EMF (electromotive force, symbol ε) is the energy per unit charge supplied by a battery or other source, measured in volts. It's the electric potential difference that pushes charge through a circuit, and for a real battery it equals terminal voltage only when no current flows.
EMF stands for electromotive force, written with the Greek letter ε (epsilon). Despite the name, it's not a force at all. It's the energy per unit charge a source like a battery provides, measured in volts. Think of it as the battery's "push rating," the maximum potential difference it can offer to drive charge around a circuit.
In the AP Physics 2 CED, charge moves in a circuit in response to an electric potential difference, which is sometimes called electromotive force or emf. That push is what creates current, the rate at which charge passes through a cross-section of wire (I = Δq/Δt). For an ideal battery, the voltage across its terminals equals ε. Real batteries have internal resistance r, so once current flows, some of that ε gets used up inside the battery itself and the terminal voltage drops below ε by an amount Ir.
EMF lives in Topic 11.1 (Electric Current) in Unit 11: Electric Circuits, supporting learning objective 11.1.A, which asks you to describe the movement of electric charges through a medium. EMF is the answer to the question "why does charge move at all?" Without a potential difference, there's no net motion of charge carriers, even though individual electrons are still jiggling around with nonzero speed. Everything else in Unit 11, from Ohm's law to Kirchhoff's rules to power dissipation, starts with a source of emf driving the circuit. If you can't tell emf apart from terminal voltage, internal resistance problems will eat you alive on the exam.
Keep studying AP® Physics 2 Unit 11
Terminal Voltage and Internal Resistance (Unit 11)
This is the pairing the exam loves most. A real battery's terminal voltage is V = ε - Ir, so the more current you draw, the bigger the gap between what the battery is rated for (ε) and what the circuit actually sees (V). When I = 0, they're equal.
Electric Current (Unit 11)
EMF is the cause and current is the effect. Charge flows because a potential difference exists, and the current's direction follows what positive charges would do, even though current itself isn't a vector.
Induced EMF and Faraday's Law (Unit 12)
Batteries aren't the only emf source. A changing magnetic flux through a loop induces an emf, which is why the same symbol ε reappears in electromagnetic induction. The 2021 long FRQ built an entire problem around an electromagnet for exactly this reason.
Power in Circuits (Unit 11)
Since emf is energy per charge and current is charge per time, εI gives the total power a battery delivers. Some of that power burns off in the internal resistance (I²r), and the rest reaches the external circuit. Maximum power transfer questions hinge on this split.
EMF shows up constantly in Unit 11 multiple choice. Classic stems give you a battery's emf and internal resistance, tell you the measured terminal voltage, and ask for the current. For example, with ε = 12 V, r = 0.5 Ω, and a terminal voltage of 11.4 V, the 0.6 V "missing" voltage drops across r, so I = 0.6/0.5 = 1.2 A. You may also see maximum power transfer questions (load resistance R equals internal resistance r) and pure symbol-recognition questions about what ε means. On FRQs, emf is the starting point for circuit analysis. The 2024 long FRQ gave "an ideal battery of emf ε" feeding four identical resistors and asked you to reason through the circuit. Note the word "ideal" there, which signals zero internal resistance. The 2023 lab-based FRQ had you analyze an unknown component in series with a 500 Ω resistor, which requires tracking how the source's emf divides across circuit elements. Know when to treat a battery as ideal (V = ε) versus real (V = ε - Ir).
EMF is what the battery can supply; terminal voltage is what it actually delivers while current flows. For an ideal battery they're identical. For a real battery, terminal voltage equals ε minus the drop across internal resistance, so V = ε - Ir. A voltmeter across a battery in an open circuit reads ε, but the moment current flows, the reading drops below ε. If an AP problem says "ideal battery," you can set them equal; if it mentions internal resistance, you can't.
EMF (ε) is the energy per unit charge supplied by a source like a battery, and it's measured in volts, not newtons, because it is not actually a force.
Charge moves in a circuit in response to a potential difference, and emf is the source of that potential difference (LO 11.1.A).
For a real battery, terminal voltage equals ε - Ir, so the terminal voltage drops below the emf whenever current flows.
When the current is zero, terminal voltage equals emf exactly, which is why an open-circuit voltmeter reading gives you ε directly.
If a problem says "ideal battery," internal resistance is zero and you can treat the terminal voltage as equal to ε.
The same symbol ε returns in Unit 12 for induced emf, where a changing magnetic flux acts as the source instead of a battery.
EMF (electromotive force, symbol ε) is the energy per unit charge provided by a battery or other source, measured in volts. It's the potential difference that drives current through a circuit, covered in Topic 11.1 (Electric Current).
No. The name is a historical leftover. EMF is energy per unit charge, so it's measured in volts (joules per coulomb), not newtons. Don't let "electromotive force" trick you into using force equations.
EMF is the battery's full rated push; terminal voltage is what's left after the drop across internal resistance, V = ε - Ir. A 12 V battery with 0.5 Ω internal resistance reading 11.4 V at its terminals is supplying 1.2 A of current.
The Greek letter ε (epsilon) stands for emf, the electromotive force of a source. When you see "a battery of emf ε" on an FRQ, like the 2024 long FRQ, treat ε as the battery's voltage.
Real batteries have internal resistance r, and current passing through it drops Ir volts inside the battery itself. So the circuit only sees ε - Ir at the terminals. The bigger the current, the bigger the loss.
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