Power — AQA GCSE Combined Science
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Power explained
Power measures how quickly energy moves or how quickly work is done, so it always involves an amount divided by a time.
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Two devices may transfer the same total energy, but the more powerful one does it in less time. For example, a 2000 W kettle transfers 2000 J every second, while a 1000 W kettle transfers 1000 J every second; after 60 s the first has transferred 120 000 J and the second 60 000 J. Because energy transferred and work done are equivalent in this context, either wording may be used. The watt is the unit: 1 W = 1 J/s. A rate compares a quantity with time, so power is not the total energy transferred, and a large energy transfer over a long time can still mean low power.
power = energy transferred / time
This equation lets you calculate power from the energy transferred and the time taken. Power is measured in watts, energy transferred in joules and time in seconds, so 1 W = 1 J/s. To use it, identify the energy transferred and the time, convert time to seconds if needed, then divide energy by time. For example, a heater transferring 36 000 J in 60 s has power 36 000 J ÷ 60 s = 600 W. Rearranged, energy transferred = power × time and time = energy transferred ÷ power. The same relationship applies when work done is used instead of energy transferred, because work done and energy transferred are equivalent in this context. Always check that time is in seconds and energy is in joules before dividing.
P = E / t
Power measures how quickly energy is transferred. The equation P = E / t links power P in watts (W) to energy transferred E in joules (J) and time t in seconds (s). One watt equals one joule per second, so a 60 W lamp transfers 60 J each second. To use the equation, identify the energy transferred and the time taken, convert time to seconds and energy to joules, then divide. For example, a heater transferring 7200 J in 60 s has P = 7200 J ÷ 60 s = 120 W. Rearranged, E = P × t finds total energy and t = E / P finds time. This relationship applies to electrical devices, mechanical work and any energy transfer, making it central to efficiency and cost calculations.
power = work done / time
This statement expresses power as the rate of doing work. Work done is the energy transferred when a force moves an object, measured in joules (J), and time is measured in seconds (s). Power is therefore in watts (W), where 1 W = 1 J/s. For example, lifting a 20 N box through 3 m does 60 J of work; if this takes 4 s, the power is 60 J ÷ 4 s = 15 W. The equation is the same relationship as P = E / t because work done and energy transferred are equivalent. To solve problems, calculate work done using W = F × s when needed, then divide by time. Rearranged, work done = power × time and time = work done / power.
P = W / t
Power is the rate at which energy is transferred or work is done. The equation P = W / t links power P in watts to work done W in joules and time t in seconds. One watt equals one joule per second, so a 60 W lamp transfers 60 J each second. To use the equation, identify the energy transferred (or work done) and the time taken, convert time to seconds and energy to joules, then divide. For example, a motor doing 3000 J of work in 20 s has P = 3000 J ÷ 20 s = 150 W. Rearranged, W = P × t and t = W ÷ P. This relationship explains why the same work done faster requires greater power, and it applies to electrical devices, machines and muscles.
power, P, in watts, W
Power P is measured in watts W, where one watt is one joule per second. This unit tells you how quickly energy is transferred or work is done. A 2000 W kettle transfers 2000 J every second, while a 10 W lamp transfers only 10 J each second. In electrical circuits, power can also be calculated using P = V I, where V is potential difference in volts and I is current in amperes, or P = I² R. The watt is the same unit whichever equation is used. When answering questions, always give power in watts unless a question asks for kilowatts or megawatts, and convert carefully: 1 kW = 1000 W and 1 MW = 1 000 000 W.
energy transferred, E, in joules, J
Energy transferred, E, is the total amount of energy moved or converted during a process, measured in joules, J. One joule equals one newton metre, so lifting a 1 N weight through 1 m transfers 1 J. In power calculations, E is the numerator: power P = E ÷ t, so a 600 J transfer in 20 s gives 30 W. Energy is conserved, so the energy transferred to a device equals the useful energy plus any energy dissipated to the surroundings, often by heating. When a device runs for a known time, E = P × t. Always convert kilojoules to joules (1 kJ = 1000 J) and megajoules to joules (1 MJ = 1 000 000 J) before substituting. Record E in J, not W or N, and check the value is sensible for the context.
