EMF and Potential Difference

Spec mapping: OCR H556 Module 4.2 — Energy, power and resistance (definition of potential difference $V = W/Q$ across a component; definition of EMF $\varepsilon = W/Q$ for a source; the volt as J C $^{-1}$ ; distinction between EMF and pd in terms of direction of energy transfer; foundations for the internal-resistance treatment in lesson 9). Refer to the official OCR H556 specification document for exact wording.

So far in this course we've talked about charge ( $Q$ ), current ( $I = dQ/dt$ ), and drift velocity ( $I = nAvq$ ) — but we haven't yet asked the central physical question: what makes the current flow in the first place? What drives the charges around the circuit?

The "push" is provided by a source of EMF — typically a battery, cell, generator, solar panel or dynamo — and the energy transfer associated with moving the charges is the quantity we measure by potential difference. EMF ( $\varepsilon$ ) is the energy put in to the charges by the source (chemical energy → electrical, in a battery), per unit charge. Potential difference ( $V$ ) is the energy taken out of the charges by a component (electrical → heat / light / motion, in a resistor or lamp or motor), per unit charge.

At first glance EMF and pd look like the same thing — they have the same units (the volt) and they're often numerically close (a fresh AA cell measured with a high-impedance voltmeter reads almost exactly $1.5$ V whether you're measuring its EMF or the terminal pd). But they describe opposite physical processes, and the distinction is one of OCR's favourite multi-mark conceptual probes. This lesson nails the definitions, the differences, and the applications — including the foreshadowing of "lost volts" you'll need for internal resistance in lesson 9.

Defining Potential Difference

Potential difference (pd) across a component is defined as the energy transferred from the electrical energy of the charges into some other form, per unit charge passing through the component.

Formally:

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

where $W$ is the energy transferred (in joules) and $Q$ is the charge (in coulombs) that has flowed through the component during that energy transfer.

The unit is the volt (V), defined as $1$ V $= 1$ J C $^{-1}$ .

So "the pd across this bulb is $6$ V" means: for every coulomb of charge that passes through the bulb, $6$ joules of electrical energy are converted into light and heat by the bulb. Pd describes the energy leaving the circuit at a component.

Worked Example 1 — pd to energy

A current of $0.30$ A flows through a $12$ V lamp for $15$ s. Calculate the energy transferred to light and heat.

Charge: $Q = It = 0.30 \times 15 = 4.5$ C.
Energy: $W = VQ = 12 \times 4.5 = 54$ J.

Equivalently $W = VIt = 12 \times 0.30 \times 15 = 54$ J. Both routes give the same answer; the first emphasises the per-coulomb interpretation, the second emphasises the power-times-time interpretation ( $P = VI$ , lesson 8).

Defining EMF

EMF (electromotive force, symbol $\varepsilon$ , Greek epsilon) is defined as the energy transferred from some other form into the electrical energy of the charges, per unit charge, by the source.

Formally:

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

The equation has exactly the same mathematical form as the pd equation, and the same unit (the volt). But the physical process is opposite: chemical energy (in a battery), mechanical energy (in a generator), or radiant energy (in a solar cell) is converted into electrical energy as charge passes through the source. EMF describes the energy entering the circuit from the source.

Important: EMF is not a force. The name "electromotive force" is a misnomer left over from the 19th century — there was no separate name for "voltage" in 1820 and "force" was used loosely for any agent that produces motion. EMF is an energy per unit charge, measured in volts (J C $^{-1}$ ), never in newtons. OCR has dropped marks for candidates who write "EMF is the force on a charge" or who try to apply $F = ma$ to it.

Worked Example 2 — EMF from energy

A battery drives $3.0$ C of charge around a circuit and in doing so transfers $18$ J of chemical energy into electrical energy. Calculate the EMF.

$\varepsilon = \frac{W}{Q} = \frac{18}{3.0} = 6.0 \text{ V}$

The battery puts $6$ J of energy into every coulomb of charge it pushes around the circuit. That energy is then available for components to dissipate further around the loop.

EMF vs PD — The Key Distinction

Here is the contrast that OCR loves to test.

