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Electricity in the home: 1.6 Power, energy transfer and electricity cost calculation practice exam questions with worked out answers

 Doc Brown's Physics exam study revision notes

Exam practice questions on calculating power, energy transfer, calculations, problem solving the costing of using electrical power, calculating current flowing through an electrical appliance,  How to calculate the energy transfers in an electrical appliance? What do we mean by a unit of electricity used?  How do we calculate the cost of running an electrical device?

INDEX for physics notes on electricity in the home

This page contains online questions only. Jot down your answers and check them against the worked out answers at the end of the page

ANSWERS to ALL the QUESTIONS at the end of the page


1.6a Power, energy transfer and electricity cost calculation QUESTIONS
  • Know how to calculate how much energy is transferred by an appliance and how much the appliance costs to run.

  • Know and appreciate examples of energy transfers that everyday electrical appliances are designed to bring about.

  • Know that the amount of energy an appliance transfers depends on how long the appliance is switched on and its power.

    • The quantity of electricity that is transferred ('used') in an appliance depends on its power and how long you use it for ie time its switched on.

    • Energy is measured in joules (E in J) and power in watts (P in W)

      • 1 watt = 1 J of energy transferred in 1 second (1 W = 1 J/s)

      • Since a joule is a very tiny amount of energy, we often quote power in kilowatts (P in kW).

      • 1 kW = 1000 W = 1000 J/s

    • A bulb might be quoted with a 50W rating (50 J/s), an iron might be quoted as having a 500 W or 0.5 kW power rating (500 J/s, 0.5kJ/s) and a three bar electric fire might have a 3kW power rating (3 kJ/s, 3000 J/s).

    • However when dealing with large amounts of electrical energy its more convenient to think and calculate in kilowatt-hours (kWh).

      • 1 kilowatt-hour = the amount of electrical energy that a 1 kW appliance uses in 1 hour.

      • In fact, in terms of the electricity use in a house, the term unit on your electricity bill means a kilowatt hour and the price will quoted as eg '9p per unit', in other words you will pay 9p for every kilowatt-hour of electrical energy you use.

  • Three formulae for calculating power, energy transferred and other things too!

    • If unsure about units of potential difference (p.d. in V), current in amperes (amps, A) or resistance in ohms (Ω) then read the first section of Ohm's Law and calculations using V = IR

    • 1.6A power P (W) = current I (A) x potential difference V (V)

      • P = IV

      • P in W or J/s,  I in amps A, V p.d. in volts.

      • The more energy transferred in a given time, the greater the power of the device.

      • The p.d. V tells you how much energy each unit of electrical charge transfers (V = E/Q, J/C.

      • See Electricity Part  3 for E = QV calculations).

      • The current I tells you how much charge passes a given point in a circuit per unit time (coulombs/second, C/s).

      • This means both p.d. and current affect the rate at which energy is transferred to an appliance from the electrical energy store to another energy store.

      • Examples of calculations using P = IV

      • It is better to be able to rearrange a formula rather than using a triangle
      • AQ1 A 2 kW electric fire is connected to a 240 V supply.

        • Calculate the current flowing through the electric fire.

        • -

      • AQ2 The current flowing through an electric motor is 12 A.

        • -

      • AQ3 What p.d. must a power supply have, to produce a power output of 2.0 kW from a machine through which 12.0 A flows?

        • -

      • AQ4 A p.d. of 12.0 V is applied across the resistor of a device with a power of 8.0 W.

    • 1.6B power = current2 x resistance

      • P = IV and since V = IR, substituting for V gives P = I2R

      • P = I2R

      • P in W or J/s,  I in amps A and p.d. in volts V, R in ohms Ω.

      • (this is useful if you don't know the p.d., but, you must know the resistance instead)

      • Examples of calculations using P = I2R

      • It is better to be able to rearrange a formula rather than using a triangle
      • BQ1 A current of 20 A passes through a resistance of 10 Ω.

        • -

      • BQ2 A 2.0 kW electric fire has 4.0 A running through this heating appliance.

        • Calculate the resistance of the heating element.

        • -

      • BQ3 A 20 Ω electrical device transfers energy with a power of 500 W.

  • 1.6C Energy transferred by device = appliance power x time

    • The total energy transferred by an electrical appliance depends on the power of the appliance (in J/S = W) and the time it is used for ... giving the simple proportionality formula ...

