW |
Power, Current & Potential Difference
The power of a device depends on:
The potential difference across the device
The current flowing through the device
The power transferred to an electrical component (or appliance) is given by the equation:
Where:
P = power measured in watts (W)
V = potential difference measured in volts (V)
I = current measured in amps (A)
Worked Example
A 12 V battery supplies a lamp. The lamp is supplied with 2880 J every minute.
The lamp manufacturer can make fuses rated to 3 A, 5 A or 15 A.
Suggest which fuse they have fitted in the lamp.
Answer:
Step 1: List the known quantities
Energy transferred, E = 2880 J
Time, t = 1 minute = 60 s
Potential difference supplied, V = 12 V
Step 2: Find the power equation from the equation sheet
Step 3: Substitute energy and time into this equation
Step 4: Rearrange the electrical power equation from the equation sheet for the current
Divide both sides by V
Step 5: Substitute power and potential difference
Step 6: Suggest a suitable fuse
A fuse should be rated to just over the normal operating current
3A is below the normal operating current of 4 A
15 A is too high above the normal operating current of 4 A
The most suitable fuse is the 5 A fuse
Energy, Potential Difference & Charge
Extension Tier only
The potential difference across a component is defined as:
The energy transferred per unit charge
In equation form, the energy transferred can be written as:
Where:
E = energy transferred measured in joules (J)
V = potential difference measured in volts (V)
Q = charge measured in coulombs (C)
This can be combined with the power definition to produce the electrical power equation:
Recall that power is defined as:
Substitute the equation above into this definition
This is equivalent to
Finally, recall the definition of the current
Substituting this into the previous equation gives the electrical power equation
Examiner Tips and Tricks
In your equation sheet, the energy transferred equation is given as
Make sure you are confident in rearranging this to the form given in this spec point.
Last updated:
You've read 0 of your 10 free revision notes
Unlock more, it's free!
Did this page help you?