ASPhysicsCIERevision Notes9. Electricity9.2 Potential Difference & PowerElectrical Power

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First exams 2025

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Electrical Power (CIE AS Physics)

Revision Note

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Katie M

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Calculating electrical power

  • Power is defined as the rate of doing work
  • Potential difference is the work done per unit charge
  • Current is the rate of flow of charge
  • Therefore, the power dissipated (delivered) by an electrical device is defined as:

P space equals space I V

  • Where:
    • P = power, measured in W
    • I = current, measured in A
    • V = potential difference, measured in V

  • Resistance is given by the equation:

R space equals space V over I space space rightwards double arrow space space V space equals space I R space space rightwards double arrow space space I space equals fraction numerator space V over denominator R end fraction

  • Where:
    • R = resistance, measured in Ω
    • V = potential difference, measured in V
    •  I = current, measured in A
  • The expression for potential difference can be substituted into the power equation to give:

P space equals space I V space equals space I open parentheses I R close parentheses space

P space equals space I squared R

  • Or the expression for current can be substituted into the power equation to give:

P space equals space I V space equals space open parentheses V over R close parentheses V

P space equals space V squared over R

  • This shows that for a given resistance, if the current or potential difference were doubled, the power would be four times greater 

Worked example

Two lamps are connected in series to a 150 V power supply.WE - power question image, downloadable AS & A Level Physics revision notesWhich 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: State the power equation

P = IV

Step 2: Rearrange for I

I space equals straight P over straight V

Step 3: Substitute in values to calculate

  • For the 41W lamp:

 I space equals fraction numerator 41 straight W over denominator 135 straight V end fraction equals 0.3 space straight A

  • For the 4.5W lamp:

 I equals fraction numerator 4.5 straight W over denominator 15 straight V end fraction equals 0.3 space straight 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
  • Therefore, the lamps will light at their normal brightness
  • This is option A

Examiner Tip

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

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.