1.7.4 Power
Learning Objective
Students will be able to:
- Define power as the rate at which work is done or energy is transferred.
- Distinguish between average and instantaneous power (extension only).
- Apply the required formulae:
- \(P = \dfrac{W}{t}\) (average mechanical power)
- \(P = \dfrac{\Delta E}{t}\) (average power using an energy change)
- \(P = IV\), \(P = I^{2}R\), \(P = \dfrac{V^{2}}{R}\) (electrical power)
- \(\displaystyle \eta = \frac{P{\text{useful}}}{P{\text{input}}}\times100\%\) (efficiency)
- Interpret power ratings, relate power to energy (kWh) and calculate energy cost.
- Identify common pitfalls and use the formulae confidently in exam questions (AO1 & AO2).
Definition
Power (P) is the scalar quantity that measures how fast work is done or how quickly energy is transferred. It is the rate of doing work (or of an energy change) with respect to time.
Key Concepts
- Average power: \(P_{\text{avg}} = \dfrac{W}{t} = \dfrac{\Delta E}{t}\). Used when the work or energy change is constant over the time interval.
- Instantaneous power (extension): \(P = \dfrac{dW}{dt}\). Gives the power at a single instant; not required for the IGCSE but useful for deeper study.
- Power is a scalar (has magnitude only, no direction).
- Always keep the time in seconds (s) to obtain power in watts (W).
Units
| Quantity | Symbol | SI Unit | Derived unit |
|---|
| Power | P | watt | W = J s⁻¹ = N m s⁻¹ = V A |
| Work / Energy | W, ΔE | joule | J = N m = V C |
| Time | t | second | s |
Relationship to Work and Energy
Work is the product of force and displacement (\(W = Fd\)). Energy is the capacity to do work. Power adds the time factor:
\[
P_{\text{avg}} = \frac{\text{work done}}{\text{time taken}} = \frac{\text{energy transferred}}{\text{time taken}}.
\]
A larger power value means the same amount of work or energy is completed in a shorter time.
Electrical Power
When the energy transfer is electrical, the same definition applies. Using Ohm’s law (\(V = IR\)) we obtain three useful forms:
- \(P = IV\) (current × voltage)
- \(P = I^{2}R\) (useful for resistive heating)
- \(P = \dfrac{V^{2}}{R}\)
Power Rating, Energy (kWh) and Cost
Efficiency
Real devices rarely convert all input power into useful output. Efficiency (\(\eta\)) quantifies the proportion of useful power:
\[
\eta = \frac{P{\text{useful}}}{P{\text{input}}}\times100\%.
\]
Worked Example 1 – Mechanical Power
- A motor lifts a 50 kg mass vertically through a height of 10 m in 5 s. Find the average power output.
- Work done = increase in gravitational potential energy:
\[
W = mgh = (50\ \text{kg})(9.8\ \text{m s}^{-2})(10\ \text{m}) = 4\,900\ \text{J}
\]
- Average power:
\[
P_{\text{avg}} = \frac{W}{t} = \frac{4\,900\ \text{J}}{5\ \text{s}} = 980\ \text{W}
\]
- Result: the motor’s average power is 980 W (≈ 1 kW).
Worked Example 2 – Electrical Power
- A 240 V electric heater has a resistance of 30 Ω. Calculate the power it consumes.
- Current from Ohm’s law:
\[
I = \frac{V}{R} = \frac{240\ \text{V}}{30\ \Omega} = 8\ \text{A}
\]
- Power:
\[
P = IV = (8\ \text{A})(240\ \text{V}) = 1\,920\ \text{W}
\]
(or \(P = V^{2}/R = 240^{2}/30 = 1\,920\ \text{W}\)).
- Result: the heater’s power rating is 1.92 kW**.
Worked Example 3 – Energy Cost (kWh)
- A 1500 W electric kettle is used for 5 minutes each day. The electricity tariff is £0.30 kWh⁻¹. Find the monthly cost (30 days).
- Convert power to kilowatts: \(P = 1.5\ \text{kW}\).
- Time per day in hours: \(t = \frac{5}{60}=0.0833\ \text{h}\).
- Daily energy: \(E_{\text{day}} = P t = 1.5 \times 0.0833 = 0.125\ \text{kWh}\).
- Monthly energy: \(E_{\text{month}} = 0.125 \times 30 = 3.75\ \text{kWh}\).
- Cost: \(\text{Cost}=3.75 \times £0.30 = £1.13\) (to 2 dp).
Common Mistakes to Avoid
- Confusing average power (\(P=W/t\)) with instantaneous power – the former is an average over the whole interval.
- Using minutes, hours, or other non‑SI time units without converting to seconds (or to hours when calculating kWh).
- Treating the watt as unrelated to the joule; remember \(1\ \text{W}=1\ \text{J s}^{-1}\).
- Neglecting efficiency when comparing input and useful power (e.g., a motor’s shaft power vs. electrical input).
- Omitting the “k” when converting watts to kilowatts for kWh calculations.
Summary Table
| Concept | Formula | Typical Units |
|---|
| Average mechanical power | \(P = \dfrac{W}{t} = \dfrac{\Delta E}{t}\) | W (J s⁻¹) |
| Instantaneous power (extension) | \(P = \dfrac{dW}{dt}\) | W |
| Electrical power (general) | \(P = IV\) | W |
| Electrical power (resistive) | \(P = I^{2}R = \dfrac{V^{2}}{R}\) | W |
| Energy (kWh) | \(E\;(\text{kWh}) = P\;(\text{kW}) \times t\;(\text{h})\) | kWh |
| Efficiency | \(\displaystyle \eta = \frac{P{\text{useful}}}{P{\text{input}}}\times100\%\) | % |
Quick Reference – Power Ratings of Everyday Devices
- LED lamp: 5–15 W
- Incandescent bulb: 40–100 W
- Hair dryer: 1 200–2 000 W
- Electric kettle: 1 500–3 000 W
- Desktop computer (average): 200–400 W
Exam Tip (AO1 & AO2)
When a question asks for “the power of a device”:
- Identify whether the problem is mechanical (use \(P=W/t\) or \(P=\Delta E/t\)) or electrical (use \(P=IV\) etc.).
- Check that all quantities are in SI units (J, s, V, A, Ω).
- If the question involves running time, decide whether you need power (W) or energy (kWh). Remember to convert minutes to hours for kWh.
- State any assumptions (e.g., neglecting friction, 100 % efficiency) and, where relevant, comment on efficiency.
Key Take‑aways
- Power measures how fast work is done or energy is transferred.
- Use \(P = W/t\) or \(P = \Delta E/t\) for average power; the instantaneous form \(P = dW/dt\) is an extension.
- Electrical power follows the same definition and can be expressed as \(IV\), \(I^{2}R\) or \(V^{2}/R\).
- Power ratings tell you the maximum average power a device can deliver; combine them with time to obtain energy (kWh) and cost.
- Always keep time in seconds (for W) or hours (for kWh) and include efficiency when comparing input and useful power.