Physics · Electric Circuits · AS Level

E.m.f. & Internal Resistance

Ever wondered why a battery reading "9 V" on the label never quite delivers 9 V once wired into a circuit? The gap between what a cell promises and what it delivers comes down to one idea: no real cell is a perfect source of energy, and this lesson is about the resistance hiding inside it.

9 min read

What emf actually means

Every cell or battery is, at heart, an energy converter — chemical energy goes in, electrical energy comes out, one coulomb of charge at a time. The electromotive force, or emf, of a source is the energy it supplies per unit charge: $\varepsilon = \dfrac{E}{Q}$, measured in volts, exactly like potential difference.

That similarity in units trips students up, because emf and potential difference describe two different things, not the same thing wearing two names. Emf describes energy being supplied — it's the property of a source, like a cell, converting one form of energy into electrical energy. Potential difference describes energy being transferred elsewhere — it's the property of a component, like a resistor, converting electrical energy into some other form, typically heat.

Why no real cell is perfect

If a cell were a perfect source of energy, every joule of chemical energy it converted would show up at its terminals ready for useful work. But cells are made of real materials — electrolytes, electrodes — none of them perfect conductors. So charge has to do some work fighting its way through the inside of the cell itself before it even reaches the external circuit.

This unavoidable internal resistance, symbol $r$, means that whenever current flows, some of the energy supplied by the cell is used up inside the cell itself, transferred as heat within its own materials, and never makes it to the external circuit at all.

A circuit diagram with a single left-hand branch containing a battery symbol labelled epsilon (ε) in series with a resistor symbol labelled lowercase r, both enclosed within a dashed-line box representing the physical boundary of the real cell. This branch connects via a top and bottom rail to two further branches in parallel: a voltmeter, drawn as a circle labelled V, and an external resistor labelled R, forming a complete circuit. — A real cell modelled as a perfect emf source ε in series with its own internal resistance r, both enclosed within the cell's boundary — the terminal pd V is only what's left after the internal resistance has taken its share.

Deriving ε = I(R + r)

Picture the cell as a perfect source of emf $\varepsilon$ in series with a small internal resistance $r$, connected to an external resistor $R$. The relationship between them falls straight out of energy conservation — energy can't simply vanish as charge moves around the loop.

Derivationε = I(R + r)

The emf $\varepsilon$ is the total energy supplied per unit charge by the source.

By definition, $\varepsilon = E/Q$.

By conservation of energy, every joule supplied per coulomb must be accounted for somewhere in the loop: some is transferred in the external resistor, and some inside the cell itself.

Energy can't disappear — it's either usefully delivered externally, or lost internally.

The energy transferred per unit charge in the external resistor is the terminal pd, $V = IR$.

The energy transferred per unit charge overcoming the internal resistance is $Ir$, using the same current $I$ since the cell and resistor are in series.

The same current flows through the internal resistance as through the external circuit — there's only one loop.

Adding these two contributions together gives the total: $\varepsilon = IR + Ir = I(R+r)$.

This is simply energy conservation written as an equation, split into "useful" and "internal" shares.

▪ derived

This equation is why a battery never quite delivers its full rated voltage once current is flowing — some emf is always "spent" internally, growing with the current drawn.

Terminal pd, and where it comes from

The voltage you'd measure across a cell's terminals with a voltmeter — the terminal pd, $V$ — is what's left of the emf once the share lost to internal resistance is subtracted. Rearranging the equation above makes this explicit.

DerivationV = ε − Ir, and the shape of the terminal-pd–current graph

Start from $\varepsilon = IR + Ir$, and recall that $V = IR$ is the terminal pd.

Substitute and rearrange to make $V$ the subject: $\varepsilon = V + Ir$, so $V = \varepsilon - Ir$.

The terminal pd is the emf minus whatever's "lost" driving current through the internal resistance.

Compare this to the equation of a straight line, $y = c + mx$: here $V$ plays the role of $y$, $I$ plays the role of $x$, the intercept $c$ is $\varepsilon$, and the gradient $m$ is $-r$.

$V = \varepsilon - Ir$ is already exactly in straight-line form, just with the terms reordered.

So a graph of terminal pd $V$ against current $I$ is a straight line, with a $y$-intercept of $\varepsilon$ and a gradient of $-r$.

