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Medium

Electromotive Force and Electric Circuits

Simple Explanation

A battery's electromotive force (EMF, Ξ΅) is the total energy it supplies per unit charge β€” but a real battery also has internal resistance, so the actual voltage available to a circuit (terminal voltage) is always a little less than the full EMF: V=Ξ΅βˆ’Ir.

Why Do We Need It?

Understanding EMF and internal resistance explains why a battery's voltage seems to "sag" under heavy load β€” a genuinely practical fact about every real power source, from AA batteries to car batteries.

See It

A simple circuit: a battery with internal resistance driving current through a resistor
Ξ΅, r (battery)IR (resistor)

A rectangular circuit loop with a battery (EMF Ξ΅ and internal resistance r) on the left side and a resistor R on the right side, with an arrow showing the direction of current flow

Formula

Terminal Voltage of a Battery

V = Ξ΅ βˆ’ Ir

A real battery's terminal voltage (what a circuit actually sees) is always slightly less than its full EMF, because some voltage is lost driving current through the battery's own internal resistance.

V
β€” terminal voltage, in volts (V)
Ξ΅
β€” electromotive force (EMF) of the battery, in volts (V)
I
β€” current flowing through the circuit, in amperes (A)
r
β€” the battery's internal resistance, in ohms (Ξ©)

When to use it: Whenever the actual voltage a circuit experiences (as opposed to the battery's ideal EMF) needs to be found, accounting for internal resistance.

Worked Example

Find a battery's terminal voltage

A battery has EMF 12 V and internal resistance 0.5 Ξ©. When it drives 2 A through a circuit, find its terminal voltage.

    Why Does This Work?

    Some of the energy the battery supplies must be used to push current through its own internal resistance β€” by Ohm's law, that "lost" voltage is Ir, so what remains available at the terminals for the rest of the circuit is Ξ΅ minus that internal voltage drop.

    Real-Life Example

    A car battery's voltage dropping while starting the engine

    A car battery's voltage visibly dips for a moment while the starter motor draws a very large current to crank the engine.

    That large starting current (I) flowing through the battery's own internal resistance (r) causes a significant voltage drop (Ir), temporarily reducing the terminal voltage below the battery's full EMF.

    Practice

    A battery has EMF 9 V and internal resistance 0.3 Ξ©. Find its terminal voltage when it drives 3 A through a circuit.

    Medium

    Common mistake

    Assuming a battery always supplies exactly its rated EMF to a circuit β€” the actual terminal voltage is always somewhat lower once current flows, due to the unavoidable internal resistance.

    Quick Review

    • V = Ξ΅ βˆ’ Ir.
    • EMF is the battery's total energy per unit charge; terminal voltage is what the circuit actually gets.
    • Terminal voltage drops further under heavier current draw.