5.4.9. Electromotive Force

Table of Contents

What Electromotive Force Means

Electromotive force, usually abbreviated as EMF, is the quantity that describes how a source of electrical energy pushes charge through a circuit. Despite its name, EMF is not actually a force measured in newtons. It is measured in volts, just like electric potential difference.

An EMF source, such as a battery or generator, supplies energy to charges. Inside the source, non-electrostatic effects move charge from lower electric potential to higher electric potential. This is what makes continuous current possible in a closed circuit.

If only electric forces acted inside a battery, positive charges would naturally move from high potential to low potential, and the battery could not maintain a current. The battery uses chemical processes to do work on the charges and separate them, creating a potential difference between its terminals.

Electromotive force is the energy supplied by a source per unit charge.
$$
\mathcal{E} = \frac{W}{q}
$$
where $\mathcal{E}$ is the EMF, $W$ is the energy given to the charge, and $q$ is the charge.

EMF as Energy per Charge

A simple way to think about EMF is to imagine a battery lifting electric charge uphill in electric potential. The battery does work on the charge, and that energy can later be delivered to the rest of the circuit.

If a source has EMF $\mathcal{E} = 12\ \text{V}$, this means it gives $12\ \text{J}$ of energy to each coulomb of charge passing through it.

This idea is very important. Voltage across a resistor tells us how much energy per charge is lost or transferred there. EMF tells us how much energy per charge is supplied by the source.

EMF and Potential Difference

EMF is closely related to potential difference, but they are not always exactly the same thing.

For an ideal source, the terminal voltage equals the EMF. In real sources, this is not always true because the source itself may have internal resistance. When current flows, some energy is lost inside the source.

Suppose a battery has EMF $\mathcal{E}$ and internal resistance $r$, and it delivers current $I$. Then the terminal voltage $V$ across the battery's external terminals is

$$
V = \mathcal{E} - Ir
$$

This means the battery gives energy to the circuit, but part of that energy is used up inside the battery itself.

For a real source delivering current,
$$
V = \mathcal{E} - Ir
$$
The terminal voltage is smaller than the EMF when current is flowing out of the source.

Internal Resistance

Real batteries and power supplies are not perfect. Charges moving inside them also face resistance. This is called internal resistance.

Because of internal resistance, not all the energy provided by the source reaches the external circuit. Some becomes thermal energy inside the source.

If the external resistance is $R$, then the total resistance in the circuit is

$$
R_{\text{total}} = R + r
$$

and the current is

$$
I = \frac{\mathcal{E}}{R + r}
$$

This shows that the current depends not only on the external circuit but also on the source itself.

Ideal and Real Sources

It is useful to compare ideal and real voltage sources.

Source type	Internal resistance	Terminal voltage when current flows
Ideal source	$0$	$V = \mathcal{E}$
Real source	$r > 0$	$V = \mathcal{E} - Ir$

An ideal source maintains the same voltage no matter how much current flows. A real source cannot do this perfectly.

Open Circuit and Closed Circuit

When no current is flowing, the circuit is open. In that case, $I = 0$, so

$$
V = \mathcal{E}
$$

Thus the terminal voltage of a battery measured with no load attached is equal to its EMF.

When the circuit is closed and current flows, the terminal voltage drops because of the internal resistance.

Physical Picture Inside a Battery

Inside a battery, chemical reactions separate charges. Positive charge is driven toward one terminal, and negative charge toward the other. This creates an electric potential difference between the terminals.

The battery therefore has two roles. It creates a separation of charge, and it continually does work to maintain that separation as current flows.

A battery is not a store of charge. It is a store of chemical energy that can be converted into electrical energy.

Rendering scientific sketch…

A battery with internal resistance connected to an external resistor

In this circuit, the source has EMF $\mathcal{E}$ and internal resistance $r$, while $R$ is the resistance of the external load.

Energy in the Complete Circuit

Each coulomb of charge gains energy $\mathcal{E}$ from the source. Then that energy is distributed in the circuit.

Part of it is lost inside the source due to the internal resistance, equal to $Ir$ volts per unit charge. The rest appears across the external circuit, equal to $V$ volts per unit charge.

So the energy balance per unit charge is

$$
\mathcal{E} = V + Ir
$$

This is another way of writing the terminal voltage relation.

For each unit charge moving through the circuit,
$$
\text{energy supplied by source} = \text{energy delivered outside} + \text{energy lost inside source}
$$
Mathematically,
$$
\mathcal{E} = V + Ir
$$

EMF from Different Sources

Although batteries are the most common example, EMF can come from many devices. The source of energy can be different in each case.

Device	Energy source producing EMF
Battery	Chemical energy
Generator	Mechanical energy
Solar cell	Light energy
Thermocouple	Thermal energy

In every case, the basic idea is the same. Some non-electric process supplies energy to charges and maintains current.

Simple Example

Consider a battery with EMF $\mathcal{E} = 9\ \text{V}$ and internal resistance $r = 1\ \Omega$, connected to an external resistor $R = 8\ \Omega$.

The total resistance is

$$
R + r = 8 + 1 = 9\ \Omega
$$

So the current is

$$
I = \frac{\mathcal{E}}{R+r} = \frac{9}{9} = 1\ \text{A}
$$

The terminal voltage is

$$
V = \mathcal{E} - Ir = 9 - (1)(1) = 8\ \text{V}
$$

This matches the voltage across the external resistor:

$$
V = IR = (1)(8) = 8\ \text{V}
$$

So the battery supplies $9\ \text{J}$ per coulomb, of which $8\ \text{J}$ per coulomb reaches the external circuit and $1\ \text{J}$ per coulomb is lost inside the battery.

Key Idea to Remember

Electromotive force is the energy per unit charge provided by a source. It is the reason charges can keep moving around a complete circuit. In ideal sources, EMF equals terminal voltage. In real sources, internal resistance causes the terminal voltage to be smaller than the EMF whenever current flows.

EMF is not a mechanical force. It is a voltage-like quantity measured in volts, representing energy supplied per unit charge.
$$
\mathcal{E} = \frac{W}{q}
$$

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