Section 1 of 5Defining Power
Electrical power comes from combining the definitions of potential difference and current.
Rate of transfer
- Power is the energy transferred per unit time, measured in watts, where one watt equals one joule transferred every second.
Link to p.d.
- Potential difference is the energy transferred per coulomb of charge, so volts can be read as joules per coulomb.
Link to current
- Current is the charge passing per second, so multiplying volts by amps gives joules per coulomb times coulombs per second.
Basic equation
- Combining these ideas gives P = IV, which applies to any component that transfers energy, whether or not it obeys Ohm's law.
Section 2 of 5Three Forms of the Equation
Three Forms of the Equation
P = IV
- Use this when the current through the component and the p.d. across it are both known or both measured.
P = I²R
- Substitute V = IR; this form suits series circuits, where the same current passes through every resistor.
P = V²/R
- Substitute I = V/R; this form suits parallel branches, where each branch has the same p.d. across it.
Doubling effects
- For a fixed resistance, doubling either the current or the p.d. makes the power four times larger.
| Known quantities | Equation | Typical use |
|---|---|---|
| I and V | P = IV | Lamp with meter readings |
| I and R | P = I²R | Heating in a series resistor |
| V and R | P = V²/R | Mains appliance of known resistance |
Section 3 of 5Power in Real Circuits
1
Identify which two of current, p.d. and resistance the question gives you, because that decides which form of the power equation to use.
2
Check that every quantity is in base units, converting milliamps to amps and kilovolts to volts before substituting values.
3
Substitute into the chosen equation and calculate, keeping the unit watts and quoting a sensible number of significant figures.
4
In a series circuit the component with the largest resistance dissipates the most power because the current is identical in each.
5
Finally compare the answer with the component's rating, since a device operating above its rated power may overheat.
Section 4 of 5Worked Example
Question
- A heating element of resistance 46 Ω is connected to a 230 V supply.
- Calculate the current in the element and the power it dissipates.
- CurrentI = V/R = 230 / 46 = 5.0 A.
- Power using P = IVP = 230 × 5.0 = 1150 W.
- Check with P = V²/R230² / 46 = 52 900 / 46 = 1150 W, so the methods agree.
Answer I = 5.0 A and P = 1.15 kW (1150 W).
Section 5 of 5Exam Checklist & Tips
4.3.1 – Key Points for the Exam
- Power The rate at which energy is transferred, measured in watts (W).
- Watt A rate of energy transfer of one joule per second.
- Potential difference Energy transferred per unit charge between two points, in volts (V).
- Resistance The ratio V/I for a component, measured in ohms (Ω).
- Dissipated power Electrical energy converted to other forms, mainly heat, each second in a component.
Common Mistakes
- Using P = I²R in a parallel circuit without checking that the current is the current through that particular resistor.
- Forgetting to square the current or p.d., which gives an answer wrong by a large factor.
- Confusing power in watts with energy in joules; power is a rate, energy is power multiplied by time.
- Leaving milliamps or kilovolts unconverted before substituting into the equation.