The lamp that would not obey Ohm's law
A student is given a 6.0 V battery of internal resistance 2.0 Ω and a small filament lamp. Using a variable supply, the student records the lamp's current and voltage after each setting has settled: (0.10 A, 1.0 V), (0.20 A, 2.2 V), (0.30 A, 3.2 V), (0.40 A, 4.0 V), (0.50 A, 4.7 V), (0.60 A, 5.2 V).
With the variable supply removed, the lamp is connected straight across the battery. From the 0.20 A row the student computes R = 2.2 V ÷ 0.20 A = 11 Ω and predicts I = 6.0 V ÷ (11 + 2.0) Ω = 0.46 A. The measured current is 0.55 A, and the student concludes that the battery's internal resistance must really be below 2.0 Ω.
The teacher asks for a written explanation of the disagreement, a corrected prediction, and a design for running the same lamp at exactly 0.30 A from the same battery using a single resistor.
1. Using the characteristic, find the actual operating point of the lamp on the battery, and state clearly where the student's method fails.
Read the worked answer
The battery fixes a straight-line relation between terminal voltage and current: V = 6.0 − 2.0 I. The lamp must sit somewhere on its measured curve and somewhere on this line; the operating point is the intersection.
- At I = 0.50 A the line gives 5.0 V but the lamp needs 4.7 V, so the line is above the curve.
- At I = 0.60 A the line gives 4.8 V but the lamp needs 5.2 V, so the line is now below the curve.
Interpolating in that 0.10 A interval, the crossing is at about I ≈ 0.54 A, V ≈ 4.9 V, which matches the measured 0.55 A within reading uncertainty.
The method fails because it treats R as a property of the lamp rather than of one operating point. Taking R = 11 Ω from the 0.20 A row builds in the resistance appropriate to a cool filament. At the true operating point the filament is far hotter, R ≈ 4.9/0.55 ≈ 9 Ω, and a smaller resistance admits a larger current. Dividing once and finishing the algebra assumes linearity the data visibly deny.
2. The student blames the battery. Test that claim against the data, and explain in microscopic terms why the lamp's resistance behaves as it does.
Read the worked answer
The battery is innocent. Check the measured state: if I = 0.55 A flows, the terminal voltage must be 6.0 − 2.0(0.55) = 4.90 V. The lamp at 0.55 A needs about 4.95 V from the table, agreeing within 0.05 V - roughly the precision of the readings. Changing the internal resistance would break this agreement, so r = 2.0 Ω is consistent with the observation.
Microscopically, the filament is a metal. The current heats it by joule heating until it glows, and at higher temperature the lattice ions vibrate with larger amplitude. Electrons then collide with the lattice more often, so the drift velocity for a given field falls and the resistance rises. A filament lamp is therefore steeply non-ohmic: doubling the current does not double the voltage, and the resistance is a function of the very current one is trying to find. This is why the characteristic must be used graphically, or solved by iteration, and why a single quoted "resistance of the lamp" is meaningless without its current.
3. Design the series-resistor solution for running the lamp at 0.30 A from the same battery. Give the resistance and the power rating you would specify, and justify both numbers, including why the nearest standard value is not simply substituted.
Read the worked answer
At 0.30 A the lamp needs 3.2 V. The internal resistance takes 0.30 × 2.0 = 0.60 V. By Kirchhoff's voltage law the resistor must take 6.0 − 3.2 − 0.60 = 2.2 V, so R = 2.2/0.30 = 7.3 Ω.
Power in the resistor: P = VI = 2.2 × 0.30 = 0.66 W. A 1 W resistor would be running at 66 % of rating, where the body gets hot and the value drifts; specifying a 2 W part gives a comfortable margin and a stable current.
Standard values are 6.8 Ω or 7.5 Ω. Substituting 7.5 Ω without thought would drop the current to about 0.29 A, but the exact answer requires re-solving with the characteristic, because the lamp's voltage also falls as the current falls, which partly compensates. The honest design procedure is to fix the resistor, draw the new load line V = 6.0 − I(2.0 + R), and read off the intersection; the lamp's non-linearity, not the tolerance of the resistor, sets the final precision.