Power — OCR A-Level Physics
Test yourself on Power with OCR A-Level practice questions.
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Power explained
Power is the rate at which work is done or energy is transferred.
Read the full explanation
It is defined as the work done W divided by the time taken t, so P = W/t. The unit of power is the watt (W), where 1 W = 1 J s⁻¹. Power can also be expressed as the rate of energy transfer, so P = E/t. For example, if a machine does 600 J of work in 3 s, its power output is 600 / 3 = 200 W. In MCQ questions, you may calculate power, compare power ratings, or relate power to work and time. Remember that power is a scalar quantity and can be calculated from the gradient of a work-time graph, while the area under a power-time graph represents the total work done or energy transferred.
(b) P = Fv
Power is the rate of doing work or transferring energy. When a constant force F acts on an object moving at constant velocity v in the direction of the force, the mechanical power delivered is P = Fv. This follows from P = W/t and W = Fs, so P = Fs/t = Fv. For example, a car engine provides a forward force of 2000 N while travelling at 15 m s⁻¹, so the useful output power is P = 2000 × 15 = 30 000 W = 30 kW. If the force and velocity are not in the same direction, only the component of force along the velocity contributes, so P = Fv cos θ. In this course, you apply P = Fv to constant-speed situations and to instantaneous power when F and v are known at a particular moment.
(c) efficiency of a mechanical system; efficiency = (useful output energy / total input energy) × 100%
Efficiency compares the useful output energy from a mechanical system with the total input energy supplied to it. It is calculated as efficiency = (useful output energy / total input energy) × 100%. Because energy is conserved, the useful output energy is always less than the total input energy in a real system, so efficiency is less than 100%. For example, an electric motor may be supplied with 500 J of electrical energy and transfer 350 J as useful kinetic energy; its efficiency is (350 / 500) × 100% = 70%. The remaining 150 J is dissipated, often as thermal energy due to friction or resistance. Efficiency can also be expressed as a decimal or as a ratio, but the specification requires the percentage form. When comparing power instead of energy, the same ratio applies if the time interval is the same.
(a) the equations P = VI, P = I²R and P = V²/R
Electrical power is the rate of energy transfer in a component. Starting from P = VI, substitute V = IR to obtain P = I²R, or substitute I = V/R to obtain P = V²/R. Each form is useful when a different pair of quantities is known. For a 12 V lamp drawing 0.50 A, P = VI = 12 × 0.50 = 6.0 W; its resistance is 24 Ω, so P = I²R = 0.50² × 24 = 6.0 W and P = V²/R = 12² ÷ 24 = 6.0 W. Choose the form matching the data: use P = I²R when current and resistance are known, and P = V²/R when voltage and resistance are known. In multiple-choice questions, check units and significant figures before selecting an option.
(b) energy transfer; W = V I t
Energy transferred by an electrical component equals power multiplied by time, so W = V I t. Power P = VI, and energy W = Pt, giving W = VIt. For a 230 V heater drawing 4.0 A for 300 s, W = 230 × 4.0 × 300 = 276 000 J, or 2.76 × 10⁵ J. The same relationship links to P = I²R and P = V²/R, so W = I²Rt and W = V²t/R are also valid when resistance is known. In multiple-choice questions, watch the time unit: seconds give joules, while hours give watt-hours. Convert minutes to seconds by multiplying by 60 before substituting.
(c) the kilowatt-hour (kW h) as a unit of energy; calculating the cost of energy.
The kilowatt-hour is a practical unit of energy: one kW h is the energy transferred by a 1 kW device operating for 1 hour. Since 1 kW = 1000 W and 1 h = 3600 s, 1 kW h = 1000 × 3600 = 3.6 × 10⁶ J. To find energy in kW h, multiply power in kilowatts by time in hours. For a 2.0 kW heater used for 3.0 hours, energy = 2.0 × 3.0 = 6.0 kW h. If electricity costs 30 p per kW h, total cost = 6.0 × 30 = 180 p, or £1.80. In multiple-choice questions, check whether power is given in watts or kilowatts and whether time is in hours or seconds before calculating.
