Electrical Power (Cambridge (CIE) IGCSE Physics)

Revision Note

Lindsay Gilmour

Written by: Lindsay Gilmour

Reviewed by: Caroline Carroll

Electrical power equation

  • The power of an appliance is defined as: 

The rate at which energy is transferred by an appliance

  • Power can be calculated in terms of energy:

P space equals fraction numerator space E over denominator t end fraction

  • 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)

  • Power can also be calculated in terms of work done:

P space equals fraction numerator space W over denominator t end fraction

  • Where:

    • W = work done, which is equivalent to energy transferred, measured in joules (J)

  • The power of an electrical device is the energy transferred per second by the device

  • The power dissipated by an electrical component can be calculated by:

P space equals space I V

  • Where:

    • P = dissipated power, measured in watts (W)

    • I = current, measured in amps (A)

    • V = potential difference, measured in volts (V)

Worked Example

Two lamps are connected in series to a 150 V power supply.

WE - power question image, downloadable AS & A Level Physics revision notes

Which statement most accurately describes what happens?

A.     Both lamps light normally

B.     The 15 V lamp blows

C.     Only the 41 W lamp lights

D.     Both lamps light at less than their normal brightness

Answer: A

Step 1: Calculate the current required for both lamps to operate

  • For the 41 W lamp, with 135 V

P space equals space I V

I space equals fraction numerator space P over denominator V end fraction

I space equals fraction numerator space 41 over denominator 135 end fraction

I space equals space 0.3 space straight A

  • For the 4.5 W lamp, with 15 V

I space equals fraction numerator space 4.5 over denominator 15 end fraction

I space equals space 0.3 space straight A

Step 2: Determine the outcome of the bulbs

  • For both bulbs to operate at their normal brightness, a current of 0.3 A is required

  • The lamps are connected in series, so the same current would flow through both

  • Therefore, the lamps will light at their normal brightness

    • This is option A

Examiner Tips and Tricks

When doing calculations involving electrical power, remember the unit is Watts W, therefore, you should always make sure that the time is in seconds

Measuring energy usage

Energy measured in joules

  • Electrical energy transferred is often calculated with units of joules

    • One joule is equivalent to one-watt second

  • Consider an average lightbulb with a power of 60 W, which is left on for 6 hours in a house

    • 1 hour is 3600 s

    • The energy transferred over this time is 1.296 × 106 J

  • This number is large and that is only one lightbulb for a single day

    • A household uses many appliances all year round; the energy transferred per month in joules would be inconveniently large

Energy measured in kilowatt-hours

  • To make these large values more relatable to daily use:

    • Power can be measured in kilowatts (kW)

    • Time can be measured in hours (h)

  • In this case, energy has units of kilowatt-hours (kW h)

    • The lightbulb from before receives 3.6 kW h of energy over the 6 hours

  • This value is much easier to understand for consumers and energy providers; thinking in terms of hours of use is more practical than seconds

Calculating with kWh

  • As has been stated previously, the equation for energy transferred is:

E space equals space P t

  • But here, different units are considered:

    • E = energy transferred, measured in kilowatt hours (kW h)

    • P = power of the appliance, measured in kilowatts (kW)

    • t = time, measured in hours (h)

  • The usual unit of energy is joules (J), which is one watt-second

  • To find the number of joules in 1 kW h, convert the power and time to watts and seconds

1 space kW space straight h space equals space 1000 space straight W space cross times space 3600 space straight s space equals space 3.6 space cross times space 10 to the power of 6 space straight J

  • 1000 watts multiplied by 3600 seconds is equal to 1000 multiplied by 3600, in watt-seconds

1000 space straight W space cross times space 3600 space straight s space equals space 1000 space cross times space 3600 space Ws space equals space 3.6 space cross times space 10 to the power of 6 space straight J

  • Therefore, 1 kWh = 3.6 × 106 J

 

  • To convert from kW h to J:

E space open parentheses kW space straight h close parentheses space space cross times space left parenthesis 3.6 space cross times space 10 to the power of 6 right parenthesis space equals space E open parentheses straight J close parentheses

  • To convert from J to kW h:

E open parentheses straight J close parentheses space space divided by space left parenthesis 3.6 space cross times space 10 to the power of 6 right parenthesis space equals space E open parentheses kW space straight h close parentheses

  • The kW h is a large unit of energy, and is mostly used for energy in homes, businesses and factories

Worked Example

A cooker transfers 1.2 × 109 J of energy electrically to the thermal store of the heating element during its use over a year.

Assume that 1 kW h costs 14.2 p. 

100 p = £1 (100 pence = 1 pound)

Calculate the cost of using the oven for the year.

Answer: 

Step 1: List the known quantities

  • Energy in joules, E open parentheses straight J close parentheses space equals space 1.2 cross times 10 to the power of 9 space straight J

  • Cost per kW h, 1 space kW space straight h space equals space 14.2 space straight p

Step 2: Convert the energy used from J to kW h

E open parentheses kW space straight h close parentheses space equals space fraction numerator 1.2 cross times 10 to the power of 9 over denominator 3.6 cross times 10 to the power of 6 end fraction space equals space 333.333 space kW space straight h

Step 3: Calculate the price

1 space kW space straight h space equals space 14.2 space straight p

333.333 space cross times space 14.2 space equals space 4733 space straight p space equals space £ 47.33

Examiner Tips and Tricks

The kilowatt hour is a tricky concept to get your head around, so make sure you are comfortable with the conversions between kW h and J well before your exam.

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Lindsay Gilmour

Author: Lindsay Gilmour

Expertise: Physics

Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.