The intercept is the pd when $I=0$ — no current means nothing is lost internally, so the terminal pd equals the full emf.

▪ derived

A straight-line graph on labelled axes: the vertical axis is terminal potential difference V, the horizontal axis is current I. The straight line starts at a clearly marked point on the vertical axis labelled epsilon (ε), representing the y-intercept where current is zero, and slopes downward from left to right in a straight diagonal line as current increases. The downward slope of the line is explicitly annotated with a label reading gradient = −r, with a small triangle or bracket alongside the line indicating rise over run to show how the gradient is measured. — Terminal pd falls linearly with current: the y-intercept gives the emf ε, and the gradient of the line gives minus the internal resistance, −r.

That intercept has a nice physical meaning: it's the terminal pd you'd read with almost no current flowing at all, when there's essentially nothing left to lose internally. As soon as real current flows, terminal pd drops below the emf, more steeply the larger the internal resistance.

The graph's other axis crossing is worth attention too. Setting $R=0$ in $\varepsilon = I(R+r)$ — a short circuit, where external resistance has effectively vanished — gives $\varepsilon = Ir$, so the short-circuit current is $I = \varepsilon/r$. On the graph, this is the $x$-intercept: the current at which $V=0$, meaning all the emf is lost driving current through the internal resistance alone. Because $r$ is normally very small, $I = \varepsilon/r$ can be very large — why short-circuiting a battery is dangerous: even a modest emf divided by a small internal resistance gives enough current to dump a lot of power ($I^2r$) inside the cell, heating it rapidly and, in the worst cases, damaging it.

Required Practical: Investigating EMF and Internal Resistance

AQA Required Practical

This is AQA's Required Practical 6: measuring a cell's emf and internal resistance.

Equipment

Equipment	Purpose
Cell or battery under test	The source whose emf and internal resistance are being found
Variable resistor (rheostat)	Varies the current drawn from the cell
Ammeter	Measures the current $I$
Voltmeter	Measures the terminal pd $V$ across the cell
Switch	Opens the circuit between readings, limiting unnecessary discharge/heating
Connecting leads	Completes the circuit

Method

MethodFinding ε and r from V against I

Connect the cell in series with an ammeter, a rheostat and a switch, with a voltmeter connected in parallel across the cell's terminals to measure terminal pd.

Close the switch and record the current $I$ and terminal pd $V$ for the current rheostat setting.

Open the switch, adjust the rheostat to a new resistance, then close the switch again and take the next reading — keeping the circuit open between readings limits unnecessary heating of the cell.

Repeat for a range of different resistances, covering as wide a range of current as practical.

Plot terminal pd $V$ against current $I$ — the $y$-intercept gives the emf $\varepsilon$ directly, and the magnitude of the gradient gives the internal resistance $r$.

▪ method

Sources of error and how to minimise them

The cell warming up during use

Leaving the circuit connected for too long lets the cell warm up, changing its internal resistance between readings. Take readings reasonably quickly, and open the switch between them where practical.

Meter resistance introducing systematic error

An ammeter with significant resistance, or a voltmeter that draws a non-negligible current, both slightly distort the readings. Use an ammeter with as low a resistance as possible and a voltmeter with as high a resistance as possible, so both stay close to the ideal assumption used in the derivation.

Loose or dirty connections

Poor connections add extra resistance that isn't part of the cell's own internal resistance. Check and secure all connections, and clean contacts if readings look inconsistent.

Combining cells: series and parallel

Real circuits rarely rely on a single cell. Torches, remote controls and battery packs combine several identical cells for either a bigger voltage or more current capability — and both the combined emf and internal resistance follow directly from the same reasoning above, applied to more than one cell at a time.

Cells in series

When cells are connected in series — positive terminal to negative terminal, around a single loop — exactly the same current $I$ flows through every cell in turn, because there's only one path for charge.

DerivationCells in series: emfs add, internal resistances add

Consider $n$ cells in series, with emfs $\varepsilon_1, \varepsilon_2, \dots, \varepsilon_n$ and internal resistances $r_1, r_2, \dots, r_n$, all connected the same way round in a single loop.

A charge $Q$ moving once around the loop passes through every cell in turn, gaining energy $\varepsilon_1 Q$ from the first, $\varepsilon_2 Q$ from the second, and so on.

Emf is energy supplied per unit charge, so each cell hands over its own share as the charge passes through it.