time, t, in seconds, s
Time, t, is the duration over which energy is transferred, measured in seconds, s. In power calculations, t is the denominator: P = E ÷ t, so 1200 J transferred in 40 s gives 30 W. Time must be in seconds before substitution, so convert minutes by multiplying by 60 and hours by multiplying by 3600. For example, a 2 kW kettle boiling for 3 minutes transfers E = 2000 W × 180 s = 360 000 J. Rearranging gives t = E ÷ P, useful when finding how long a device must run. Record t in s, not minutes or hours, and check the answer is sensible: a low-power device takes longer to transfer the same energy.
work done, W, in joules, J
Work done measures the energy transferred when a force moves an object along its line of action. The equation is W = F s, where W is work done in joules (J), F is force in newtons (N) and s is distance moved in metres (m). One joule equals one newton metre, so lifting a 2 N apple 0.5 m transfers W = 2 × 0.5 = 1 J. Work is a scalar; only the component of force along the displacement does work. If force and motion are perpendicular, no work is done. Energy transferred and work done are numerically equal, so the joule also measures energy. In power calculations, divide work done by time taken, giving P = W ÷ t. Always convert centimetres to metres and grams to kilograms before substituting, and check the unit of every quantity.
An energy transfer of 1 joule per second is equal to a power of 1 watt.
Power measures how quickly energy is transferred or work is done. One watt is defined as a transfer of one joule each second, so 1 W = 1 J/s. The equation is P = E ÷ t or P = W ÷ t, where P is power in watts (W), E or W is energy or work in joules (J) and t is time in seconds (s). For example, a lamp transferring 300 J in 20 s has power 300 ÷ 20 = 15 W. A 60 W bulb transfers 60 J every second. Rearranging gives E = P t, so a 2 kW heater running for 30 s transfers 2000 × 30 = 60 000 J. Convert kilowatts to watts by multiplying by 1000, and minutes to seconds by multiplying by 60, before substituting. Higher-tier students may also combine P = W ÷ t with W = F s to obtain P = F v.
Students should be able to give examples that illustrate the definition of power eg comparing two electric motors that both lift the same weight through the same height but one does it faster than the other.
Power is the rate of energy transfer, or the rate of doing work: P = E ÷ t, where E is energy transferred in joules and t is time in seconds, giving power in watts. Two motors lifting the same weight through the same height transfer the same energy, because energy transferred equals weight × height, so the slower motor takes longer and has lower power. The faster motor transfers the same energy in less time, so its power is greater. This shows power depends on both energy and time, not on energy alone. For example, lifting 600 N through 2 m transfers 1200 J; in 4 s that is 300 W, but in 2 s it is 600 W.
Students should be able to explain how the power transfer in any circuit device is related to the potential difference across it and the current through it, and to the energy changes over time:
Power is the rate of energy transfer. In a circuit device, power P in watts equals potential difference V across the device multiplied by current I through it (P = V I). Since V is energy transferred per unit charge and I is charge flow per second, their product gives energy per second. For a resistor, combining P = V I with V = I R gives P = I² R, showing power increases with the square of the current. Energy transferred over time is E = P t; a 60 W lamp on for 120 s transfers E = 60 × 120 = 7200 J. In a resistor, energy is transferred to the thermal store of the surroundings; in a motor, it is transferred mechanically to a kinetic store; in a lamp, it is transferred by radiation (light). These equations apply to any circuit device, whether energy is transferred usefully or by heating.
power = potential difference × current
Electrical power is the rate at which a component transfers energy. When a current I flows through a component with potential difference V across it, each coulomb of charge carries energy V and I coulombs pass each second, so the energy transferred each second is V × I. This gives power = potential difference × current, with power in watts (W), potential difference in volts (V) and current in amperes (A). For example, a 12 V lamp drawing 0.50 A has power 12 × 0.50 = 6.0 W, so it transfers 6.0 J every second. Rearranged, V = P ÷ I and I = P ÷ V, which lets you find an unknown when the other two quantities are known.
P = V I
P = V I is the symbolic form of the power equation: power equals potential difference multiplied by current. P is power in watts, V is potential difference in volts and I is current in amperes. The equation shows that a component transfers more energy each second when either the potential difference across it or the current through it increases. For example, a heater on a 230 V supply drawing 4.0 A has power P = 230 × 4.0 = 920 W. The same relationship rearranges to V = P ÷ I and I = P ÷ V. Because the equation is given in symbols, you must match each symbol to its quantity and unit before substituting, and you should quote the unit W with a numerical answer.
power = current ² × resistance
This relationship lets you calculate the power dissipated by a component when you know the current through it and its resistance. Power is the rate of energy transfer, measured in watts (W), where 1 W = 1 J/s. Current is measured in amperes (A) and resistance in ohms (Ω). Because the current is squared, doubling the current quadruples the power for the same resistance. For example, a 4 Ω resistor carrying 3 A dissipates P = 3² × 4 = 9 × 4 = 36 W. This is useful for heating effects in wires and resistors, and it links to P = VI and V = IR. Rearranged, I² = P ÷ R and R = P ÷ I². Always square the current before multiplying by resistance, and check units are A and Ω to obtain W.