Quantity	Where the energy transfer happens	Direction of transfer	Per unit
EMF ( $\varepsilon$ )	Inside a source (battery, cell, generator, solar cell)	Other form $\to$ electrical	Charge that passes through the source
Potential difference ( $V$ )	Across a component (resistor, bulb, motor)	Electrical $\to$ other form	Charge that passes through the component

Both have units of volts (J C $^{-1}$ ), but they describe energy moving in opposite directions through opposite kinds of element. A gold-standard definition pair: "EMF is the energy transferred from chemical (or other non-electrical) energy to electrical energy per unit charge by the source; potential difference is the energy transferred from electrical energy to other forms per unit charge by a component."

The distinction becomes operationally critical in lesson 9, where we'll see that the terminal pd of a battery is less than its EMF whenever the battery is delivering current — because some of the electrical energy is dissipated inside the battery itself due to its internal resistance. The "lost volts" are exactly $\varepsilon - V_{\text{terminal}} = Ir$ where $r$ is the internal resistance.

The Volt — A Deeper Look

From $V = W/Q$ we get $1$ V $= 1$ J / $1$ C $= 1$ J C $^{-1}$ . So a volt is really "joules per coulomb" — an energy per unit charge. This is a far more physical way to think about voltage than the vague "how much push", and it makes many derivations much easier.

Some useful consequences:

The energy gained by a charge $Q$ accelerated through a pd $V$ (e.g. an electron in a CRT or a capacitor) is $W = QV$ .
The energy lost by a charge $Q$ passing through a resistor with pd $V$ is $W = QV$ .
The power dissipated by a component is then $P = W/t = VQ/t = VI$ (since $Q/t = I$ ).
The work done per electron crossing a pd of $V$ volts is $W_{\text{e}} = eV$ , exactly $1.60 \times 10^{-19}$ J per volt. This is so useful that physicists define the electronvolt $1$ eV $= 1.60 \times 10^{-19}$ J as a natural energy unit for atomic and particle physics (Module 4.5 photoelectric effect; Module 6.4 nuclear physics).

Worked Example 3 — Energy and the Electronvolt

An electron is accelerated from rest through a potential difference of $200$ V. Calculate:

(a) The energy gained by the electron, in J and in eV.

(b) Its final speed (non-relativistic).

(a) Energy: $W = eV = (1.60 \times 10^{-19}) \times 200 = 3.2 \times 10^{-17}$ J. In eV: $W = 200$ eV exactly (since $V$ in volts gives $W$ in eV directly when applied to a single electron).

(b) From $\tfrac{1}{2}m_e v^2 = W$ with $m_e = 9.11 \times 10^{-31}$ kg:

$v = \sqrt{\frac{2W}{m_e}} = \sqrt{\frac{2 \times 3.2 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 8.4 \times 10^6 \text{ m s}^{-1}$

About $3\%$ of $c$ — still non-relativistic, just. The eV is a convenient unit precisely because you can read it off the accelerating voltage; this is why particle physicists work in keV, MeV, GeV.

Measuring pd — The Voltmeter

To measure the potential difference across a component you use a voltmeter, connected in parallel across the component. An ideal voltmeter has infinite resistance, so that no current flows through it; inserting the meter does not disturb the circuit it's measuring.

Real voltmeters have a very large but finite input impedance. A typical digital lab voltmeter has $10$ M $\Omega$ input impedance, which is "effectively infinite" for most A-Level circuits (the meter draws less than $1$ $\mu$ A even at $10$ V). Older moving-coil voltmeters had lower impedance ( $\sim 1$ k $\Omega$ /V) and could noticeably disturb high-impedance circuits — which is why the "voltmeter disturbance" effect is examinable.

flowchart LR
  B[Battery] --> A((Ammeter)) --> R[Resistor]
  R --> B
  R -.- V1((V))
  V1 -.- W1["parallel across R"]

Common Exam Mistake: Connecting a voltmeter in series with a component. Because its resistance is huge, almost no current flows — the circuit effectively breaks, and the reading is meaningless. Voltmeter $\Rightarrow$ parallel; ammeter $\Rightarrow$ series.

Measuring EMF

The EMF of a cell can be measured by connecting a high-resistance voltmeter directly across its terminals with no current drawn from the cell (or with current negligible).

Why "no current"? Because internal resistance (lesson 9) causes the measured terminal pd to drop below $\varepsilon$ whenever current is drawn. With $I = 0$ , the voltage drop across the internal resistance ( $Ir$ ) is also zero, and the voltmeter reads exactly $\varepsilon$ .

