What is potential difference?

"Potential" refers to electrical potential energy, and "difference" is a reminder that you always compare two points. The pd across a lamp is the energy each coulomb gives to the lamp as it passes through. For a wider introduction, with the water analogy and everyday voltages, see our guide to what voltage is.

Potential difference definition and formula

At A-level, the AQA specification defines potential difference as work done per unit charge:

At GCSE, AQA gives the same idea rearranged, as "energy transferred = charge flow × potential difference":

Both say the same thing: the volt is a joule per coulomb. AQA also notes that questions are set using the term potential difference, but you get credit for correctly using either potential difference or voltage.

The equations you need with potential difference

Potential difference in series and parallel circuits

• Series: in AQA's words, the total potential difference of the power supply is shared between the components. Add up the pds across each component and you get the supply pd.
• Parallel: the potential difference across each component is the same, and equals the pd across the supply.

In energy terms, each coulomb leaves the supply with a fixed amount of energy. In series, it hands some to each component in turn until it is all used. In parallel, each coulomb goes through only one branch, so it hands all its energy to that branch.

How to measure potential difference

Potential difference is measured with a voltmeter connected in parallel with the component, one lead on each side of it. That is because pd is a difference between two points: the meter compares the two ends. A good voltmeter has a very high resistance, so almost no current flows through it and it barely changes the circuit. An ammeter, by contrast, goes in series.

Try placing the meters correctly in a circuit, then press Mark to check in our Electricity course.

Direct and alternating potential difference

AQA's GCSE asks you to explain the difference between direct and alternating potential difference.

• Direct potential difference stays in one direction, as from a cell or battery. It produces a direct current (dc).
• Alternating potential difference keeps reversing direction, as from the mains. It produces an alternating current (ac). UK mains alternates 50 times a second (50 Hz), at about 230 V.

In a plug, the potential difference between the live wire and earth is about 230 V. The earth wire sits at 0 V and stays idle unless a fault sends current through it.

Potential difference vs emf

At A-level you need to tell these apart. Both are measured in volts and both are energy per unit charge, but the energy goes in opposite directions.

Real batteries have internal resistance, r. Some of the emf is used pushing current through the battery itself, so the pd across its terminals, the terminal pd, is less than the emf whenever current flows. The difference, Ir, is often called the lost volts:

Worked examples

Example 1: energy from potential difference

15 C of charge flows through a heater with a pd of 6 V across it. How much energy is transferred? E = QV = 15 × 6 = 90 J.

Example 2: finding the potential difference

A motor transfers 360 J of energy when 30 C of charge passes through it. What is the pd across it? V = E ÷ Q = 360 ÷ 30 = 12 V.

Example 3: two equations together

A current of 3 A flows for 2 minutes through a lamp with a pd of 12 V. How much energy does the lamp transfer?

1. Convert the time: 2 minutes = 120 s.
2. Charge: Q = It = 3 × 120 = 360 C.
3. Energy: E = QV = 360 × 12 = 4,320 J.

A quicker route is E = Pt with P = IV = 36 W, giving 36 × 120 = 4,320 J. Getting the same answer two ways is a good check.

Example 4: sharing pd in series

A 4 Ω resistor and an 8 Ω resistor are in series with a 12 V supply. Total resistance is 12 Ω, so the current is 12 ÷ 12 = 1 A. The pds are V = IR = 1 × 4 = 4 V and 1 × 8 = 8 V, which add up to the supply's 12 V. The bigger resistor gets the bigger share.

Example 5: emf and internal resistance (A-level)

A battery has an emf of 9.0 V and an internal resistance of 0.5 Ω. It is connected to a 4.0 Ω resistor. Find the current, the terminal pd and the lost volts.

1. Current: I = ε ÷ (R + r) = 9.0 ÷ 4.5 = 2.0 A.
2. Terminal pd: V = IR = 2.0 × 4.0 = 8.0 V.
3. Lost volts: Ir = 2.0 × 0.5 = 1.0 V. Check: 8.0 + 1.0 = 9.0 V, the emf.

Example 6: accelerating an electron (A-level)

An electron is accelerated through a pd of 2,000 V. How much energy does it gain? The charge of an electron is 1.60 × 10⁻¹⁹ C, so E = QV = 1.60 × 10⁻¹⁹ × 2,000 = 3.2 × 10⁻¹⁶ J. This is why physicists use the electronvolt: the energy one electron gains through 1 V, so here 2,000 eV.

Exam tips

• Learn the definition word for word: energy transferred (or work done) per unit charge. Saying "the energy in the circuit" or "the force of the current" loses the mark.
• Write the equation first, then substitute, then give the unit. If an arithmetic slip creeps in, you can still earn method marks.
• Watch units: minutes to seconds before using Q = It, and mA to A before using V = IR.
• Say "across", not "through": pd is across a component, current flows through it.

Common mistakes

• Connecting the voltmeter in series. It must go in parallel, across the component.
• Thinking pd is shared in parallel. It is the same across each branch. Current is what splits.
• Mixing up emf and pd. Emf is energy given to charge by a source. Pd is energy transferred from charge to a component.
• Forgetting internal resistance. When current flows, a battery's terminal pd is less than its emf.
• Confusing charge and current. Charge is in coulombs. Current is coulombs per second, in amps.

Read next

Physics · Fields and gravityElectric potential: V = kQ/r, equipotentials and examplesElectric potential for A-level Physics: what it means, V = Q/4πε₀r for a point charge, potential difference and work done, equipotentials and examples.