Electrical Energy & Power (AQA A Level Physics): Revision Note

Exam code: 7408

Lindsay Gilmour

Written by: Lindsay Gilmour

Reviewed by: Timothy Sit

Updated on

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Understand this topic with deeper explanations and understanding checks.

Energy & power

  • In mechanics, power P is defined as the rate of doing work

    • The potential difference is the work done per unit charge

    • Current is the rate of flow of charge

  • Therefore, the electrical power is defined as the rate of change of work done:

P = Et = Wt

  • Where:

    • P = power (W)

    • E = energy transferred (J)

    • t = time (s)

    • W = work done (J)

  • The work done is the energy transferred so the power is the energy transferred per second in an electrical component

  • The power dissipated (produced) by an electrical device can also be written as

P = IV

  • Where:

    • P = power (W)

    • I = current (A)

    • V = potential difference (V)

  • Using Ohm's Law V = IR to rearrange for either V or I and substituting into the power equation, means power can be written in terms of resistance R

P = I2R

P = V2R

  • Where:

    • R = resistance (Ω)

  • This means for a given resistor if the current or voltage doubles the power will be four times as great

    • Which equation to use will depend on whether the value of current or voltage has been given in the question

  • Rearranging the energy and power equation, the energy can be written as:

E = VIt

  • Where:

    • E = energy transferred (J)

    • V = voltage (V)

    • I = current (A)

    • t = time (s)

Worked Example

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

Circuit diagram of two lamps in series across a 150 V supply. A voltmeter across the 41 W lamp reads 135 V, and the voltmeter across the 4.5 W lamp reads 15 V.

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

[1]

Answer:

  • Calculate current needed for both lamps to operate

P = IV

I = PV

I = 41135 = 0.3 A

I = 4.515 = 0.3 A

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

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

  • This means that the lamps will light at their normal brightness

  • Therefore, the correct answer is A [1 mark]

Examiner Tips and Tricks

You can use the mnemonic “Twinkle Twinkle Little Star, Power equals I squared R” to remember whether to multiply or divide by resistance in the power equations. When doing calculations involving electrical power, remember the unit is Watts W, therefore, you should always make sure that the time is in seconds

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Lindsay Gilmour

Author: Lindsay Gilmour

Expertise: Physics Content Creator

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.

Timothy Sit

Reviewer: Timothy Sit

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.