The Filament Lamp: A Straight Line That Should Not Be Straight
A student investigates a 6.0 V torch bulb using a variable power supply, an ammeter and a voltmeter. She records seven pairs: (0.50 V, 0.100 A), (1.00 V, 0.160 A), (2.00 V, 0.240 A), (3.00 V, 0.310 A), (4.00 V, 0.360 A), (5.00 V, 0.410 A), (6.00 V, 0.450 A).
Her partner draws a single straight line from the origin through the 6.0 V point, takes its gradient, and writes: 'The resistance of the bulb is 13.3 Ω.' The pair must submit one value of resistance, labelled with the condition under which it applies, together with a short justification.
The teacher has written on the board only: 'A filament lamp is not an ohmic conductor.' The student also notes that the ammeter was on the 1 A range with a resolution of 0.01 A, that the voltmeter read to 0.01 V, and that the bulb is rated 6.0 V, 0.45 A.
1. Compute the resistance at each of the seven points and describe the pattern.
Read the worked answer
Using R = V/I at each measured pair:
0.50/0.100 = 5.0 Ω; 1.00/0.160 = 6.25 Ω; 2.00/0.240 = 8.33 Ω; 3.00/0.310 = 9.68 Ω; 4.00/0.360 = 11.1 Ω; 5.00/0.410 = 12.2 Ω; 6.00/0.450 = 13.3 Ω.
The resistance rises monotonically, roughly tripling between the lowest and highest points, and the increments are largest at low voltage. That is the signature of a filament whose temperature, and therefore resistivity, climbs with current: the lamp is non-ohmic, and no single resistance describes it. The gradient of a V-I graph only equals resistance when the graph is a straight line through the origin; here the gradient itself changes from point to point. Note also that the ratio R = V/I defines resistance at each operating point regardless of linearity, so the values are meaningful individually even though they are not constant.
2. Explain why the straight-line fit is wrong and what the 13.3 Ω figure actually represents.
Read the worked answer
The fit asserts V is proportional to I, which the data contradict: a line through the origin and the 6.0 V point predicts 0.075 A at 1.00 V, whereas 0.160 A was measured, an error of more than a factor of two. It also discards six of the seven points, and it treats the two lowest-voltage points, where the curve bends most, as noise.
What the partner has actually calculated is V/I at a single operating point, not a line's gradient: 6.00/0.450 = 13.3 Ω, the resistance of the filament when it is white-hot at its rated 6.0 V and 0.45 A. That number is correct as a statement about one condition and wrong as a statement about the bulb. Reporting it unlabelled implies a constant property, and that implication is what the teacher's board note rules out.
3. State the value the pair should submit for operation at 3.0 V, with an uncertainty justified from the stated resolution.
Read the worked answer
At 3.00 V the current is 0.310 A, so R = 3.00/0.310 = 9.68 Ω.
Uncertainty: the ammeter resolves 0.01 A, so the current carries about ±0.005 A, i.e. 0.005/0.310 = 1.6%. The voltmeter at 3.00 V contributes about 0.005/3.00 = 0.17%, negligible by comparison. Combining in quadrature or simply taking the dominant term gives about ±0.16 Ω, so the reported value is 9.7 Ω ± 0.2 Ω at an operating point of 3.0 V.
The report should read: 'The resistance of the bulb when operated at 3.0 V is 9.7 Ω ± 0.2 Ω. The resistance is not constant; it rises with temperature, reaching 13.3 Ω at 6.0 V.' Submitting 13.3 Ω for a 3.0 V operating point would be wrong by 37% and would misrepresent the device.