P = I² R
This is the symbolic form of the power equation linking power P in watts, current I in amperes and resistance R in ohms. It applies to any component obeying Ohm's law and describes the rate at which electrical energy is transferred to thermal energy in a resistor. Because current is squared, the equation shows that increasing current has a much larger effect on power than increasing resistance. For example, if I = 2 A and R = 5 Ω, then P = 2² × 5 = 4 × 5 = 20 W. The equation can be rearranged as I = √(P ÷ R) and R = P ÷ I². It is consistent with P = VI and V = IR, since substituting V = IR into P = VI gives P = I²R. Use SI units throughout and remember that the square applies only to the current.
potential difference, V, in volts, V
Potential difference, symbol V, is the energy transferred per unit charge as charge moves between two points in a circuit. It is measured in volts, symbol V, so the quantity and its unit share the same letter. One volt means one joule of energy is transferred per coulomb of charge, linking V = W ÷ Q to the power equation P = V I. In a series circuit the supply potential difference is shared between components; in parallel, components across the same supply have equal potential differences. A voltmeter is connected in parallel across the component being measured, because it must compare the energy at two points. For example, a 12 V supply driving 2 A through a lamp transfers 24 J each second, since P = V I = 12 V × 2 A = 24 W.
current, I, in amperes, A (amp is acceptable for ampere)
Electric current, symbol I, is the rate of flow of electric charge, measured in amperes, symbol A; 'amp' is an acceptable word for ampere. One ampere is one coulomb of charge passing a point each second, so I = Q ÷ t. In the power equation P = V I, current must be in amperes before substitution. An ammeter measures current and is connected in series with the component, because the same current must pass through both the meter and the component. In a series circuit the current is the same at every point, while in parallel branches the currents sum to the total current from the supply. For example, if 30 C of charge passes a point in 10 s, then I = Q ÷ t = 30 C ÷ 10 s = 3 A.
resistance, R, in ohms, Ω
Resistance, measured in ohms (Ω), opposes the flow of electric current. In the context of electrical power, resistance is a key factor in determining the rate at which energy is transferred. The power P dissipated by a resistor is calculated using the equation P = I²R, where I is the current in amperes and R is the resistance in ohms. For example, if a current of 3 A flows through a 4 Ω resistor, the power dissipated as heat is 3² × 4 = 9 × 4 = 36 W. This equation highlights that power loss is proportional to the square of the current, which is crucial for understanding heating effects in circuits. Students must be able to recall and apply this equation, rearrange it to find I or R, and use the correct units.
Your focus
- Define power as the rate of energy transfer or the rate of doing work.
- State that the watt is the unit of power and relate 1 W to 1 J/s.
- Compare the power of two devices that transfer different amounts of energy in different times.
Show all 57 objectives
- Use power = energy transferred ÷ time to calculate power in watts.
- Convert time to seconds and energy to joules before substituting into the equation.
- Rearrange the equation to calculate energy transferred or time from the other two quantities.
- Recall and apply the equation P = E / t to calculate power, energy or time.
- Convert between units of time and energy correctly before substitution.
- Explain that a higher power means the same energy is transferred in less time.
- Apply power = work done / time to calculate power, work done or time.
- Calculate work done from force and distance and then use it in the power equation.
- Explain the difference between work done in joules and power in watts.
- Recall and use the equation P = W / t to calculate power, work done or time.
- Convert between units of time and energy correctly before substitution.
- Explain that a watt is one joule per second and interpret power as a rate.
- Recall that power is measured in watts and that 1 W = 1 J/s.
- Convert between watts, kilowatts and megawatts accurately.
- Use the watt correctly when calculating power in mechanical and electrical contexts.
- Define energy transferred, E, and state its unit as the joule, J.
- Calculate energy transferred using E = P × t with consistent SI units.
- Convert energy values between joules, kilojoules and megajoules accurately.
- Define time, t, and state its unit as the second, s.
