EMF and Internal Resistance: Terminal Voltage and Lost Volts

By Engr. Henry Akpan

Why a Battery's Voltage Drops Under Load

Measure a car battery with a voltmeter and you get about 12.6 V. Turn the key to start the engine and the reading drops sharply, and the headlights dim. A new AA cell reads 1.5 V, but connect it to a powerful motor and the voltage across it falls. The reason is that every real source of electricity has internal resistance. This tutorial explains e.m.f. and internal resistance, terminal potential difference and "lost volts", how to measure them with a simple experiment and graph, cells in series and parallel, and the maximum power transfer theorem, with WAEC and university-level worked examples.

Electromotive Force (e.m.f.)

The electromotive force, E, of a cell is the energy it converts from chemical (or other) energy into electrical energy per coulomb of charge passing through it. It is measured in volts (1 V = 1 J/C). Despite its name, it is not a force.

The e.m.f. equals the potential difference across the cell's terminals when no current flows (an open circuit). A high-resistance voltmeter connected alone across a cell reads very nearly its e.m.f.

Internal Resistance and Lost Volts

Charge flowing through a cell must pass through the cell's own materials (the electrolyte and electrodes), which have resistance. This is the internal resistance, r. When a current I flows, part of the e.m.f. is used up driving the current through r. This is the lost volts, Ir, which becomes heat inside the cell.

What is left appears across the terminals and drives current through the external circuit. This is the terminal potential difference, V:

E = V + Ir, and since V = IR for an external resistance R, E = I(R + r).

So the current is I = E/(R + r), and the terminal p.d. is V = E − Ir.

Worked Examples on e.m.f. and Internal Resistance

Example 1. A cell of e.m.f. 1.5 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. Find the current, the terminal p.d. and the lost volts.

I = 1.5/(2.5 + 0.5) = 0.5 A. V = IR = 0.5 × 2.5 = 1.25 V. Lost volts = Ir = 0.5 × 0.5 = 0.25 V. Check: 1.25 + 0.25 = 1.5 V ✓.

Example 2. The terminal p.d. of a battery is 9.0 V when no current flows and 8.4 V when it supplies 2.0 A. Find r.

E = 9.0 V. Lost volts = 9.0 − 8.4 = 0.6 V = Ir, so r = 0.6/2.0 = 0.3 Ω.

Example 3. A car battery has E = 12.6 V and r = 0.02 Ω. The starter motor draws 150 A. Find the terminal p.d.

V = 12.6 − 150 × 0.02 = 12.6 − 3.0 = 9.6 V. This large drop is why the headlights dim while the engine is cranking. A car battery must have a very low r to deliver such large currents.

Example 4 (two readings). When a cell is connected to a 4 Ω resistor the current is 0.5 A; with a 9 Ω resistor the current is 0.25 A. Find E and r.

E = 0.5(4 + r) and E = 0.25(9 + r). So 2 + 0.5r = 2.25 + 0.25r, giving 0.25r = 0.25 and r = 1 Ω. Then E = 0.5 × 5 = 2.5 V.

Measuring e.m.f. and Internal Resistance by Experiment

Connect the cell in series with an ammeter and a variable resistor (rheostat), with a voltmeter across the cell's terminals. Change the resistance and record pairs of readings of V and I.

Rearranging E = V + Ir gives V = −rI + E, which has the form y = mx + c. So a graph of V (y-axis) against I (x-axis) is a straight line:

  • the intercept on the V-axis is the e.m.f., E (the terminal p.d. when I = 0);
  • the gradient is −r, so r is minus the gradient;
  • the intercept on the I-axis is E/r, the short-circuit current that would flow if the terminals were joined directly.

For example, if the line meets the V-axis at 1.50 V and falls to 1.10 V at 0.80 A, the gradient is −0.40/0.80 = −0.5, so r = 0.5 Ω and E = 1.50 V. Take readings quickly and switch off between them, because a cell's e.m.f. and internal resistance change as it warms up and runs down.

Cells in Series and Parallel

  • In series (positive to negative): the e.m.f.s add and the internal resistances add. Four 1.5 V cells, each with r = 0.2 Ω, give E = 6.0 V and r = 0.8 Ω.
  • In parallel (identical cells): the e.m.f. stays the same, but the internal resistance is divided by the number of cells. Four identical 1.5 V, 0.2 Ω cells in parallel give E = 1.5 V and r = 0.05 Ω. Parallel cells can supply a larger current and last longer.
  • A cell connected the wrong way round in a series string subtracts its e.m.f. but still adds its internal resistance.

Power and the Maximum Power Transfer Theorem

The power delivered to the external resistance R is

P = I²R = E²R / (R + r)².

If R is very small, the current is large but almost all the power is wasted in r. If R is very large, little current flows. In between there is a maximum. Differentiating P with respect to R and setting the result to zero gives:

Maximum power is delivered to the load when R = r.

The maximum power is then P(max) = E²/(4r).

Example 5. A source has E = 12 V and r = 3 Ω. Find the load resistance for maximum power and that power.

R = 3 Ω, I = 12/6 = 2 A, P = 2² × 3 = 12 W, equal to E²/(4r) = 144/12 = 12 W ✓. With R = 1 Ω: I = 3 A, P = 9 W; with R = 9 Ω: I = 1 A, P = 9 W. Both are less than 12 W.

Efficiency at maximum power. When R = r, the same power is wasted in the source as is delivered to the load, so the efficiency is only 50%. In general, efficiency = R/(R + r).

So where efficiency matters, as in power stations and batteries, the load is made much larger than r. Matching R to r is used where getting the most power from a weak signal matters more than efficiency, such as matching a loudspeaker to an amplifier or an aerial to a radio receiver.

Why Internal Resistance Matters in Practice

  • As batteries age or run down, their internal resistance rises, so they can no longer deliver large currents even if their open-circuit voltage still looks normal. Battery testers therefore check the voltage under load.
  • Cold weather increases the internal resistance of car batteries, which is why cars are harder to start on cold mornings.
  • High-voltage supplies such as electron-beam equipment are deliberately given a large internal resistance, so that if someone touches the terminals the current is limited.
  • Lithium-ion cells have low internal resistance, which is why they can power drones and power tools, and also why short-circuiting them is dangerous: the short-circuit current E/r is very large.

Summary

  • e.m.f. E is the energy converted per coulomb; it equals the terminal p.d. when no current flows.
  • E = V + Ir = I(R + r); the lost volts Ir increase with current.
  • A graph of V against I has intercept E and gradient −r.
  • Series cells add e.m.f.s and internal resistances; identical parallel cells keep E but divide r.
  • Maximum power is delivered when R = r, but the efficiency is then only 50%.

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