    • E = Pt (but focussing on its 'electrical' connection, not platinum!)

      • E energy in J, P power in W or J/s, t time in s

    • rearrangements: P = E/t and t = E/P

    • It is better to be able to rearrange a formula rather than using a triangle
    • Application: power of appliance = electrical energy transferred / time used

    • Formula connection: Since P = IV, substituting for P gives energy transferred E = IVt

      • When electrical charge moves through a potential difference energy is transferred as work done against the electrical resistance (p.d.).

      • The energy of the charge comes from the power source (dc battery, ac mains electricity) which raises the potential energy of the electrons.

      • The charge, (usually electrons), 'falls'  through the p.d. across the components of a circuit, giving up its electrical potential energy to another energy store e.g. thermal, or other form of energy e.g. sound or light.

      • The energy transferred in an electrical device can be calculated from the formula:

      • energy transferred in joules = current in amps x potential difference in volts x time in seconds

      • E (J) = I (A) x V (V) x t (s),   E = IVt

      • If you increase the p.d. or the current flowing through a circuit, the more energy you can transfer in a given time.

    • You can use two different sets of units

    • (1st) The usual and familiar J, W and s.

      • E is energy transferred in joules, J

      • P is power in watts, W = J/s

      • t is time in seconds, s

      • Examples of calculations

      • CQ1 An 800 watt oven is used for one and a half hours.

        • How much energy in MJ is transferred to the thermal energy store of the oven?

        • -

      • CQ2 An electric heater transfers 1.5 MJ of energy every minute.

        • Calculate the power of the electric fire in kW.

        • -

      • CQ3 A rechargeable battery can deliver a total of 8.0 MJ of energy to a device.

        • If the device delivers a power output of 25 W, to the nearest hour, how long can it be used for?

        • -

      • CQ4 The p.d. across a resistor is 24.0 V. If a current of 3.0 A is flowing, how much energy is transferred in 5 minutes?

        • -

      • CQ5 A 1200 W toaster is used for a total of 10 minutes.

        • How much energy is transferred in this time?

        • -

      • CQ6 An appliance transfers 180 000 J of energy in two minutes.

        • Calculate the power of the appliance.

      • ANSWERS to CQ QUESTIONS at the end of the page

  • 1.6D The kilowatt-hour and cost of electricity

    • The practical everyday units e.g. on an appliance or electricity bill.

    • E is energy transferred in kilowatt-hours, kWh

    • P is power in kilowatts, kW (1 kW = 1000 J/s)

    • t is time in hours, h

    • Power equation: P = E/t,  E = P x t,  t = E/P

  • It is better to be able to rearrange a formula rather than using a triangle
  • The power formula triangle for the units of power in kilowatts (kW), units of energy in kilowatt-hours (kWh) and units of time in hours (h).

    • units of electricity are measured in kilowatt-hours e.g. for an appliance

    • kilowatt-hours = power in kW x time appliance used in hours

    • It is a measure of the energy transferred, and since 1 W = 1 J/s

    • 1 kWh = 1000 W x 3600 seconds = 3.6 X 106 J = 3.6 MJ

  • Be able to calculate the cost of using mains electricity given the cost per kilowatt-hour.

    • You should know this includes both the cost of using individual appliances and the interpretation of electricity meter readings to calculate total cost over a period of time.

      • 1 unit of electrical energy used = 1 kilowatt hour (kWh)

      • (i) Units of electricity used = power (kW) x time (hours)

      • (ii) Cost of electricity used = units x cost per unit

    • For an individual appliance/device, combining (i) and (ii) gives for example

      • cost (p) = power (kW) x time (hours) x electricity unit cost (p/unit)

    • DQ1 Examples of cost calculations:

      • (for the sake of argument we'll call the price of electricity 12p per unit)

      • (a) What is the cost per week of using a 40W light bulb for 36 hours in a week?

        • -

      • (b) What is the cost of doing a single wash in a machine rated at 2.5 kW if it takes 30 minutes to complete the washing and spin drying cycle?

        • -

      • (c)(i) Ignoring the standing charge, what is the quarterly bill cost to a household that uses 650 units of electrical energy in three months?

        • -

      • (d) How long could you run a 500 W plasma TV screen for 20p?

