EMF and internal resistance of a cell

What just happened?

You shorted the cell — the load resistance fell almost to zero, so the current surged and the entire voltage "sat down" on the internal resistance. Almost nothing was left at the terminals.

EMF, current and terminal voltage:

— the EMF ε is the source voltage with no load (electromotive force). r — the internal resistance of the cell (Ω), R — the load resistance (Ω), I — the current (A), U — the terminal voltage (V).

Why does a battery "sag" under load?

Every cell has an internal resistance r. When current flows, part of the voltage is already lost inside the cell — the drop $I\cdot r$. That is why under load we measure less than the EMF at the terminals:

Two extreme cases:

Open circuit ($R\to\infty$): the current $I=0$, so the drop $I\cdot r=0$ and U = ε (the full EMF). Short circuit ($R\to 0$): maximum current $I=\varepsilon/r$, and U ≈ 0 — the entire voltage across the internal resistance.

Why you need this for the exam:

Cell problems are usually the pair $I=\varepsilon/(R+r)$ and $U=\varepsilon-Ir$. The trick: a voltmeter at the terminals measures U, not ε — you read the EMF only with an open circuit. Knowing U and I at two loads, you can work out ε and r.

Remember: U = ε − I·r. Under load the terminal voltage is less than the EMF by the drop across the internal resistance; at open circuit U = ε, in a short circuit U ≈ 0.