AP Syllabus focus: 'A nonideal battery is modeled as an ideal battery in series with internal resistance; current reduces its terminal potential difference.'
Real batteries do not behave like perfect energy sources. In AP Physics 2, they are modeled with a simple internal resistor that explains why the voltage available to a circuit falls when current increases.
Modeling a Real Battery
A nonideal battery is treated as two parts combined into one device: an ideal battery, which provides energy to charges, and an internal resistance, which represents energy lost inside the battery itself. In circuit diagrams, this model is equivalent to an ideal source placed in series with a small resistor inside the battery.

Equivalent-circuit model of a real battery: an ideal emf source () in series with an internal resistance (). The labeled terminals emphasize that the external circuit sees a reduced voltage when current flows because some potential drop occurs inside the battery across . Source
This simple model captures a major difference between real sources and idealized ones.
When the battery pushes charge through a circuit, the same current must also pass through this internal resistor. That is the key reason a real battery does not always provide its full labeled voltage to the external circuit.
Internal resistance: The effective resistance inside a battery that causes some energy to be dissipated within the battery as current passes through it.
This model does not mean there is literally a single tiny resistor sitting inside the battery. Instead, it is a useful representation of the battery’s chemical and structural limits. For AP Physics 2, the internal resistance is usually treated as a constant value called .
Terminal Potential Difference
The terminal potential difference is the actual potential difference measured across the battery’s external terminals. This is the voltage that the rest of the circuit receives, so it is often smaller than the battery’s emf when current is flowing.
Terminal potential difference: The potential difference across the two external terminals of a battery; it is the energy transferred per unit charge to the external circuit.
A real battery can therefore be described in two related ways. Its emf is the maximum energy per unit charge the battery can provide ideally, while its terminal potential difference is the part that remains available outside the battery after internal losses are accounted for.
For a battery that is delivering current to a circuit, the internal resistance produces a potential drop of inside the battery.
= terminal potential difference across the battery, V
= emf of the ideal battery, V
= current supplied by the battery, A
= internal resistance of the battery,
This equation shows the central idea of the topic: as current increases, the term increases, so the terminal potential difference decreases.

Voltage profile around a simple series circuit showing the battery’s emf rise and the internal potential drop inside the source. The plot makes it clear that the terminal potential difference equals , with the remaining drop occurring across the external load. Source
If the battery supplies very little current, the internal voltage drop is small, and the terminal potential difference stays close to the emf.
An important special case occurs when the circuit is open and no current flows. Then , so there is no voltage drop across the internal resistance. Under that condition, the terminal potential difference equals the emf. This is why a battery can appear to have its full voltage when tested with almost no load, yet provide less voltage when connected to a device that draws significant current.
How Current Changes the Available Voltage
The amount of voltage “lost” inside the battery depends directly on current.
Zero current: no internal voltage drop, so .
Small current: a small internal drop occurs, so terminal potential difference is only slightly less than emf.
Large current: a larger internal drop occurs, so the terminal potential difference can be much lower than emf.
Because of this, the voltage supplied by a real battery is not fixed by the battery alone. It also depends on how much current the external circuit draws. A device that demands more current causes a greater loss inside the battery, reducing the voltage available at the terminals.
This explains why batteries in high-current situations may seem weaker than expected. The battery is still providing energy, but a larger fraction of that energy is being used up inside the battery rather than being delivered to the external circuit.
Interpreting the Series Model
Thinking of the battery as an ideal source in series with an internal resistor helps with both circuit reasoning and physical interpretation. Since the elements are in series, the current through the ideal source and the internal resistance is the same. The ideal source raises electric potential, while the internal resistance lowers part of it before charge reaches the external circuit.
The model also explains why terminal voltage is sometimes called the loaded voltage of the battery. When a load is attached and current flows, the battery’s internal resistance matters. A larger internal resistance causes a larger internal drop for the same current, so two batteries with the same emf can produce different terminal potential differences in the same situation.
In AP Physics 2 problems, the most important reasoning is usually qualitative:
A larger current means a smaller terminal potential difference for a discharging battery.
A larger internal resistance means a smaller terminal potential difference at the same current.
A battery with no current has terminal potential difference equal to its emf.
These ideas connect the battery’s internal properties to what the external circuit actually experiences. The terminal potential difference is the useful voltage available to the rest of the circuit, while the difference between emf and terminal potential difference represents the energy per unit charge lost inside the battery because of internal resistance.
FAQ
A common method is to measure the battery’s emf with almost no current drawn, then connect a load and measure both the terminal voltage and the current.
You can then use $r=\dfrac{\mathcal{E}-V_{terminal}}{I}$.
Another method is to take several pairs of current and terminal voltage data, then analyze how $V_{terminal}$ changes as $I$ changes.
As a battery discharges, its chemical reactants are used up and the internal chemical processes become less effective at moving charge.
This can happen because:
ion movement becomes less efficient
reaction products build up
the battery’s internal materials no longer support charge flow as well
The result is a larger internal voltage drop for the same current.
Internal resistance causes energy to be dissipated inside the battery itself.
The heating effect increases with current because the internal power loss is $P=I^2r$.
That means:
small current produces little heating
large current can produce significant heating
very large current can reduce performance and damage the battery
This is one reason high-current devices are demanding on batteries.
During charging, current is forced through the battery in the opposite direction from normal discharge.
In that case, the terminal potential difference can be greater than the emf because the charger must both overcome the battery’s emf and drive current through its internal resistance.
A simple charging model is $V_{terminal}=\mathcal{E}+Ir$.
Two batteries can have the same emf but very different internal resistances.
A battery with lower internal resistance:
has a smaller internal voltage drop at high current
delivers a larger terminal voltage under load
wastes less energy inside itself
That is why some batteries are better suited for motors, camera flashes, and starting systems, even when their rated emf is the same.
Practice Questions
A battery has an emf of 12.0 V and an internal resistance of . It supplies a current of 2.0 A.
Calculate the terminal potential difference of the battery. [2 marks]
Uses or equivalent. (1 mark)
Substitutes correctly and gets 11.0 V. (1 mark)
A nonideal battery has emf 9.0 V and internal resistance .
(a) Explain why the terminal potential difference is less than 9.0 V when the battery is delivering current. [2 marks]
(b) The battery supplies a current of 2.0 A. Calculate the terminal potential difference. [2 marks]
(c) The current increases to 3.0 A. State and explain what happens to the terminal potential difference. [1 mark]
(a) States that current passes through the internal resistance. (1 mark)
(a) States that this causes an internal potential drop, so less potential difference is available at the terminals. (1 mark)
(b) Uses . (1 mark)
(b) Calculates V. (1 mark)
(c) States that terminal potential difference decreases because the internal drop becomes larger when current increases. (1 mark)
