A steady current in a resistive circuit dissipates electrical energy. To maintain that current, a source such as a battery must use non-electrostatic processes to move charge from lower electric potential back toward higher electric potential inside the source.

The electromotive force (EMF), written $\mathcal E$, is the energy supplied by the source per unit charge:

$$\boxed{\mathcal E=\frac{W_{\rm source}}{q}}.$$

Its SI unit is the volt. Despite the name, EMF is not a force; it is an energy-per-charge quantity.

For an ideal voltage source with no internal losses, the terminal voltage equals the EMF.

Real source with internal resistance

A simple model of a real battery is an ideal EMF $\mathcal E$ in series with an internal resistance $r$.

When the source delivers current $I$ to an external load, the internal resistance produces an internal voltage drop

$$Ir.$$

The terminal voltage is therefore

$$\boxed{V_{\rm term}=\mathcal E-Ir}$$

for the discharging sign convention.

The larger the current drawn, the farther the terminal voltage falls below the open-circuit EMF.

When

$$I=0,$$

no internal voltage drop occurs and

$$\boxed{V_{\rm term}=\mathcal E}.$$

This is why measuring a battery with a very high-resistance voltmeter approximates its open-circuit EMF.

Battery connected to a resistor

Suppose a battery with EMF $\mathcal E$ and internal resistance $r$ drives an external resistor $R$.

Kirchhoff's loop law gives

$$\mathcal E-Ir-IR=0,$$

so

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

The terminal voltage across the external resistor is

$$V_{\rm term}=IR =\mathcal E\frac{R}{R+r}.$$

Worked example

A battery has

$$\mathcal E=12.0,\mathrm V$$

and

$$r=0.50,\Omega.$$

It is connected to

$$R=5.5,\Omega.$$

The current is

$$I=\frac{12.0}{5.5+0.50}=2.0,\mathrm A.$$

The terminal voltage is

$$V_{\rm term}=12.0-(2.0)(0.50)=\boxed{11.0,\mathrm V}.$$

The external resistor also has

$$IR=(2.0)(5.5)=11.0,\mathrm V,$$

as required.

Power supplied and lost internally

The ideal source converts other forms of energy into electrical energy at rate

$$\boxed{P_{\rm source}=\mathcal E I}.$$

The internal resistance dissipates

$$P_{\rm int}=I^2r,$$

while the external circuit receives terminal power

$$P_{\rm term}=V_{\rm term}I.$$

For the simple discharging model,

$$\mathcal E I=V_{\rm term}I+I^2r.$$

This is an energy balance: some source power reaches the external circuit and some becomes internal heating.

Charging a source

If an external circuit drives current into the source's positive terminal strongly enough, energy flows into the source rather than out of it. In the same series-resistance model, the terminal voltage during charging can exceed the EMF because the current direction through $r$ reverses relative to the discharging convention.

Real batteries have chemistry-dependent behavior that is more complicated than one constant internal resistance. The model is nevertheless useful for separating a source's energy-per-charge capability from the voltage losses that appear when current flows.