Define potential difference as energy per unit charge, connect it to resistance through Ohm's law, and calculate power dissipated in resistors.
AQA A-Level Physics · Unit 5: ElectricityWhen charge flows through a component, work is done (energy is transferred). The potential difference (p.d.) across a component is defined as the work done per unit charge in moving charge through it:
| Symbol | Quantity | Unit |
|---|---|---|
| V | Potential difference (voltage) | V (volts) |
| W | Work done (energy transferred) | J |
| Q | Charge | C |
Rearranging: W = QV = IVt. This shows that the energy transferred (work done) depends on the p.d., the current, and the time.
The concept of potential difference is related to electric potential: the p.d. between two points equals the difference in electric potential between them. Potential is measured relative to an arbitrary zero (often earth).
The resistance of a conductor measures how much it opposes the flow of current. It is defined as the ratio of p.d. to current:
| Symbol | Quantity | Unit |
|---|---|---|
| R | Resistance | Ω (ohms) |
| V | Potential difference across component | V |
| I | Current through component | A |
An ohmic conductor has a constant resistance — its V-I graph is a straight line through the origin. Examples: metallic conductors at constant temperature, resistors.
A non-ohmic conductor has a resistance that changes with conditions. Examples: filament lamp (resistance increases with temperature), diode (one-way), thermistor (resistance changes with temperature).
The rate at which energy is dissipated in a component is the electrical power. Starting from P = W/t = VQ/t = VI:
| Symbol | Quantity | Unit |
|---|---|---|
| P | Power dissipated | W (watts) |
| V | Potential difference | V |
| I | Current | A |
| R | Resistance | Ω |
The three forms are obtained by substituting V = IR or I = V/R:
Energy transferred in time t: E = Pt = VIt = I²Rt
When current flows through a resistor, electrical energy is converted to thermal energy (heat). This is the heating effect of current. The rate of heating is P = I²R — proportional to the square of the current.
Applications exploiting the heating effect:
In power transmission, energy is lost as heat in cables. Since P_loss = I²R_cable, transmission at high voltage (and therefore low current for the same power P = VI) minimises these losses. This is why the National Grid uses very high voltages (~400 kV) for long-distance transmission.
The kilowatt-hour (kWh) is a practical unit of energy used in electricity billing: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 MJ.
Q1. A resistor has a p.d. of 6.0 V across it and a current of 0.3 A through it. What is its resistance?
Q2. What is the power dissipated in a 100 Ω resistor carrying a current of 0.2 A?
Q3. Why does the National Grid transmit electricity at very high voltages (hundreds of kV)?
Q4. A 60 W light bulb is run for 10 hours on a 230 V supply. Calculate the energy used in joules and in kWh.
Q5. A resistor obeys Ohm's law. If the p.d. across it is doubled, by what factor does the power dissipated change?
Challenge Q1. A student uses a 12 V battery to heat a 5.0 Ω resistor for 2 minutes. (a) Calculate the energy dissipated. (b) If the battery has an internal resistance of 0.5 Ω, what fraction of the total energy is wasted in the battery?
Challenge Q2. Derive the formula P = I²R starting from P = VI and Ohm's law (V = IR). Then derive P = V²/R. State clearly which formula to use in each situation.
Challenge Q3. A 100 W, 230 V rated light bulb burns out after 1000 hours. Another 100 W, 230 V rated bulb is accidentally connected to a 115 V supply. What power does it now dissipate? Assume resistance is constant.