2.4.1 Electrical power
Power: how fast a device transfers energy
Power
The rate at which energy is transferred, measured in watts (W\text{W}W).
- Every circuit device transfers energy from one store to another; a lamp transfers energy electrically from the supply and raises the thermal energy store of the lamp and its surroundings, while also transferring some energy by light.
- The power of a device tells you how quickly it makes that transfer.
- A power of one watt means energy is transferred at a rate of one joule per second.
- So a higher-power device transfers more energy every second than a lower-power one.
Power from potential difference and current
- The power transferred by an electrical device depends on the potential difference across it and the current through it.
- These are linked by P=VIP = VIP=VI, where PPP is power in watts (W\text{W}W), VVV is potential difference in volts (V\text{V}V), and III is current in amperes (A\text{A}A).
- Current is the rate of flow of charge, so a larger current means more charge passes each second.
- Potential difference is the energy per unit charge, so a larger potential difference means each unit of charge carries more energy.
- Increasing either quantity therefore increases the power transferred.
A motor has a potential difference of 12 V12\ \text{V}12 V across it and a current of 3.5 A3.5\ \text{A}3.5 A through it. Calculate its power.
- Write the equation: P=VIP = VIP=VI.
- Substitute the values: P=12×3.5P = 12 \times 3.5P=12×3.5.
- Calculate, with the unit: P=42 WP = 42\ \text{W}P=42 W.
- The motor transfers energy at a rate of 42 J42\ \text{J}42 J each second.
Power from current and resistance
- Power can also be found from the current through a device and its resistance: P=I2RP = I^2RP=I2R, where PPP is power in watts (W\text{W}W), III is current in amperes (A\text{A}A), and RRR is resistance in ohms (Ω\OmegaΩ).
- The current is squared, so changing the current has a large effect on the power.
- If the current doubles while the resistance stays the same, the power becomes four times larger because 22=42^2 = 422=4.
A heating element has a resistance of 20 Ω20\ \Omega20 Ω and a current of 2.0 A2.0\ \text{A}2.0 A through it. Calculate the power.
- Write the equation: P=I2RP = I^2RP=I2R.
- Substitute the values: P=(2.0)2×20P = (2.0)^2 \times 20P=(2.0)2×20.
- Square the current first: (2.0)2=4.0(2.0)^2 = 4.0(2.0)2=4.0.
- Complete the calculation: P=4.0×20=80 WP = 4.0 \times 20 = 80\ \text{W}P=4.0×20=80 W.
- The element transfers energy at a rate of 80 J80\ \text{J}80 J each second.
- Do not confuse power with energy: energy is how much is transferred, power is how quickly.
- In P=I2RP = I^2RP=I2R, only the current is squared, so work out I2I^2I2 before multiplying by the resistance, and never square the resistance.
- Use the potential difference across the device and the current through the device.
- Pick the equation that matches the quantities you are given: use P=VIP = VIP=VI for potential difference and current, and P=I2RP = I^2RP=I2R for current and resistance.
- Write the equation, substitute with correct units, show each step, and give the answer in watts. Both equations must be recalled and may need rearranging.
- What does the power of an electrical device tell you?
- Write the equation linking power, potential difference and current.
- Write the equation linking power, current and resistance.
- State the units of power, potential difference, current and resistance.
- If the current doubles at constant resistance, by what factor does the power change?