Free Pearson Edexcel International A Level Physics lesson
Real cells and internal resistance
Separate source energy per charge from the voltage available outside.
What you’ll learn
- Distinguish e.m.f. from terminal potential difference.
- Determine e.m.f. and internal resistance from a graph.
- Evaluate cell loading and the limits of the simple source model.
A source has an internal energy cost
The electromotive force ε of a cell is the energy it supplies per unit charge through its chemical processes. Despite its name, e.m.f. is measured in volts and is not a mechanical force. A real cell also transfers some energy internally when current flows. Representing that effect by an internal resistance r gives an internal potential drop Ir while the cell supplies the external circuit.
For current I leaving the positive terminal, energy conservation gives ε = Vterminal + Ir, so Vterminal = ε − Ir. With no current through an ideal infinite-resistance voltmeter, the terminal voltage approaches the e.m.f. Connecting a finite load produces current and generally lowers terminal voltage. A cell’s quoted e.m.f. therefore does not guarantee the same terminal voltage for every load.
The relation assumes ε and r remain approximately constant over the measured current range and time. Temperature, state of charge, chemical polarisation and ageing can alter them. If the cell is being charged, the current direction reverses and the external terminal voltage can exceed its e.m.f.; keep the sign convention explicit. For this lesson, numerical examples use a supplying cell unless the question says otherwise.
Supplying cell: ε = Vterminal + Ir; I = ε/(Rexternal + r); Vterminal = IRexternal
Symbols and units
- ε is emf in V; Vterminal is terminal voltage in V; I is supplied current in A; r is internal resistance in Ω; R is external resistance in Ω.
How the equation explains the idea
- For a supplying cell, energy per charge balances ε = Vterminal + Ir. Combining with Vterminal = IR gives I = ε/(R + r).
- At open circuit I = 0 and terminal voltage equals emf. Increasing supplied current increases the lost volts Ir and lowers terminal voltage in this constant-r model.
Worked example
Account for each volt
- A cell has e.m.f. 1.50 V and internal resistance 0.50 Ω.
- At current 0.40 A, the internal drop is Ir = 0.20 V.
- Terminal voltage = 1.50 − 0.20 = 1.30 V.
The cell supplies 1.50 J per coulomb, of which 1.30 J per coulomb is available externally in this model.
Keep this idea. Use an energy-per-charge loop balance and a stated current direction to distinguish ε from terminal voltage.
Core practical 8: use a voltage–current line
Connect a cell to a variable external resistance, with an ammeter in series and a voltmeter across the cell’s terminals. Include a switch and a suitable current-limiting resistance so the cell is not short-circuited. Choose several safe load settings, close the switch briefly, allow the readings to settle and record current and terminal potential difference together. Open the switch between measurements to reduce heating and unnecessary discharge.
Plot terminal voltage vertically against current horizontally. The model V = ε − rI has vertical intercept ε and gradient −r. Use a best-fit line over the region where the data support an approximately constant source model. Two well-separated points on the fitted line give a more robust gradient than adjacent raw points. A zero-current intercept can be extrapolated without physically drawing a dangerously large short-circuit current.
Repeat some settings in a different order to check whether the cell changes during the experiment. Curvature, drift or hysteresis may indicate temperature or chemical effects rather than merely random meter scatter. Check voltmeter and ammeter ranges, contact quality and resolution. The experiment determines effective parameters for the measured conditions, not a universal lifelong value for the cell. Record the cell type and its condition sufficiently clearly for another group to interpret the result.
V = ε − rI; gradient of V versus I = −r; vertical intercept = ε
Symbols and units
- V is terminal voltage in V, I is supplied current in A, ε is emf in V and r is internal resistance in Ω.
How the equation explains the idea
- Compare V = ε − rI with a straight line y = c + mx. With V vertical and I horizontal, intercept c = ε and gradient m = −r.
- A line dropping from 1.5 V at 0 A to 1.1 V at 0.8 A has slope −0.5 V A⁻¹, so r = 0.5 Ω. The negative slope does not mean a negative resistance.
Conditions. Constant-emf, constant-internal-resistance supplying-cell models with ε = 1.5 V. Current is treated as a controlled operating parameter; the range is illustrative, not a safe-current rating.
Supplied current / A; Terminal voltage / V.
Both intercept at 1.5 V. For r = 0.5 Ω, terminal voltage at 0.8 A is 1.1 V; for r = 1 Ω it is 0.7 V. Internal resistance is minus the gradient.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Two points on a fitted line
- A fitted line passes through (I, V) = (0.20 A, 1.42 V) and (0.80 A, 1.18 V).
- Gradient = (1.18 − 1.42)/(0.80 − 0.20) = −0.40 Ω, so r = 0.40 Ω.
- ε = V + Ir = 1.42 + 0.20 × 0.40 = 1.50 V.
The negative slope encodes voltage sag; internal resistance is the positive magnitude.
Keep this idea. Fit terminal voltage against current safely, then read ε from the intercept and r from the negative gradient.
Power inside and outside a real source
A supplying cell transfers total power εI. In the effective internal-resistance model, internal heating is I²r and power delivered to an external resistance is I²R. Their sum equals εI because ε = I(R + r). The external share of this idealised electrical power is R/(R + r), so a large external resistance can improve that fraction while also reducing current and total output power.
For a fixed ε and r, external power is ε²R/(R + r)². It reaches a maximum when R equals r, but then half the source power is dissipated internally. This mathematical result is not a recommendation to operate a cell at that load: current or heating may be unsafe, and the constant-r model may fail. Distinguish maximum output power from maximum efficiency and from the manufacturer’s safe operating conditions.
A near short circuit gives a small terminal voltage yet can produce a large current and substantial internal heating. A low voltage is therefore not evidence that an overloaded cell is harmless. Use suitable loads and current limits in practical work rather than attempting to test the model by directly shorting a cell. Changes caused by depletion or heating should be reported as physical limitations, not silently absorbed into a single unexplained resistance value.
Psource = εI; Pinternal = I²r; Pexternal = I²R = ε²R/(R + r)²
Symbols and units
- ε is emf in V; I is current in A; r and R are internal and external resistances in Ω; powers are in W.
How the equation explains the idea
- Multiply ε = IR + Ir by I: εI = I²R + I²r. Source power is split between useful external power and internal heating.
- Insert I = ε/(R + r) to get Pexternal = ε²R/(R + r)². This is largest at R = r for a fixed positive r, but efficiency R/(R + r) is then only 50%.
- Maximum-power matching is not maximum efficiency; it can also exceed a real cell’s safe current, so the model is not an instruction to short or overload a cell.
Conditions. Ideal constant ε = 3 V and r = 1 Ω model, Pexternal = ε²R/(R + r)². The R = 0 endpoint is a mathematical short-circuit limit, not a recommended experiment or a safe operating condition.
External resistance / Ω; External power / W.
External power peaks at R = r = 1 Ω with 2.25 W. Half the source power is then lost internally. At R = 0 external power is zero while the model still predicts internal heating; at very large R power tends towards zero because current becomes small.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A power balance under load
- A 6.0 V source with r = 1.0 Ω supplies R = 5.0 Ω, giving I = 1.0 A.
- Source power is 6.0 W; internal power is 1.0² × 1.0 = 1.0 W.
- External power is 1.0² × 5.0 = 5.0 W, and 1.0 + 5.0 = 6.0 W.
Power transferred internally remains part of the complete energy balance.
Keep this idea. Balance source, internal and external power, and distinguish output power from efficiency and safe operation.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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