DQ2 More examples of cost calculations:

One unit of electricity is equal to using a 1000W (a power of 1kW) appliance for 1 hour.

The cost  of electricity (in 2022) has rocketed since I first designed these questions in 2002)

(a) What is the cost of using a 1.5kW heater for 2 hours if the cost of electricity is 7p/unit?

-

(b) What is the cost of using a 3 kW heater for 6 hours if the cost of electricity is 6p/unit?

-

(c) What is the cost of using a 60W light bulb for 20 hours if the cost of electricity is 5p/unit?

-

(d) What is the cost of using a 700W iron for 2.5 hours if the cost of electricity is 8p/unit?

-

(e) What is the cost of using a 200W hair dryer for 10 minutes if the cost of electricity is 6p/unit?

-

(f) What is the cost of using a 4kW oven for 5 hours if the cost of electricity is 7p/unit?

-

ANSWERS to DQ2 QUESTIONS at the end of the page

INDEX of ELECTRICITY Notes 1. Electricity in the home


Keywords, phrases and learning objectives for selected electricity calculations

Be able to do questions on calculating power, energy transfer, unit costs of electricity.

Know how to solve problems on costing of electrical power e.g. the cost of using an appliance for a given time.

Know how to calculate the current flowing through an appliance.

Check out your practical work you did or teacher demonstrations you observed, all of this is part of good revision for your module examination context questions and helps with 'how science works'.

  • Reading the electricity meter at home on a daily or weekly basis where you may have looked for trends in usage and try to explain these, eg in terms of weather conditions - extra usage of heating systems in cold weather,

  • Plan and carry out an investigation using an electrical joulemeter to measure the energy transferred by low voltage bulbs of different powers, low voltage motors and low voltage immersion heaters.


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INDEX of ELECTRICITY Notes 1. Electricity in the home

1.6b ANSWERS to power, energy transfer and electricity cost calculation question

AQ1 A 2 kW electric fire is connected to a 240 V supply.

  • Calculate the current flowing through the electric fire.

  • P = 2 kW = 2000 J/s

  • P = IV, I = P/V = 2000/240 = 8.33 A (3 sf)

AQ2 The current flowing through an electric motor is 12 A.

  • If it operates of a 24 V battery, what power does the electric motor generate?

  • P = IV = 12 x 24 = 288 W

AQ3 What p.d. must a power supply have, to produce a power output of 2.0 kW from a machine through which 12.0 A flows?

  • P = IV, V = P/I = 2000/12 = 167 V (3 sf)

AQ4 A p.d. of 12.0 V is applied across the resistor of a device with a power of 8.0 W.

  • (a) Calculate the current flowing through the device.

    • P = IV, I = P/V = 8.0/12.0 = 0.667 = 0.67 A (2 sf)

  • (b) Calculate the resistance of the device.

    • From Ohm's Law equation: V = IR,  so  R = V / I =12.0 / 0.667 = 18 Ω (2 sf)

BQ1 A current of 20 A passes through a resistance of 10 Ω.

  • What power is the circuit generating.

  • P = I2R = 202 x 10 = 4000 W = 4 kW

BQ2 A 2.0 kW electric fire has 4.0 A running through this heating appliance.

  • Calculate the resistance of the heating element.

  • P = I2R, R = P/I2 = 2000/42 = 2000/16 = 125 Ω

BQ3 A 20 Ω electrical device transfers energy with a power of 500 W.

  • Calculate the current flowing through the device.

  • P = I2R,  I = √(P/R) = √(500/20) = 5.0 A (3 sf)

CQ1 An 800 watt oven is used for one and a half hours.

  • How much energy in MJ is transferred to the thermal energy store of the oven?

  • 800 W = 800 J/s, time = 1.5 x 60 x 60 = 5400 s

  • E = Pt = 800 x 5400 = 4 320 000 J = 4 320 kJ = 4.32 MJ

CQ2 An electric heater transfers 1.5 MJ of energy every minute.

  • Calculate the power of the electric fire in kW.

  • 5.0 MJ = 5.0 x 106 J, time = 5 x 60 = 300 s

  • E = Pt,  P = E/t = 1.5 x 106/300 = 5000 W = 5.0 kW

CQ3 A rechargeable battery can deliver a total of 8.0 MJ of energy to a device.

