electromotive force and internal resistance

A real source of current is characterized by its electromotive force $\varepsilon$ and its internal resistance $r$. When a current $I$ flows, part of the EMF is "lost" inside the source, so the terminal voltage (which is also the voltage across the load) is

$$U = \varepsilon - I r.$$

For a closed circuit with external resistance $R$, Ohm's law takes the form

$$I = \frac{\varepsilon}{R + r}.$$

With an open circuit ($I \to 0$) we get $U \to \varepsilon$: that is why EMF is measured with a voltmeter of very large resistance. The useful power is $P_{\text{useful}} = I^2 R = UI$, the losses inside the source are $P_{\text{internal}} = I^2 r$, the total power is $\varepsilon I$, and the efficiency is $\eta = \frac{R}{R + r}$. The useful power is maximal when $R = r$, but then the efficiency is only $50\%$.

A battery with EMF $\varepsilon = 12$ V and internal resistance $r = 0.5\ \Omega$ is connected to a resistor $R = 5.5\ \Omega$. Find the current in the circuit and the terminal voltage of the battery.

A source with EMF $\varepsilon = 9$ V and internal resistance $r = 1.2\ \Omega$ is connected to an unknown resistor $R$. A voltmeter across the source terminals reads $U = 7.5$ V. Find the current, the resistance $R$ and the power lost inside the source.

A source with EMF $\varepsilon = 24$ V is connected to a resistor $R = 6\ \Omega$, and the terminal voltage is $U = 18$ V. Find the internal resistance $r$, the power losses and the efficiency. Then determine at which load resistance the useful power is maximal, and calculate this maximum value.

A self-study investigation in a simulation (for example PhET "Circuit Construction Kit" or any DC circuit simulator).

1. Build a circuit: a source with given $\varepsilon$ and internal resistance $r$, in series with it a variable resistor $R$, an ammeter in the circuit and a voltmeter across the source terminals.
2. Set several values of $R$ (e.g. $1, 2, 4, 8, 16\ \Omega$) and record the pair $(I, U)$ for each.
3. Plot $U$ against $I$ and check that the points fall on the straight line $U = \varepsilon - I r$.
4. Read $\varepsilon$ as the intercept of the line with the $U$ axis, and $r$ as the magnitude of the line's slope.
5. Repeat for a different value of the EMF and compare the values of $r$ obtained.

Why simulation only: a real battery is easy to short-circuit ($R \to 0$); the current $I = \varepsilon / r$ is then large, the source overheats and may catch fire. Do not perform this experiment with real batteries.

Show that the terminal voltage of a real source falls linearly as the current grows, and link the parameters of that line to the EMF and the internal resistance.

On the left, a circuit: a source drawn as an ideal EMF $\varepsilon$ in series with internal resistance $r$, connected to an external resistance $R$ with an ammeter and a voltmeter. On the right, a $U(I)$ graph: the straight line $U = \varepsilon - I r$ with intercept $(0, \varepsilon)$ on the voltage axis and slope $-r$; below the current axis the corresponding values $I = \varepsilon / (R + r)$ are marked, and a dashed line shows the drop $I r$.

A smartphone's Li-ion cell has $\varepsilon = 3.85\ \text{V}$ and an internal resistance $r \approx 0.08\ \Omega$ while it is new. Recording 4K video makes the phone draw $I = 1.6\ \text{A}$, so the terminal voltage is $U = \varepsilon - Ir = 3.85 - 1.6 \times 0.08 = 3.72\ \text{V}$, and $P = I^2 r = 0.20\ \text{W}$ is dissipated as heat inside the cell. After two years of charging, $r$ can grow to $0.3\ \Omega$; the same load then gives $U = 3.85 - 0.48 = 3.37\ \text{V}$, low enough for the phone to switch itself off although the charge indicator still shows 40 %.

This exercises the meaning of $\varepsilon$ as the source's total potential rise and of $r$ as the resistance inside the source: the drop $Ir$ subtracts from $\varepsilon$ and never reaches the external circuit — it heats the cell instead.

A car battery is marked $\varepsilon = 12\ \text{V}$ and has an internal resistance $r \approx 0.02\ \Omega$. The starter motor draws $I = 250\ \text{A}$ for about $1.5\ \text{s}$. Terminal voltage while cranking: $U = 12 - 250 \times 0.02 = 12 - 5 = 7\ \text{V}$. That 5 V drop is what makes the headlights dim and the dashboard flicker; inside the battery $P = I^2 r = 1250\ \text{W}$ heats the plates. With everything switched off and no current flowing, a voltmeter across the terminals reads $U \approx \varepsilon = 12\ \text{V}$, because $I = 0$.

The example separates the open-circuit reading ($I = 0$, the voltmeter shows $\varepsilon$) from the loaded terminal voltage $U = \varepsilon - Ir$; it also shows that the lost power $I^2 r$ grows with the square of the current, which is why a starter current of hundreds of amperes matters so much.

An e-bike pack is rated $\varepsilon = 36\ \text{V}$ with an internal resistance $r = 0.3\ \Omega$ (it contains many cells in series and parallel). On a steep climb the motor controller draws $I = 15\ \text{A}$. The terminal voltage sags to $U = 36 - 15 \times 0.3 = 31.5\ \text{V}$: the motor receives $UI = 472.5\ \text{W}$, while $P = I^2 r = 67.5\ \text{W}$ warms the pack. On the flat the same rider needs only $I = 3\ \text{A}$, so $U = 36 - 0.9 = 35.1\ \text{V}$ — a sag of less than $1\ \text{V}$, and the display hardly changes.

The scenario tests $U = \varepsilon - Ir$ as a function of the load current: the same source gives different terminal voltages at $3\ \text{A}$ and at $15\ \text{A}$, and the internal loss $I^2 r$ limits how much power can ever be taken from the pack.