10.1 - Internal resistance and terminal voltage

Energy transferred inside a source

Why might a cell read $1.5\,\mathrm{V}$ with no lamp connected but only $1.2\,\mathrm{V}$ while lighting the lamp? A cell can supply energy and transfer some energy internally at the same time. The p.d. available to the external circuit is then smaller than its e.m.f.

Model a real source as an ideal source of e.m.f. $E$ in series with an internal resistance $r$. The resistor is a model for internal energy transfer; it need not be a separate physical resistor hidden inside the cell. Current passes through both the source and its internal resistance.

[DIAGRAM: asset_name: A real source and its load; asset_slug: circuits-internal-source-model; caption: Internal and external resistances are in the same current path; description: A complete circuit contains an ideal cell of e.m.f. E and internal resistor r inside a dashed source boundary, connected to external load R. Terminal dots A and B bound the entire real source. Labels distinguish the internal p.d. Ir from the terminal p.d. V across the load.]

For a charge $Q$, the source supplies energy $EQ$. Some is transferred internally and the rest is delivered to the external circuit. The source's terminal p.d. $V$ describes only the external energy per charge. “Lost volts” is shorthand for an internal p.d.; it does not mean that energy or charge has vanished.

A $1.5\,\mathrm{V}$ source supplying $0.20\,\mathrm{A}$ through $1.5\,\Omega$ of internal resistance has an internal p.d. of $Ir=0.30\,\mathrm{V}$. For every coulomb, $1.5\,\mathrm{J}$ is supplied, $0.30\,\mathrm{J}$ is transferred internally and $1.2\,\mathrm{J}$ reaches the external circuit. That accounts for the two readings in the opening question without using up current.

The energy account also gives power: the source supplies $EI$, the external circuit receives $VI$ and the internal resistance transfers $I^2r$, all in watts when SI units are used. The power relation $P=I^2R$ was taught earlier and is not on the data sheet: learn it. A hot or depleted source may therefore behave differently during a long experiment.

Terminal p.d. and the complete resistance

Conservation of energy per charge connects the three voltages: the e.m.f. supplies the internal p.d. plus the external terminal p.d. Current therefore affects the external voltage even if the e.m.f. remains unchanged.

If the load is a resistor $R$, its p.d. is $V=IR$. Substitution gives $E=I(R+r)$, hence $I=E/(R+r)$. This is a derived complete-circuit relation, not a separate printed data-sheet formula. Include internal resistance when calculating current; use only the external resistance to find the p.d. delivered to the load.

You can work backwards from a load measurement too.