EMF and Internal Resistance

Connect a fresh 1.5\ \text{V} cell to a torch bulb and put a good voltmeter across the cell's terminals. You might expect a steady 1.5\ \text{V} — but the reading sags, maybe to 1.3\ \text{V}, the moment the bulb lights. Draw more current — short a heavy motor across it — and the terminal reading collapses further. Where did the missing voltage go?

It went inside the cell. A real source is not the perfect, effortless energy pump we pretend it is in a first circuits lesson. Its chemicals, its electrodes and its electrolyte all resist the very current the cell is trying to push, so some of the cell's own energy is dissipated within it before the charge ever reaches the outside world. That built-in resistance is the cell's internal resistance, and this page is about the tug-of-war between the full voltage a source could supply and the smaller voltage a circuit actually gets.

Two voltages: emf and terminal p.d.

We need two clearly different quantities. The first is the electromotive force, or emf, written \varepsilon. Despite the name it is not a force at all — it is an energy per charge, measured in volts just like any pd:

\varepsilon = \dfrac{\text{energy converted by the source}}{\text{charge that passed}} = \dfrac{E}{Q}.

The emf is the full energy the source hands to every coulomb — the "no-load" voltage you would measure across the terminals when no current flows (an ideal voltmeter, open circuit). A 1.5\ \text{V} AA cell has \varepsilon = 1.5\ \text{V}: each coulomb is loaded with 1.5\ \text{J} by the chemistry inside.

The second quantity is the terminal potential difference V — the voltage that the external circuit actually receives, measured across the cell's terminals while current is flowing. Because some energy is spent driving charge through the internal resistance r, the terminal pd is always less than the emf whenever a current is drawn.

The lost volts: \varepsilon = V + Ir

Picture the single loop: an emf \varepsilon in series with an internal resistance r, driving a current I around an external load of resistance R. The emf is the total energy given to each coulomb, and that energy is spent in exactly two places — in the external load, and inside the cell itself. Applying Kirchhoff's voltage law around the loop, the emf equals the sum of the pds across every resistance:

\varepsilon = IR + Ir = I(R + r).

But IR is just the pd delivered to the external load — which is the terminal pd V itself. So the same statement can be written in the form you will use most:

\varepsilon = V + Ir \qquad\Longrightarrow\qquad V = \varepsilon - Ir.

The term Ir is the "lost volts": the slice of the emf dropped inside the cell, dissipated as heat in its internal resistance and never delivered to the load. The terminal pd is what's left of the emf after that internal toll is paid.

Read V = \varepsilon - Ir as a story about current. When I = 0 (open circuit, no load) the lost volts vanish and V = \varepsilon — the terminal reading equals the full emf. As the load R shrinks, the current I rises, the lost volts Ir grow, and the terminal pd drops. Demand a huge current and the terminal pd can collapse dramatically.

The emf \varepsilon is the energy converted by the source per unit charge, \varepsilon = E/Q, equal to the terminal pd on open circuit (no current).
For a source of internal resistance r driving current I through an external load R, \varepsilon = I(R + r) = V + Ir.
The terminal p.d. is V = \varepsilon - Ir, always less than \varepsilon when I > 0; the lost volts Ir are dissipated inside the source.

Worked examples

Example 1 — terminal pd from the lost volts. A battery of emf \varepsilon = 12\ \text{V} and internal resistance r = 0.5\ \Omega drives a current of 4\ \text{A}. What is its terminal pd?

V = \varepsilon - Ir = 12 - (4)(0.5) = 12 - 2 = 10\ \text{V}.

The lost volts are Ir = 2\ \text{V}, so the external circuit receives only 10\ \text{V} of the battery's 12\ \text{V}.

Example 2 — current from the whole loop. A cell of emf 1.5\ \text{V} and internal resistance r = 0.5\ \Omega is connected to a resistor R = 2.5\ \Omega. Find the current, then the terminal pd. Use \varepsilon = I(R + r):

I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{2.5 + 0.5} = \dfrac{1.5}{3} = 0.5\ \text{A}.

V = \varepsilon - Ir = 1.5 - (0.5)(0.5) = 1.5 - 0.25 = 1.25\ \text{V}.

Check: the load pd is IR = 0.5 \times 2.5 = 1.25\ \text{V} — the same V, as it must be, since the load is what the terminals feed.