- Convert times from minutes and hours into seconds accurately.
- Calculate time taken using t = E ÷ P with consistent SI units.
- Calculate work done using W = F s with correct units.
- Convert distances and masses into SI units before substitution.
- Explain why work done equals energy transferred and how it links to power.
- Define the watt as one joule per second.
- Calculate power, energy transferred or time using P = E ÷ t.
- Convert between watts and kilowatts and between seconds and minutes in power problems.
- Define power as the rate of energy transfer or the rate of doing work.
- Calculate power using P = E ÷ t with consistent units.
- Explain, using a worked example, why the faster of two motors lifting the same load has the greater power.
- State and use the equation P = V I to calculate power in a circuit device.
- Explain how P = V I relates to energy transfer per second and how E = P t gives energy transferred over time.
- Use P = I² R to explain how power transfer in a resistor depends on current and resistance.
- State the equation power = potential difference × current and identify the unit of each quantity.
- Calculate power from given values of potential difference and current, including after converting units.
- Rearrange the equation to determine an unknown potential difference or current from power and the other quantity.
- Match each symbol in P = V I to its quantity and unit.
- Use P = V I to calculate power, potential difference or current from the other two quantities.
- Interpret a calculated power value as the rate of energy transfer in joules per second.
- Select and apply power = current² × resistance to calculate power in a circuit.
- Rearrange the equation to determine current or resistance from power and one other quantity.
- Explain the effect of changing current on power dissipated at constant resistance.
- Apply P = I² R to calculate power, current or resistance in electrical circuits.
- Derive P = I² R from P = VI and V = IR.
- Interpret how current and resistance affect the rate of energy transfer in a component.
- Define potential difference as energy transferred per unit charge and state its unit, the volt (V).
- Apply V = W ÷ Q and P = V I to calculate potential difference, energy transferred, charge or power.
- Explain how a voltmeter is connected and interpret potential difference values in series and parallel circuits.
- Define electric current as the rate of flow of charge and state its unit, the ampere (A).
- Apply I = Q ÷ t and P = V I to calculate current, charge, time, power or potential difference.
- Explain how an ammeter is connected and interpret current values in series and parallel circuits.
- Recall and apply the equation P = I²R.
- Rearrange the power equation to solve for resistance or current.
- Explain the heating effect of current in a resistor using the power equation.
Power exam tips
Marking Points
- States that power is the rate of energy transfer or the rate of doing work, linking an amount of energy or work to the time taken.
- Recognises that the watt is the unit of power and that 1 W equals 1 J transferred each second.
- Explains that two devices transferring equal energy can have different powers if the times taken differ.
- Uses the idea of rate to compare devices, for example a 2000 W heater transfers energy twice as fast as a 1000 W heater.
- Distinguishes power from energy: energy is the total transferred, while power describes how quickly that transfer happens.
- Selects the equation power = energy transferred ÷ time and substitutes the correct values.
- Converts time to seconds and energy to joules before dividing, for example 3 minutes becomes 180 s.
- Calculates power correctly, such as 36 000 J ÷ 60 s = 600 W.
- Rearranges the equation to find energy transferred or time when those are the unknown quantities.
- States the unit of the answer as W and checks that the value is reasonable for the device described.
- State that power is the rate of energy transfer, measured in watts (W), where 1 W = 1 J/s.
- Select P = E / t and substitute the energy transferred in joules and the time in seconds correctly.
- Convert units before calculating, for example minutes to seconds or kilojoules to joules.
- Rearrange the equation to E = P × t or t = E / P when the question asks for energy or time.
- Give the answer with the correct unit, usually W, and a sensible number of significant figures.
- Define work done as energy transferred when a force moves an object, measured in joules (J).
- State that power is the rate of doing work and is measured in watts (W), where 1 W = 1 J/s.
- Use power = work done / time, substituting work done in joules and time in seconds.
- Calculate work done first using W = F × s when the question gives force and distance rather than energy.
- Rearrange to work done = power × time or time = work done / power when required.
- State that power is the rate of energy transfer or work done, measured in watts where 1 W = 1 J/s.
- Select and apply P = W / t, substituting work done in joules and time in seconds.
- Rearrange the equation correctly to W = P × t or t = W ÷ P when the question requires it.
- Convert units before substitution, for example minutes to seconds or kilojoules to joules, and give the answer with the correct unit.
- Interpret a numerical answer as the energy transferred each second, linking the value to the rate of transfer.
