Electrical Power (Cambridge O Level Physics)

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Electrical Power Equation

  • 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 space equals fraction numerator space E over denominator t end fraction space equals fraction numerator space W over denominator t end fraction

  • Where:
    • P = power in watts (W)
    • E = energy in joules (J)
    • t = time in seconds (s)
    • W = work done in (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 space equals space I V

  • Where:
    • P = power in watts (W)
    • I = current in amps (A)
    • V = potential difference in volts (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 space equals space I squared R

P space equals fraction numerator space V squared over denominator R end fraction

  • Where:
    • P = power in watts (W)
    • I = current in amps (A)
    • R = resistance in ohms (Ω)
    • V = potential difference in volts (V)

  • 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 space equals space V I t

  • Where:
    • E = energy transferred in joules (J)
    • V = potential difference in volts (V)
    • I = current in amps (A)
    • t = time in seconds (s)

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

  • Calculate the current needed for both lamps to operate

P space equals space I V

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

    • For the 41 W lamp:

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

    • For the 4.5 W lamp

I space equals fraction numerator space 4.5 over denominator 15 end fraction space equals space 0.3 space straight A

  • For both lamps 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.

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

The Kilowatt Hour (kWh)

  • Energy usage in homes and businesses is calculated and compared using the kilowatt hour
  • The kilowatt hour is defined as:

A unit of energy equivalent to one kilowatt of power expended for one hour

   

  • Appliances are given power ratings, which tell consumers:

The amount of energy transferred (by electrical work) to the device every second

Power Rating for a Kettle

power-rating, IGCSE & GCSE Physics revision notes

This kettle uses between 2500 and 3000 W of electrical energy

  • This energy is commonly measured in kilowatt-hour (kW h), which is then used to calculate the cost of energy used

Calculating with kWh

  • The kilowatt hour can also be defined using an equation:

E space equals space P t

  • Where
    • E = energy (kWh)
    • P = power (kW)
    • t = time (h)
      • This equation is unusual because S.I. unit are not used, both energy and power are × 103, and time is in hours, not seconds

  • Since the usual unit of energy is joules (J), this is the 1 W in 1 s
    • Therefore:

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

    • Since 1 kW = 1000 W and 1 h = 3600 s

    

  • To convert between Joules and kW h:

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

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

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

Worked example

A cooker transfers 1.2 × 109 J of electrical energy to heat up a meal.

Calculate the cost of cooking the meal if 1 kW h costs 14.2p

Answer:

Step 1: Convert from J to kW h

left parenthesis 1.2 space cross times space 10 to the power of 9 right parenthesis space divided by space left parenthesis 3.6 space cross times space 10 to the power of 6 right parenthesis space equals space 333.333 space kW space straight h

Step 2: 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

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Leander

Author: Leander

Expertise: Physics

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.