Example 3 — finding r from two readings. On open circuit a battery reads \varepsilon = 9.0\ \text{V}. When it delivers 3\ \text{A} the terminal pd falls to V = 8.1\ \text{V}. What is its internal resistance? Rearrange V = \varepsilon - Ir:

r = \dfrac{\varepsilon - V}{I} = \dfrac{9.0 - 8.1}{3} = \dfrac{0.9}{3} = 0.3\ \Omega.

The 0.9\ \text{V} that vanished between the two readings is precisely the lost volts Ir at 3\ \text{A}.

See it: watch the terminal pd sag

Below is a real cell — an emf \varepsilon with its internal resistance r tucked inside the dashed box — driving an external load R, with a voltmeter reading the terminal pd V. Turn the load down: as R falls the current I climbs, the lost volts Ir grow, and the terminal pd V visibly sags below the emf. Turn the internal resistance r up and the same current costs far more lost volts — a "tired" cell.

Push the load R to its largest (almost open circuit) and the terminal pd creeps back up towards the full emf, because I \to 0 makes the lost volts vanish. That is exactly why a voltmeter on open circuit reads \varepsilon, but the reading dips the instant a real load draws current.

Measuring \varepsilon and r from a graph

Here is the experiment every A-level student meets. Vary the load, and for each setting record the current I and the terminal pd V. Then plot V (vertical) against I (horizontal). Rearrange the emf equation into the form of a straight line y = c + mx:

V = \varepsilon - Ir \qquad\Longleftrightarrow\qquad V = \underbrace{\varepsilon}_{\text{intercept}} + \underbrace{(-r)}_{\text{gradient}}\,I.

the y-intercept (where the line meets the V-axis, at I = 0) is the emf \varepsilon;
the gradient of the line is -r, so the internal resistance is minus the slope — a steeper downward line means a larger r;
the line also cuts the I-axis at the short-circuit current I_{\text{sc}} = \varepsilon / r, where R = 0 and all the emf is lost internally.

Drag the two sliders and watch the line move: the emf slides the whole line up and down (its intercept), while r tilts it — set r = 0 and the line goes flat, the mark of a perfect, ideal cell whose terminal pd never sags.

Why batteries go "flat" — and headlights dim

Internal resistance quietly explains a handful of everyday electrical mysteries.

Why a battery goes "flat" and dim. As a cell ages, its chemistry degrades and its internal resistance r climbs. Even if the emf is still nearly its nominal value, the growing Ir eats more and more of it, so the terminal pd delivered to the bulb collapses under load and the light fades.
Why car headlights dim when you start the engine. A starter motor demands an enormous current — hundreds of amps — for a second or two. Through the battery's small internal resistance that vast I makes a large Ir, so the terminal pd briefly plunges and every other load on the same battery, including the headlights, momentarily dims. Once the engine catches and the starter disengages, I drops, the lost volts shrink, and the lights brighten again.

The terminal pd is always less than the emf when current flows. They are equal only on open circuit (I = 0, no load). If a question gives you the "voltage of the battery" while current is being drawn, that number is the terminal pd V, not the emf \varepsilon.
The lost volts Ir are dissipated inside the cell, not in the wires or the load. That energy warms the battery itself. A load drawing a big current makes a bigger internal drop — the whole reason terminal pd sags under heavy load.
Gradient of the V–I graph is -r, not r. The line slopes down, so its slope is negative; the internal resistance is the magnitude of that slope. The intercept — not the gradient — gives the emf.

Take a battery that a torch has given up on, and clip a digital voltmeter across it — it may cheerily read almost the full 1.5\ \text{V}. Has the shop sold you a lie about it being flat? No: a voltmeter has an enormous resistance, so it draws a current of only microamps. With I essentially zero, the lost volts Ir are essentially zero whatever r is, so the meter sees the emf regardless of how tired the cell is.

A torch bulb is a completely different customer. It demands a real current of, say, 0.3\ \text{A}. If the aged cell's internal resistance has ballooned to r = 4\ \Omega, the lost volts are Ir = 0.3 \times 4 = 1.2\ \text{V} — nearly all of the emf — leaving a terminal pd of just 0.3\ \text{V}, far too little to make the filament glow. High internal resistance is the difference between a battery that looks full and one that can do anything. It is also why jump-starting a car works: a healthy battery with very low r can source the huge starter current the flat one cannot.