Emf vs Voltage: What Is the Difference?

The name is wrong, and the units prove it

Electromotive force is measured in volts. A force is measured in newtons. Whatever emf is, it is not a force, and the name survives only as a historical label that the CED keeps as an alias for a potential difference. Essential knowledge 11.1.A.1.i puts it that way directly: electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or emf, written with a script $\mathcal{E}$.

So emf and voltage are not rival quantities. Emf is a particular potential difference, the one a source maintains in order to drive charge around a loop, and it is best read as the energy each coulomb picks up on its way through the source. Volts are joules per coulomb, and a $9 \ \mathrm{V}$ cell hands each coulomb $9 \ \mathrm{J}$.

The comparison that matters is therefore not emf against voltage in general. It is emf against the voltage you can measure, the potential difference across the battery's own terminals with the circuit running. AP gives that second quantity a name and a symbol, $\Delta V_{terminal}$, and the two are equal in exactly one circumstance.

Essential knowledge 11.5.B.1.iii is the definition to quote: the potential difference a battery would supply if it were ideal is the potential difference measured across the terminals when there is no current in the battery, and is sometimes referred to as its emf. Read it slowly. The emf is defined by a measurement you take with the current switched off. Everything else on this page is what happens once you switch it on.

Side by side

The row worth staring at is the second-to-last. Neither the AP Physics 2 sheet nor the AP Physics C: Electricity and Magnetism sheet prints $\Delta V_{terminal} = \mathcal{E} - Ir$; both CEDs label it a derived equation, which means you are expected to produce it from Kirchhoff's loop rule rather than look it up. On the AP Physics 2 sheet the symbol $\mathcal{E}$ appears only in the Magnetism box, in $\lvert \mathcal{E} \rvert = \lvert \Delta \Phi_B / \Delta t \rvert$ and $\mathcal{E} = B\ell v$. The Electricity box does not use it at all.

The case that separates them: the same cell, three loads

Take one real cell with $\mathcal{E} = 6.0 \ \mathrm{V}$ and internal resistance $r = 0.50 \ \Omega$. Nothing about the cell changes across the three rows below. Only what you connect it to changes, and only one of the two quantities responds.

Check the middle row: $I = 6.0/(11.5 + 0.50) = 6.0/12.0 = 0.50 \ \mathrm{A}$, so $\Delta V_{terminal} = 6.0 - (0.50)(0.50) = 5.75 \ \mathrm{V}$, and independently $IR = (0.50)(11.5) = 5.75 \ \mathrm{V}$. The last row: $I = 6.0/1.5 = 4.0 \ \mathrm{A}$, so $\Delta V_{terminal} = 6.0 - (4.0)(0.50) = 4.0 \ \mathrm{V}$, and $IR = (4.0)(1.0) = 4.0 \ \mathrm{V}$. Both routes agree in both rows, which is the check to run every time.

The emf column never moves. It is a property of the chemistry inside the cell, not of what the cell happens to be attached to. The terminal column drops by $0.25 \ \mathrm{V}$ and then by $2.0 \ \mathrm{V}$, because the current has to pass through the source's own resistance on its way out, and the volts it spends doing so never reach the terminals.

That is what makes the two genuinely different quantities rather than two names for one. One is fixed and unmeasurable while the circuit runs. The other is measurable and depends on what you asked the cell to do.

Where the missing volts go

Essential knowledge 11.5.B.2 gives the model: the internal resistance of a nonideal battery may be treated as the resistance of a resistor in series with an ideal battery and the remainder of the circuit. Draw it that way and the rest is the loop rule.

Walk once round a single loop containing the ideal source, the internal resistance $r$ and an external resistance $R$, with current $I$. The loop rule requires the potential differences to sum to zero:

The terminals sit either side of the ideal source and its internal resistor, so what a voltmeter reads across them is $\mathcal{E} - Ir$, which is also $IR$. Rearranged, that is the CED's derived equation at 11.5.B.3, stated in words there as: when there is current in a nonideal battery with internal resistance $r$, the potential difference across the terminals of the battery is reduced relative to the potential difference when there is no current in the battery.

Follow the energy rather than the volts and the picture sharpens. Multiply the loop equation by $I$:

The left side is the rate at which the source supplies energy. The first term on the right is dissipated inside the source, warming the battery. Only the second term reaches the circuit. The emf measures what the source produces; the terminal voltage measures what escapes.

Two consequences follow that questions like to test:

• A cell can have a healthy emf and still perform badly. A worn cell usually has a grown $r$, not a collapsed $\mathcal{E}$, which is why it reads near its rated value on an unloaded meter and then sags under load.
• The larger the current, the larger the gap. The shortfall $Ir$ is proportional to the current, so a torch bulb barely exposes it and a starter motor exposes it dramatically.

Both courses treat this identically, and both default to ideal

This is one of the few blocks that AP Physics 2 and AP Physics C: Electricity and Magnetism print word for word. Compare the two CEDs at Topic 11.5 and 11.5.B.1, 11.5.B.1.i to 11.5.B.1.iii, 11.5.B.2 and 11.5.B.3 are the same statements with the same numbering, including the same derived equation. The calculus course does not extend the treatment; internal resistance is handled at the same algebraic level in both.

The default is also the same, and it is stated as a boundary statement under Topic 11.5 in both CEDs:

> Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed.