- State that power is measured in watts, W, and that 1 W = 1 J/s.
- Use the unit W correctly in calculations and final answers, including kW and MW conversions.
- Link the watt to rate of energy transfer, explaining that a higher power means more energy transferred each second.
- Apply electrical power relationships such as P = V I or P = I² R and give the answer in watts.
- Compare the power ratings of appliances and interpret what the values mean in terms of energy transferred per second.
- State that E is energy transferred, measured in joules, J, and that 1 J = 1 N m.
- Use E = P × t to calculate energy transferred when power in watts and time in seconds are known.
- Convert kJ to J by multiplying by 1000, and MJ to J by multiplying by 1 000 000, before substitution.
- Recognise that energy transferred to a device equals useful energy plus energy dissipated, usually by heating.
- Substitute values into P = E ÷ t correctly, keeping E in joules and t in seconds.
- Give the unit J with the numerical answer and reject answers left in W or N.
- State that t is time in seconds, s, and that it measures the duration of an energy transfer.
- Convert minutes to seconds by multiplying by 60, and hours to seconds by multiplying by 3600, before substitution.
- Use t = E ÷ P to find the time taken when energy transferred and power are known.
- Substitute time in seconds into P = E ÷ t, keeping E in joules and P in watts.
- Give the unit s with the numerical answer and reject answers left in minutes or hours.
- Interpret a longer time as a smaller power for the same energy transferred.
- States that work done is the energy transferred when a force moves an object through a distance along the force's line of action.
- Recalls and applies W = F s, substituting force in newtons and distance in metres to give work done in joules.
- Recognises that 1 J = 1 N m and that work done and energy transferred are numerically equal.
- Uses work done in power calculations by dividing by time, for example P = W ÷ t.
- Handles unit conversions correctly, such as 50 cm = 0.50 m, before calculating.
- Explains that a force perpendicular to the displacement does no work on the object.
- States that one watt equals one joule transferred per second, linking power to the rate of energy transfer.
- Applies P = E ÷ t or P = W ÷ t, using energy in joules and time in seconds to obtain power in watts.
- Rearranges the equation to find energy transferred, E = P t, or time taken, t = E ÷ P.
- Converts kilowatts to watts and minutes to seconds correctly before calculating.
- Interprets a power rating, such as 60 W, as 60 J transferred each second.
- Combines P = W ÷ t with W = F s to derive P = F v for a constant speed.
- States that power is the rate of energy transfer or the rate of doing work, with the equation P = E ÷ t.
- Explains that both motors transfer the same energy because they lift the same weight through the same height.
- Compares the times taken, identifying that the faster motor transfers the same energy in less time.
- Concludes that the faster motor has the greater power because power is energy transferred per second.
- Uses a numerical example, such as 1200 J in 4 s = 300 W compared with 1200 J in 2 s = 600 W.
- Power is the rate of energy transfer, measured in watts, where 1 W = 1 J/s.
- For any circuit device, power transfer is given by P = V I, where V is the potential difference across the device and I is the current through it.
- Because V is energy per unit charge and I is charge per second, the product V I gives energy transferred per second.
- For a resistor, P = I² R follows from P = V I and V = I R, showing that power depends on the square of the current.
- Energy transferred over a time interval is E = P t, so doubling the time doubles the energy transferred for a constant power.
- Energy may be transferred to thermal or kinetic stores, or transferred by pathways such as radiation (light), depending on the device.
- State that power is the rate of energy transfer, measured in watts, where 1 W = 1 J/s.
- Substitute the correct values into power = potential difference × current, keeping V in volts and I in amperes.
- Calculate correctly, for example 12 V × 0.50 A = 6.0 W, and give the unit W.
- Rearrange the equation to find potential difference (V = P ÷ I) or current (I = P ÷ V) when required.
- Recognise that for a fixed potential difference, increasing the current increases the power transferred.
- Identify P as power in watts, V as potential difference in volts and I as current in amperes.
- Substitute values into P = V I correctly, for example 230 V × 4.0 A = 920 W.
- Rearrange to V = P ÷ I or I = P ÷ V and calculate the unknown accurately.
- Give the answer with the correct unit, converting to kilowatts if the question requires it.
- Interpret the result as the rate of energy transfer, for example 920 W means 920 J transferred each second.
- State that power is the rate of energy transfer and is measured in watts (W), where 1 W = 1 J/s.