  • If the device delivers a power output of 25 W, to the nearest hour, how long can it be used for?

  • P = 25 J/s, E = 8.0 x 106 J

  • E = Pt,  t = E/P = 8.0 x 106/25 = 3.2 x 105 s

  • 1 hour = 60 x 60 = 3600 s

  • 3.2 x 105/3600 = 89 hours

CQ4 The p.d. across a resistor is 24.0 V. If a current of 3.0 A is flowing, how much energy is transferred in 5 minutes?

  • time = 5 x 60 = 300 seconds.

  • E (J) = I (A) x V (V) x t (s),   E = IVt

  • E = 3.0 x 24.0 x 300 = 21 600 J

CQ5 A 1200 W toaster is used for a total of 10 minutes.

  • How much energy is transferred in this time?

  • P = E/t,  E = P x t, W = J/s

  • E = 1200 x 10 x 60 = 720 000 J = 7.2 x 105 J

CQ6 An appliance transfers 180 000 J of energy in two minutes.

  • Calculate the power of the appliance.

  • E = P x t, so P = E / t = 180 000 / (2 x 60) = 1500 W (1.5 kW)

DQ1 Examples of cost calculations:

(a) What is the cost per week of using a 40W light bulb for 36 hours in a week?

  • power = 40/1000 = 0.04 kW, units = kWh = 0.04 x 36

  • cost = 0.04 x 36 x 12 = 17.3p (3 sig. figs.)

  • It doesn't seem a lot, but throughout a house it soon adds up, so always switch unwanted lights!

(b) What is the cost of doing a single wash in a machine rated at 2.5 kW if it takes 30 minutes to complete the washing and spin drying cycle?

  • power 2.5 kW, time = 30/60 = 0.5 hour

  • units = 2.5 x 0.5

  • cost = 2.5 x 0.5 x 12 = 15p

(c)(i) Ignoring the standing charge, what is the quarterly bill cost to a household that uses 650 units of electrical energy in three months?

  • cost = 650 x 12 = 7800p or 78

  • (c)(ii) If the family wants to cut the cost of the quarterly bill to 60, what is the maximum number of units of electrical energy they can use?

  • cost = units x unit cost, so units = cost/unit

  • cost = 60/12p = 6000/12 = 500 units

(d) How long could you run a 500 W plasma TV screen for 20p?

  • cost (p) = power (kW) x time (hours) x electricity unit cost (p/unit)

  • power = 500/1000 = 0.5 kW

  • rearranging the equation

  • time = cost / (power x unit cost)

  • time = 20 / (0.5 x 12) = 20 / 6 = 3.33 hours (3 hours 20 minutes)

DQ2 More examples of cost calculations:

One unit of electricity is equal to using a 1000W (a power of 1kW) appliance for 1 hour.

The cost  of electricity (in 2022) has rocketed since I first designed these questions in 2002)

(a) What is the cost of using a 1.5kW heater for 2 hours if the cost of electricity is 7p/unit?

Answer: cost = power x time x cost per unit = 1.5 x 2 x 7 = 21p

(b) What is the cost of using a 3 kW heater for 6 hours if the cost of electricity is 6p/unit?

Answer: cost = power x time x cost per unit = 3 x 6 x 6 = 108p (1.08)

(c) What is the cost of using a 60W light bulb for 20 hours if the cost of electricity is 5p/unit?

Answer: cost = power x time x cost per unit = (60/1000) x 20 x 5 = 21p

(don't forget to change W into kW)

(d) What is the cost of using a 700W iron for 2.5 hours if the cost of electricity is 8p/unit?

Answer: cost = power x time x cost per unit = (700/1000) x 2.5 x 8 = 14p

(don't forget to change W into kW)

(e) What is the cost of using a 200W hair dryer for 10 minutes if the cost of electricity is 6p/unit?

Answer: cost = power x time x cost per unit = (200/1000) x (10/60) x 6 = 0.2p

(don't forget to change W into kW and minutes into hours)

(f) What is the cost of using a 4kW oven for 5 hours if the cost of electricity is 7p/unit?

Answer: cost = power x time x cost per unit = 4 x 5 x 7 = 140p (1.40)

INDEX of ELECTRICITY Notes 1. Electricity in the home

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