Current, potential difference and resistanceEdexcel International A Level Physics: Revision notes

Section 1

Electric current and charge

Electric current is the rate of flow of charged particles. In a metal the charge carriers are conduction electrons; in an electrolyte or a gas they are ions. The relationship is

$I = \frac{Q}{t}$

where $I$ is the current in amperes (A), $Q$ the charge in coulombs (C) and $t$ the time in seconds (s). So 1 A = 1 C s⁻¹. For a varying current, $I = \Delta Q/\Delta t$ .

Charge is quantised: $Q = Ne$ , where $e = 1.60 \times 10^{-19}$ C is the elementary charge and $N$ the number of charge carriers. Conventional current is drawn from the positive to the negative terminal, opposite to the direction of electron flow.

Key termscurrentcoulombcharge carrier

Common mistake

Converting time: t must be in seconds. A current of 0.50 A for 3.0 minutes transfers 0.50 × 180 = 90 C, not 0.50 × 3.0 = 1.5 C.

Section 2

Potential difference

The potential difference (p.d.) between two points is the energy transferred per unit charge when charge moves between them:

$V = \frac{W}{Q}$

where $W$ is the energy transferred (J) and $Q$ the charge (C). The unit is the volt: 1 V = 1 J C⁻¹. In a component, electrical energy is transferred to other forms (thermal, kinetic, light), so $W = VQ$ . Combining with $Q = It$ gives $W = VIt$ .

Key termspotential differencevolt

Exam tip

If a question gives a p.d. and a current for a time, find Q = It first, then W = VQ. Check the time is in seconds.

Section 3

Resistance

The resistance of a component is the ratio of the p.d. across it to the current through it:

$R = \frac{V}{I}$

The unit is the ohm (Ω): 1 Ω = 1 V A⁻¹. This is a definition, so it applies to every component at every p.d., whether or not the current is proportional to the p.d. A large resistance means a small current for a given p.d.

Resistance arises because the charge carriers collide with the vibrating ions of the material and lose energy, which transfers thermal energy to the surroundings.

Key termsresistanceohm

Common mistake

R = V/I at a point on an I–V graph is not the gradient of the graph unless the line is straight through the origin.

Section 4

Ohm's law

Ohm's law states that the current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature is constant.

So an ohmic conductor has a straight-line I–V graph through the origin, and V/I (the resistance) is constant. Ohm's law is therefore a special case of the definition $R = V/I$ : it describes conductors where $R$ stays constant. A filament lamp is non-ohmic because the current heats the filament, the ions vibrate more, the electrons collide more often and the resistance rises.

Key termsOhm's lawohmic conductor

Exam tip

Always include the phrase 'at constant temperature' when stating Ohm's law.

Section 5

Worked example

A 12 V supply drives a current of 0.40 A through a resistor for 2.0 minutes.

Charge: $Q = It = 0.40 \times 120 = 48$ C
Energy: $W = VQ = 12 \times 48 = 576$ J
Resistance: $R = V/I = 12/0.40 = 30$ Ω

If the p.d. is doubled and the resistor is ohmic (temperature constant), the current doubles to 0.80 A.

Must Know

I = Q/t, so 1 A = 1 C s⁻¹; Q = Ne
V = W/Q, so 1 V = 1 J C⁻¹; W = VQ = VIt
R = V/I defines resistance for every component
Ohm's law: I ∝ V for a metallic conductor at constant temperature
Ohmic: straight I–V line through the origin; filament lamp: R rises with temperature

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Current, potential difference and resistance

A student connects a small electric motor to a 6.0 V battery of negligible internal resistance. The motor draws a steady current of 0.50 A for 3.0 minutes, and the whole 6.0 V is across the motor.
Calculate the number of electrons that pass through the motor in this time. (Elementary charge e = 1.60 × 10⁻¹⁹ C.)2 marks
A student investigates a length of nichrome wire kept at a constant temperature in a water bath. She measures the current for different potential differences across the wire and records: 0.10 A at 0.50 V, 0.20 A at 1.00 V, 0.30 A at 1.50 V and 0.40 A at 2.00 V.
Use the student's data to decide whether the wire obeys Ohm's law. Justify your answer.2 marks
During a thunderstorm a lightning flash lasts 2.0 ms. A total charge of 24 C is transferred between the cloud and the ground through a potential difference of 1.5 × 10⁸ V. A metal lightning conductor on a nearby building carries the discharge and has a resistance of 0.015 Ω.
Calculate the mean current during the flash and the energy transferred.3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).