- Identify the equation power = current² × resistance and use it with current in amperes and resistance in ohms.
- Square the current value before multiplying by resistance, for example 3² × 4 = 36 W.
- Rearrange the relationship to find current or resistance when power is known, such as R = P ÷ I².
- Explain that for a fixed resistance, doubling the current increases power by a factor of four because current is squared.
- Identify each symbol: P is power in watts, I is current in amperes and R is resistance in ohms.
- Substitute numerical values into P = I² R, square the current and multiply by resistance.
- Rearrange the equation to make current or resistance the subject, for example I = √(P ÷ R).
- Show that P = I² R follows from combining P = VI with V = IR.
- Use the equation to compare power dissipation in components with different currents or resistances.
- State that potential difference is the energy transferred per unit charge between two points in a circuit.
- Recall that the unit of potential difference is the volt, symbol V, and that one volt equals one joule per coulomb.
- Use the equation V = W ÷ Q, or P = V I, correctly when potential difference, energy, charge or power values are given.
- Describe connecting a voltmeter in parallel across a component, and explain that this is needed to measure the difference in energy per charge between two points.
- Interpret circuit values: in series the supply potential difference is shared, while in parallel components across the same supply have the same potential difference.
- State that current is the rate of flow of electric charge, symbol I, measured in amperes, symbol A (or amps).
- Recall that one ampere equals one coulomb per second, linking I = Q ÷ t to the definition.
- Use I = Q ÷ t and P = V I correctly, converting time to seconds and charge to coulombs where needed.
- Describe connecting an ammeter in series with a component so that the current being measured passes through the meter.
- Interpret current values in series and parallel circuits, including that series current is the same throughout and parallel branch currents add to the supply current.
- State that resistance is measured in ohms (Ω) and opposes current flow.
- Recall and apply the equation P = I²R to calculate power, current, or resistance.
- Rearrange the equation correctly, for example, R = P ÷ I² or I = √(P ÷ R).
- Explain that power dissipated by a resistor is transferred as thermal energy to the surroundings.
- Recognise that doubling the current increases the power dissipated by a factor of four due to the I² term.
Examiner Tips
- 💡Define power using the word rate and name both energy transferred and work done as acceptable quantities.
- 💡When comparing appliances, quote their power values and state the time interval, such as energy per second.
- 💡Check that any calculation answer has the unit W and that the time used is in seconds.
- 💡Write down the equation, then substitute values with units before doing the arithmetic.
- 💡Convert minutes to seconds by multiplying by 60, and check the conversion before calculating.
- 💡If asked for energy or time, rearrange the equation first and then substitute to reduce errors.
- 💡Write the equation, then substitute values with units before calculating to reduce errors.
- 💡Check the unit asked for; if the answer is in kilowatts, divide watts by 1000.
- 💡Use the rearranged forms E = P × t and t = E / P when the question gives power and asks for energy or time.
- 💡Underline the quantity the question asks for so you choose the correct rearrangement.
- 💡Show the work-done calculation separately if force and distance are given, then divide by time.
- 💡Include the unit W with your final answer and check it is reasonable for the device described.
- 💡Write the equation, then substitute values with units before calculating so the examiner can follow your method.
- 💡Check that time is in seconds and energy is in joules; convert first, then calculate.
- 💡If the question asks for time or work done, rearrange the equation before substituting numbers to reduce errors.
- 💡Include the unit W with every power answer unless the question specifies kW or MW.
- 💡Show conversions clearly, for example 2.5 kW = 2500 W, before using the value in an equation.
- 💡When comparing appliances, refer to energy transferred per second rather than total energy used.
- 💡Underline the quantity asked for and its unit before calculating, so you know whether to find E in J or P in W.
- 💡Write the equation, substitute numbers with units, then give the answer with the unit J.
- 💡Check the size of your answer: a 2 kW heater running for 30 s transfers 60 000 J, not 60 J.
- 💡Convert all times to seconds before substituting into any power equation.
- 💡Show the conversion step, such as 4 min × 60 = 240 s, so the examiner can award method credit.
- 💡Check the answer is sensible: a 100 W lamp transferring 5000 J takes 50 s, not 50 min.
- 💡Write the equation, substitute numbers with units, then give the answer with the correct unit, for example W = 15 N × 3 m = 45 J.
- 💡Check whether the question asks for work done or power; if it gives a time, you may need to divide by time.