So the ideal-battery assumption is not a simplification you choose. It is the stated exam default, and it is repeated in the exam conventions box on the AP Physics 2 equation sheet, which lists strings, springs, batteries, wires and meters as ideal. An ideal battery has negligible internal resistance (11.5.B.1), so its terminal voltage equals its emf at every current and the distinction on this page collapses.

The distinction is switched on by a specific trigger, and you should learn the trigger rather than the topic. A question is about internal resistance if, and only if, one of these appears:

• an internal resistance $r$ is named for the source;
• the battery is described as nonideal or real;
• you are given terminal-voltage readings at two or more different currents;
• a voltmeter reading across the battery is smaller than the stated battery rating.

Absent all four, use the emf as the potential difference across the source and do not invent an $r$.

The second sentence of the boundary statement is a separate gift. Circuits with batteries of different potential differences connected in parallel will not be assessed, so a schematic that would need that analysis is not going to appear.

When it costs a mark

Calling emf a force. It will not lose a calculation, but it does lose the justification point on a question that asks what emf is. The safe sentence: emf is the energy supplied per unit charge by the source, measured in volts, equal to the terminal potential difference when there is no current.

Using the emf where the terminal voltage belongs. Given a nonideal cell and asked for the potential difference across the external resistor, $IR$ and $\mathcal{E} - Ir$ both work and $\mathcal{E}$ alone does not. This inflates every downstream answer: the current, the power in the load, the brightness of a bulb.

Dividing the emf by the external resistance alone. The current is $\mathcal{E}/(R + r)$, not $\mathcal{E}/R$. Internal resistance is in series with everything else (11.5.B.2), so it joins the sum.

Reading a graph's slope as $+r$. Plot $\Delta V_{terminal}$ against $I$ and the line falls: the gradient is $-r$, and the internal resistance is the magnitude of that slope. The vertical intercept is the emf, because that is the reading extrapolated to zero current, which is the CED's definition.

Inventing an internal resistance. The exam default is ideal. If no $r$ is given and nothing in the stem says nonideal, adding one is not caution, it is a wrong model.

Assuming a voltmeter across a battery reads the emf. It reads the terminal voltage. It reads the emf only when the current in the battery is zero, which for an ideal voltmeter of infinite resistance (11.5.C.2.ii) is the case when nothing else is connected.

When they coincide, and why that hides the distinction

For most of a circuits unit the two numbers are the same number, and that is by design rather than by accident.

Whenever the battery is ideal. Which, by boundary statement, is whenever the question has not said otherwise. Almost every practice circuit you meet has $r = 0$, so $\Delta V_{terminal} = \mathcal{E}$ identically and there is nothing to distinguish.

Whenever the current is zero. An open switch, a broken loop, or a battery on a shelf: no current in the battery means no $Ir$ term, so the terminals show the full emf. This is the case the CED uses as the definition, and it is why a multimeter across a loose battery reads close to its rating.

Whenever the internal resistance is negligible compared with the load. With $\mathcal{E} = 6.0 \ \mathrm{V}$, $r = 0.50 \ \Omega$ and a $600 \ \Omega$ load, the current is about $0.010 \ \mathrm{A}$ and the shortfall $Ir$ is about $0.005 \ \mathrm{V}$. Two significant figures cannot see it. The distinction is real and quantitatively invisible.

Those three cases cover so much of the course that many students finish it believing emf is simply a fancier word for the battery's voltage. It is a fancier word for the battery's voltage at zero current, and the two readings part company the instant charge starts moving. The exam exposes that on a specific kind of question: one that hands you terminal voltage at two different currents and asks for the emf. There is no way to answer it if the two words mean the same thing.

Where this sits on the AP exam

The pairing is Topic 11.5, Compound Direct Current (DC) Circuits, inside Unit 11, Electric Circuits, which carries 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods. Learning objective 11.5.B is stated as: describe a circuit with resistive wires and a battery with internal resistance. The same objective and the same essential knowledge statements appear in AP Physics C: E&M at its Topic 11.5.

The word emf itself enters much earlier, at 11.1.A.1.i in Topic 11.1, as an alias for the potential difference that makes charge move. That is worth knowing, because a question can use the word without any intention of raising internal resistance.

Emf gets its equations somewhere else entirely: electromagnetic induction. On the AP Physics 2 sheet the script $\mathcal{E}$ appears in the Magnetism box, in $\lvert \mathcal{E} \rvert = \lvert \Delta \Phi_B/\Delta t \rvert$ and $\mathcal{E} = B\ell v$. On the AP Physics C: E&M sheet it appears in $\mathcal{E} = \oint \vec E \cdot d\vec \ell = -\frac{d\Phi_B}{dt}$, in the solenoid form with $N$ loops, and in $\mathcal{E} = -L \, dI/dt$ for an inductor. An induced emf is the same kind of quantity as a battery's: joules per coulomb driving charge round a loop. Only the mechanism differs, chemistry in one case and a changing magnetic flux in the other. See Faraday's law for that side.

The experimental route shows up in the Experimental Design and Analysis free-response question, where a graph's slope and intercept are asked to yield physical quantities. Terminal voltage against current is the archetype for a cell: intercept gives $\mathcal{E}$, slope magnitude gives $r$.

For the neighbouring distinction between what is measured across an element and what is measured through it, see voltage vs current. For reducing the rest of the network once the source is settled, use the series and parallel circuits guide.