Your focus
- Define power as the rate of doing work or transferring energy.
- Recall and apply the equation P = W/t to calculate power.
- State the unit of power (watt) and express it in terms of joules per second.
Show all 18 objectives
- Recall and apply the equation P = Fv to calculate mechanical power.
- Convert between units of speed and power correctly in calculations.
- Explain the conditions under which P = Fv applies, including constant force and velocity in the same direction.
- Calculate the efficiency of a mechanical system using energy or power values.
- Distinguish between useful output energy and total input energy in a given system.
- Explain why efficiency is always less than 100% in real mechanical systems.
- State and use the three power equations P = VI, P = I²R and P = V²/R.
- Derive P = I²R and P = V²/R from P = VI and V = IR.
- Choose the most efficient equation form for a given set of known quantities.
- State and apply W = VIt to calculate energy transferred.
- Convert time units correctly before substitution.
- Distinguish between energy in joules and power in watts.
- Define the kilowatt-hour and convert it to joules.
- Calculate energy transferred in kW h from power and time.
- Determine the cost of electrical energy from a unit price.
Power exam tips
Marking Points
- Power is defined as the rate of doing work or the rate of energy transfer.
- The equation for power is P = W/t, where W is work done in joules and t is time in seconds.
- The unit of power is the watt (W), which is equivalent to one joule per second (J s⁻¹).
- Power can also be calculated using P = E/t, where E is energy transferred.
- Power is a scalar quantity and its unit can be expressed in base units as kg m² s⁻³.
- Power is the rate of energy transfer or work done, measured in watts (W), where 1 W = 1 J s⁻¹.
- For a constant force F acting in the direction of motion at constant velocity v, the mechanical power is given by P = Fv.
- The equation is derived from P = W/t and W = Fs, giving P = Fs/t = Fv when v = s/t is constant.
- If force and velocity are not parallel, use the component of force along the velocity: P = Fv cos θ.
- Typical application: a vehicle moving at constant speed against resistive forces, where the driving force equals the total resistive force.
- Efficiency is the ratio of useful output energy to total input energy, expressed as a percentage.
- The equation is efficiency = (useful output energy / total input energy) × 100%.
- Useful output energy is the energy transferred to the intended form, such as kinetic energy or gravitational potential energy.
- Total input energy is the total energy supplied to the system, including energy eventually dissipated as heat or sound.
- Efficiency is always less than 100% for real systems because some energy is dissipated to the surroundings.
- Efficiency can also be calculated using power values if the time interval is the same: efficiency = (useful output power / total input power) × 100%.
- States that P = VI gives power as the product of potential difference and current.
- Derives P = I²R by substituting V = IR into P = VI.
- Derives P = V²/R by substituting I = V/R into P = VI.
- Selects the appropriate form for the quantities given in a calculation.
- Applies correct units: watts for power, volts for potential difference, amperes for current and ohms for resistance.
- States that energy transferred W equals power multiplied by time.
- Uses W = VIt with potential difference in volts, current in amperes and time in seconds.
- Recognises that W = I²Rt and W = V²t/R follow from substituting the power equations.
- Converts time to seconds before calculating energy in joules.
- Interprets the result as energy in joules, not power in watts.
- Defines the kilowatt-hour as the energy transferred by a one-kilowatt device in one hour.
- Converts between kW h and joules using 1 kW h = 3.6 × 10⁶ J.
- Calculates energy in kW h by multiplying power in kW by time in hours.
- Calculates cost by multiplying energy in kW h by the unit price.
- Handles unit conversions between watts and kilowatts and between minutes and hours.
Examiner Tips
- 💡In multiple-choice questions, check the units of the given quantities; if time is in minutes, convert to seconds before using P = W/t.
- 💡Remember that power can be calculated from the gradient of a work-time graph, and the area under a power-time graph gives the work done.
- 💡If a question asks for the power of a device, ensure you use the work done by the device, not the total energy input if efficiency is involved.