- 💡Show conversion steps clearly, such as 250 g = 0.25 kg, so the examiner can award method credit even if the final value slips.
- 💡Write the unit equation 1 W = 1 J/s at the start of a power calculation to keep the meaning clear.
- 💡Convert all values to watts, joules and seconds before substituting, and carry units through the working.
- 💡For a rearranged question, state the rearranged equation first, for example t = E ÷ P, then substitute.
- 💡Write the equation, substitute the values with units, then state the conclusion in words.
- 💡When comparing two situations, quote both calculated powers so the comparison is explicit.
- 💡Check that the unit of power is the watt, where 1 W = 1 J/s.
- 💡Write the equation you are using, substitute values with units, and give the unit with your answer.
- 💡When a question asks about energy changes over time, calculate E = P t and state the store or stores that gain energy, or the pathway used.
- 💡Write the equation, then substitute values with units before calculating, so an examiner can follow your method even if the final number slips.
- 💡Check the unit asked for; if the answer is required in kilowatts, divide the value in watts by 1000 at the end.
- 💡For rearrangement questions, rearrange the equation symbolically first, then substitute, to reduce arithmetic errors.
- 💡Underline the quantities given in the question and label them P, V or I before choosing the rearrangement you need.
- 💡Show the substituted equation with units, then the answer with its unit, so method and final value are both clear.
- 💡Estimate first: 230 × 4 is about 900, so an answer of 920 W is reasonable and a value of 92 W signals a slip.
- 💡Write the equation, substitute values with units, then calculate and give the answer in watts.
- 💡If the current is given in mA, divide by 1000 before squaring to avoid a large error.
- 💡Use the squared relationship to compare two cases: doubling current gives four times the power at constant resistance.
- 💡Quote the equation in symbol form, then substitute values clearly before evaluating.
- 💡Check whether the question asks for power, current or resistance and rearrange before substituting.
- 💡Use the equation to justify why high-current circuits need thicker wires to limit heating.
- 💡Always write the unit after a calculated value, for example 6.0 V, and check that the unit matches the quantity asked for.
- 💡When using P = V I, rearrange before substituting so the unknown is isolated, then substitute values with units.
- 💡In circuit questions, sketch the circuit and mark where the voltmeter connects before choosing a calculation method.
- 💡Check that current is in amperes before substituting into P = V I, converting milliamperes by dividing by 1000 if necessary.
- 💡Show the rearranged equation, the substitution and the answer with its unit to make each step clear.
- 💡Write down P = I²R before substituting any numbers to ensure you do not forget the squared term.
- 💡When finding current from power and resistance, do not forget the final step of taking the square root.
Common Mistakes
- Treating power as a total amount of energy rather than a rate; correct this by always dividing energy or work by time.
- Writing the unit of power as J or J s; correct this by using W, where 1 W = 1 J/s.
- Assuming a more powerful device always transfers more energy; correct this by noting that a low-power device running for longer can transfer more total energy.
- Dividing time by energy instead of energy by time; correct this by writing the equation with energy on top and time underneath.
- Using minutes or hours directly in the calculation; correct this by converting all times to seconds first.
- Forgetting to include the unit W with the final answer; correct this by writing the unit after every calculated power.
- Using time in minutes instead of seconds: always multiply minutes by 60 before dividing.
- Confusing power with energy: power is the rate in watts, while energy is the total in joules.
- Dividing time by energy instead of energy by time: check that P = E / t gives a larger power for a shorter time.
- Forgetting to calculate work done from force and distance before dividing by time: first find W = F × s.
- Mixing up work done with power: work done is in joules, power is in watts.
- Using distance instead of time in the denominator: the denominator must be time in seconds.
- Dividing time by work instead of work by time: correct by checking that P = W / t and that a larger time gives a smaller power for the same work.
- Forgetting to convert minutes to seconds: correct by multiplying minutes by 60 before substituting into the equation.
- Mixing up joules and watts in the final answer: correct by remembering that power is measured in watts and work done or energy in joules.
- Writing power in joules: correct by remembering that joules measure energy or work done, while watts measure the rate of energy transfer.
- Treating kW and W as interchangeable: correct by multiplying kW by 1000 to convert to W.
- Confusing power with energy in a comparison: correct by explaining that power is the rate, so a higher wattage transfers more energy each second.
- Writing the unit as J/s or W: correct this by remembering J is energy, while J/s and W are power.
- Substituting kilojoules directly into E = P × t: correct this by converting kJ to J first, for example 2.4 kJ = 2400 J.
