When a cell supplies current, its terminal potential difference is its electromotive force minus the potential difference across its internal resistance. A smaller external resistance draws more current in the simple cell model, so the terminal potential difference falls.
The 2027 H2 Physics 9478 syllabus includes how internal resistance affects terminal potential difference and output power. This worked example compares two loads on the same source and then uses those two readings to estimate the internal resistance.
- Treat the internal resistance and the external load as resistors in series.
- Use the total resistance to find the current.
- Calculate terminal potential difference with V = E − Ir or V = IR.
- Use the current and voltage changes to check the internal resistance.
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Separate source voltage from terminal voltage
Use E for the electromotive force, r for the internal resistance and R for the external load. For a source delivering current to a single load, E = I(R + r). The terminal potential difference V is the potential difference across the load, so V = IR = E − Ir.
The model takes E and r as constant for the comparison. With an open circuit, an ideal voltmeter reads E because there is no current and the internal potential difference Ir is zero. Once a load is connected, the same voltmeter reads the voltage available at the terminals.
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Worked example | One cell, two external resistances
Take a cell with E = 1.50 V and r = 0.50 Ω. First connect a 2.50 Ω load. The total resistance is 3.00 Ω and the current is 1.50 ÷ 3.00 = 0.50 A. The terminal potential difference is 0.50 × 2.50 = 1.25 V.
Now replace the load with a 1.00 Ω resistor. The total resistance is 1.50 Ω, so the current is 1.00 A. The terminal potential difference is 1.00 × 1.00 = 1.00 V. The smaller resistor draws more current, increasing the drop inside the cell from 0.25 V to 0.50 V.
The output powers are 1.25 × 0.50 = 0.625 W and 1.00 × 1.00 = 1.00 W. In this comparison the load receives more power even though its terminal voltage is lower, because the current has increased enough to outweigh the voltage drop.
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Work backwards from two measurements
Suppose you were given only the two pairs of readings, (0.50 A, 1.25 V) and (1.00 A, 1.00 V). On a graph of V against I, the slope is (1.00 − 1.25) ÷ (1.00 − 0.50) = −0.50 V A⁻¹. Compare V = E − Ir with a straight-line equation. The slope is −r and the vertical intercept is E.
The internal resistance is therefore 0.50 Ω. Find E using either point. From the first, E = 1.25 + (0.50 × 0.50) = 1.50 V. Substitute the second pair to check that it gives the same intercept.
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Try a prediction before calculating
What terminal potential difference would you expect at 0.75 A if the model still holds? That current lies halfway between the two earlier readings, so the voltage should also lie halfway between 1.25 V and 1.00 V. Using V = 1.50 − (0.75 × 0.50) gives 1.125 V, consistent with the graph.
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Frequently asked questions
Why is the gradient negative if resistance is positive?
Increasing current increases the internal potential difference Ir, leaving less terminal potential difference. The gradient is −r, so the resistance is the magnitude of that gradient.
Why does the current calculation include r?
In the model the same current flows through the internal and external resistances. The electromotive force drives current through their combined resistance.
Can I determine the internal resistance from just one loaded reading?
If E is known independently and the current is non-zero, use r = (E − V) ÷ I. If E is unknown, the two readings let you determine both E and r under the constant-resistance model.