A more precise method — used for high-stakes metrology — is the potentiometer (lesson 12). The potentiometer balances the unknown EMF against a precisely known fraction of a reference EMF, so that at the balance point no current is drawn from the cell under test. The accuracy can be parts-per-million; modern reference cells based on Josephson junctions achieve parts-per-billion.

Worked Example 4 — Battery Energy Audit

A $12.0$ V car battery delivers a steady current of $4.0$ A to its headlamps for $90$ s. Calculate:

(a) The total charge that flows.

(b) The total electrical energy supplied by the battery.

(a) Charge: $Q = It = 4.0 \times 90 = 360$ C.

(b) Energy supplied: $W = \varepsilon Q = 12.0 \times 360 = 4320$ J $\approx 4.3$ kJ.

(We treat the battery pd as equal to its EMF here, ignoring internal resistance.)

(c) Energy per electron: $W_e = \varepsilon e = 12.0 \times 1.60 \times 10^{-19} = 1.92 \times 10^{-18}$ J $= 12.0$ eV.

That tiny energy is the work done on each electron in its one traverse of the circuit. Multiply by the vast number of electrons per second ( $I/e \approx 2.5 \times 10^{19}$ ) and you get the macroscopic power $P = \varepsilon I = 48$ W.

Worked Example 5 — Foreshadowing Internal Resistance

A torch bulb with a pd of $2.5$ V across it carries a current of $0.30$ A. In $60$ s:

Charge through the bulb: $Q = It = 0.30 \times 60 = 18$ C.
Energy transferred in the bulb: $W_{\text{bulb}} = VQ = 2.5 \times 18 = 45$ J.

Meanwhile the battery (EMF $\varepsilon = 3.0$ V, delivering the same $0.30$ A for $60$ s):

Energy supplied by battery: $W_{\text{battery}} = \varepsilon Q = 3.0 \times 18 = 54$ J.

The difference $W_{\text{battery}} - W_{\text{bulb}} = 54 - 45 = 9$ J is dissipated as heat inside the battery due to its internal resistance. We return to this in lesson 9. The key takeaway: the battery delivers $54$ J, the bulb consumes $45$ J, and energy is conserved overall. The "lost" $9$ J is not gone — it's dissipated as heat in the internal resistance of the battery itself.

This is why the terminal pd (= $V$ measured across the battery's external terminals while it delivers current) is less than the EMF by an amount $\varepsilon - V = Ir$ , where $r$ is the internal resistance. In numbers: $\varepsilon - V = 3.0 - 2.5 = 0.5$ V at $I = 0.30$ A, giving $r = 0.5/0.30 \approx 1.7$ $\Omega$ .

Energy Conservation Around a Loop

The principle underlying Kirchhoff's second law (lesson 11) is energy conservation:

$\sum \text{EMFs around a loop} = \sum \text{pds around the same loop}$

i.e. the total energy supplied per unit charge as you traverse the loop equals the total energy dissipated per unit charge. This is the loop-rule analogue of Kirchhoff-1 (which expresses charge conservation at a node).

Concretely, for a loop containing a single battery ( $\varepsilon$ ) and several components with pds $V_1, V_2, \dots, V_k$ :

$\varepsilon = V_1 + V_2 + \dots + V_k$

You've already used this rule implicitly whenever you've added series-component voltages. We make it explicit and general in lesson 11.

Worked Example 6 — Two Bulbs in Series

Two identical bulbs are connected in series across a $9.0$ V battery (assume negligible internal resistance). A current of $0.40$ A flows. Calculate:

(a) The pd across each bulb.

(b) The energy dissipated by each bulb in $30$ s.

(a) By symmetry, each bulb has the same pd. By energy conservation, $V_1 + V_2 = \varepsilon = 9.0$ V, so $V_1 = V_2 = 4.5$ V.

EMF and Potential Difference

EMF and Potential Difference

Defining Potential Difference

Worked Example 1 — pd to energy

Defining EMF

Worked Example 2 — EMF from energy

EMF vs PD — The Key Distinction

The Volt — A Deeper Look

Worked Example 3 — Energy and the Electronvolt

Measuring pd — The Voltmeter

Measuring EMF

Worked Example 4 — Battery Energy Audit

Worked Example 5 — Foreshadowing Internal Resistance

Energy Conservation Around a Loop

Worked Example 6 — Two Bulbs in Series

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