- 💡Always write the equation P = Fv, substitute values with units, and state the unit of power as W or J s⁻¹.
- 💡If a velocity is given in km h⁻¹, convert to m s⁻¹ by dividing by 3.6 before using P = Fv.
- 💡In multiple-choice questions, check whether the force and velocity are in the same direction; if not, use the component of force along the velocity.
- 💡Link P = Fv to efficiency questions: useful output power may be less than input power, so calculate efficiency separately if needed.
- 💡Write the efficiency equation clearly and substitute values with units before calculating.
- 💡If the question gives power rather than energy, check that the time interval is the same for both input and output; if so, use power values directly.
- 💡Express the final answer as a percentage, and include the % symbol.
- 💡In multiple-choice questions, eliminate options greater than 100% immediately for real systems.
- 💡Write the chosen equation before substituting numbers so the examiner can follow your reasoning.
- 💡Check that the final unit is the watt and that the magnitude is sensible for the component described.
- 💡When two forms could be used, pick the one requiring the fewest rearrangements to reduce arithmetic slips.
- 💡Write the equation, substitute values with units, then evaluate to reduce errors.
- 💡Check whether the question asks for energy in joules or in kilowatt-hours before finalising the answer.
- 💡Estimate the expected order of magnitude to catch decimal-point slips.
- 💡Convert power to kilowatts and time to hours before multiplying to obtain kW h.
- 💡Keep the unit price consistent with the energy unit used in the calculation.
- 💡Check whether the answer should be in pence or pounds and convert if necessary.
Common Mistakes
- Confusing power with energy: power is the rate of energy transfer, not the total energy. Correction: check the units; energy is in joules, power in watts.
- Using time in minutes or hours without converting to seconds. Correction: always convert time to seconds before calculating power.
- Forgetting that the watt is a derived unit: 1 W = 1 J s⁻¹ = 1 kg m² s⁻³. Correction: use consistent units in calculations.
- Assuming power is always constant: in reality, power can vary with time. Correction: for varying power, use average power or calculus if required.
- Mistaking power for force or energy: power is the rate of energy transfer, not the total energy transferred. Correct by using P = Fv only when a rate is required.
- Using P = Fv with v as an average speed when the force is not constant: the equation applies to constant force and constant velocity, or to instantaneous values. Correct by checking that F and v are constant or by using instantaneous values.
- Forgetting to convert units such as km h⁻¹ to m s⁻¹ before substituting into P = Fv. Correct by converting all quantities to SI units first.
- Assuming P = Fv always gives the total power input: it gives the useful mechanical power output for the force considered. Correct by distinguishing input power from useful output power.
- Confusing useful output energy with total input energy: the useful output is always smaller. Correct by identifying which energy transfer is intended and which is wasted.
- Forgetting to multiply by 100% and leaving the answer as a decimal. Correct by converting the ratio to a percentage as required by the equation.
- Using the wrong units or mixing energy and power in the same calculation. Correct by ensuring both quantities are either energies in joules or powers in watts.
- Assuming efficiency can be greater than 100% if the output seems larger. Correct by recognising that energy is conserved and some is always dissipated in real systems.
- Squaring the wrong quantity: the error is writing P = IR²; the correction is P = I²R, where only the current is squared.
- Using P = V²/R when resistance is unknown: the correction is to use P = VI if current is known instead.
- Forgetting to square the current before multiplying by resistance: the correction is to evaluate I² first, for example 0.50² = 0.25, then multiply by R.
- Using time in minutes without converting: the correction is to multiply minutes by 60 to obtain seconds before substituting into W = VIt.
- Confusing energy with power: the correction is that power is the rate of transfer in watts, while energy is the total transferred in joules.
- Omitting one factor from the product: the correction is to multiply all three quantities V, I and t together, not just two of them.
- Treating kW h as a unit of power: the correction is that kW h measures energy, while kW measures power.
- Using power in watts directly with time in hours: the correction is to convert watts to kilowatts by dividing by 1000 first.
- Forgetting to convert pence to pounds when the question asks for cost in pounds: the correction is to divide the pence total by 100.