- Confusing E with power P: correct this by checking whether the quantity is total energy in J or energy per second in W.
- Substituting minutes directly into P = E ÷ t: correct this by converting to seconds, for example 5 min = 300 s.
- Writing the unit as min or h: correct this by converting the final answer to seconds, or by stating the unit clearly as s.
- Confusing t with the number of repetitions or cycles: correct this by using the total duration of the energy transfer in seconds.
- Using the distance travelled rather than the distance moved along the line of the force; correct by resolving the force or using the component of displacement parallel to the force.
- Forgetting to convert centimetres to metres, giving an answer 100 times too large; correct by dividing centimetres by 100 before substituting.
- Confusing work done in joules with power in watts; correct by remembering work is energy transferred, while power is the rate of transfer, P = W ÷ t.
- Treating watts as a unit of energy rather than power; correct by stating that watts measure joules per second.
- Substituting time in minutes without converting to seconds, giving a power 60 times too small; correct by multiplying minutes by 60 first.
- Using P = E t instead of P = E ÷ t; correct by checking that dividing a larger energy by a longer time gives a smaller power.
- Saying the faster motor transfers more energy: correct this by noting both transfer the same energy because weight and height are unchanged, and only the time differs.
- Confusing power with force or energy: correct this by defining power as energy transferred per second, measured in watts.
- Using P = E × t instead of P = E ÷ t: correct this by checking that a shorter time must give a larger power for the same energy.
- Using P = V I for a device but forgetting that V must be the potential difference across that device, not the supply potential difference: correct this by identifying the device and using the potential difference across it.
- Confusing power and energy: correct this by stating that power is the rate of energy transfer in watts, while energy transferred is power multiplied by time in joules.
- Assuming all electrical energy is transferred usefully: correct this by stating that some energy is often transferred by heating to the surroundings, so the useful energy output is less than the total energy input.
- Multiplying current by time instead of by potential difference: power is V × I, while energy transferred is V × I × t, so check which quantity the question asks for.
- Using milliamperes or kilovolts directly in the equation: convert 250 mA to 0.25 A and 2 kV to 2000 V before substituting.
- Writing the unit of power as joules: power is measured in watts, and joules describe energy transferred, not the rate of transfer.
- Confusing the symbol I with the number 1 or with current in milliamperes: I stands for current in amperes, so convert mA to A first.
- Substituting power for potential difference or current: check that the value placed in V is in volts and the value placed in I is in amperes.
- Forgetting to square or otherwise adjust when using related equations: P = V I applies directly, while P = I²R and P = V² ÷ R are different forms for different known quantities.
- Multiplying current by resistance before squaring, giving I × R instead of I² × R. Correction: square the current first, then multiply by resistance.
- Using current in milliamperes or resistance in kilohms without converting. Correction: convert to A and Ω before substituting.
- Treating the equation as P = I × R² and squaring resistance instead of current. Correction: the squared term is current, so write I² R.
- Squaring the whole product IR instead of only the current. Correction: apply the square to I only, so P = I² × R.
- Forgetting to take the square root when rearranging for current. Correction: I = √(P ÷ R), not P ÷ R.
- Mixing units, such as using mA with Ω and expecting W. Correction: convert current to A before calculating.
- Writing the unit as 'v' or confusing the quantity symbol V with the unit symbol V; correction: both are capital V, but the quantity is potential difference and the unit is the volt.
- Connecting a voltmeter in series with a component; correction: a voltmeter is connected in parallel across the component so it compares two points.
- Treating potential difference and current as the same quantity; correction: current is the rate of flow of charge in amperes, while potential difference is energy transferred per unit charge in volts.
- Connecting an ammeter in parallel with a component; correction: an ammeter is connected in series so the current to be measured flows through it.
- Using minutes directly in I = Q ÷ t; correction: convert time to seconds before dividing, since one ampere is one coulomb per second.
- Forgetting to convert milliamperes (mA) to amperes (A) before calculating power or charge; correction: divide the mA value by 1000 to get amperes.
- Forgetting to square the current when calculating power. Correction: always calculate I² first before multiplying by R.
- Incorrectly rearranging to find current, such as I = P ÷ R. Correction: remember to take the square root, so I = √(P ÷ R).
- Using incorrect units for substitution. Correction: ensure current is in amperes (A) and resistance is in ohms (Ω) before calculating power in watts (W).