The total energy gained per unit charge over the whole loop is therefore $\varepsilon_1 + \varepsilon_2 + \dots + \varepsilon_n$ — which is by definition the combined emf: $\varepsilon_{\text{total}} = \varepsilon_1 + \varepsilon_2 + \dots + \varepsilon_n$.

Because the same current $I$ flows through every internal resistance in turn, one after another, they behave exactly like ordinary resistors in series: $r_{\text{total}} = r_1 + r_2 + \dots + r_n$.

Series resistances always add, whether they belong to separate components or sit inside separate cells.

For $n$ identical cells, each with emf $\varepsilon$ and internal resistance $r$, this simplifies to $\varepsilon_{\text{total}} = n\varepsilon$ and $r_{\text{total}} = nr$.

▪ derived

Watch the orientation

This only holds if every cell is connected the same way round. If one cell is reversed relative to the others, its emf subtracts from the total instead of adding — a cell inserted backwards into a torch fights against the rest of the pack, which is why the total emf of two identical opposing cells in series is zero, not double.

Identical cells in parallel

Connecting cells in parallel instead — all positive terminals joined to one node, all negative terminals to another — changes the picture. Every cell shares the same pair of terminals, so every cell must have the same terminal pd, but the current can now split between them.

Derivationn identical cells in parallel: emf unchanged, internal resistance divides by n

Consider $n$ identical cells, each with emf $\varepsilon$ and internal resistance $r$, connected in parallel and supplying a common external circuit.

Because all $n$ cells are identical and connected between the same two nodes, they must share current equally: if any two carried different currents, they'd have different terminal pds despite being connected to the same two points, which is impossible.

Two points in a circuit can only be at one potential difference at a time.

So if the total current drawn from the combination is $I$, each individual cell supplies $i = \dfrac{I}{n}$.

Applying $V = \varepsilon - ir$ to a single cell in the combination: $V = \varepsilon - \dfrac{I}{n}r = \varepsilon - I\left(\dfrac{r}{n}\right)$.

Each cell obeys its own emf equation using its own share of the current.

Comparing this to the general single-source equation $V = \varepsilon_{\text{eq}} - Ir_{\text{eq}}$ shows $\varepsilon_{\text{eq}} = \varepsilon$ and $r_{\text{eq}} = \dfrac{r}{n}$.

Same trick as before — matching coefficients in a straight-line-shaped equation.

▪ derived

So $n$ identical cells in parallel behave as a single source with the same emf as one cell, but with internal resistance shared out $n$ ways, exactly as ordinary resistors combine in parallel. That's the point of wiring cells in parallel: it doesn't raise the available voltage, but it lowers the effective internal resistance, so the combination can supply more current before its terminal pd sags noticeably. Series combinations do the opposite — they raise the available emf, at the cost of adding the internal resistances too.

Key takeaways

Emf, $\varepsilon = E/Q$, is the energy supplied per unit charge by a source — it describes energy being supplied, not transferred elsewhere.
Potential difference describes energy being transferred or converted by a component; emf and pd are measured in the same unit but describe different roles in a circuit.
Internal resistance $r$ is an unavoidable property of any real cell, causing some energy to be transferred inside the cell itself whenever current flows.
Energy conservation around the loop gives $\varepsilon = I(R+r)$, splitting the emf into a share delivered externally ($IR$) and a share lost internally ($Ir$).
Rearranged, $V = \varepsilon - Ir$ shows the terminal pd falls as current increases, at a rate set by the internal resistance.
A graph of terminal pd against current is a straight line with y-intercept $\varepsilon$ and gradient $-r$ — this is exactly how emf and internal resistance are found experimentally.
Setting $R=0$ (a short circuit) gives $\varepsilon = Ir$, so the short-circuit current is $I=\varepsilon/r$ — the graph's $x$-intercept — and because $r$ is usually tiny, this current can be dangerously large.
Cells in series: emfs add ($\varepsilon_{\text{total}} = \varepsilon_1+\varepsilon_2+\dots$) and internal resistances add ($r_{\text{total}} = r_1+r_2+\dots$) — but a reversed cell subtracts its emf instead.
$n$ identical cells in parallel keep the same emf $\varepsilon$ as a single cell, but the combined internal resistance drops to $r/n$, letting the combination supply more current with less terminal-pd sag.

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