The Bulb That Broke Ohm's Law
A lab group is measuring the current–voltage relationship for a small incandescent bulb and a 100 Ω carbon resistor. Using a variable DC supply, they record for the bulb: 1.5 V, 0.15 A; 3.0 V, 0.22 A; 6.0 V, 0.32 A. The resistor behaves as expected, with current proportional to voltage. A student concludes that 'Ohm's law is false for the bulb.' During the first trial, the ammeter was accidentally connected in parallel with the bulb, and the supply's internal resistance is not negligible at higher currents. The voltmeter has a finite resistance, and the connecting leads add a small resistance. The teacher asks the group to decide whether the bulb's behavior disproves Ohm's law, to compute the bulb's resistance at each point, and to design a cleaner test. The group must also explain why the bulb's resistance changes and how to report the result without overgeneralizing.
1. Calculate the bulb's resistance at each of the three points. Does the changing resistance mean the bulb violates Ohm's law? Explain carefully.
Read the worked answer
R = V/I:
- 1.5 V / 0.15 A = 10 Ω
- 3.0 V / 0.22 A ≈ 13.6 Ω
- 6.0 V / 0.32 A ≈ 18.8 Ω
The resistance rises as current and temperature rise. This does not mean the bulb violates Ohm's law in the strict sense that V = IR fails at a given instant. V = IR is the definition of resistance at each operating point, and it still holds. What fails is the assumption that R is constant, which is what 'ohmic' means. The bulb is non-ohmic because its filament temperature changes with current; the resistivity of tungsten increases with temperature. The student should say 'the bulb is non-ohmic' rather than 'Ohm's law is false.' Also, the measured R includes the effects of the ammeter, voltmeter, and leads, so the true filament resistance is slightly different.
2. The ammeter was initially connected in parallel with the bulb. Explain what this would do to the circuit and the readings, and why those first-trial data should be discarded.
Read the worked answer
An ammeter is designed to have very low resistance so that it does not change the current it measures. Connected in parallel with the bulb, it provides a low-resistance path that effectively shorts the bulb. Most of the supply current flows through the ammeter, not the filament. The bulb may not light or may be very dim; the ammeter reading is the short-circuit current, not the bulb current; and the voltage across the bulb is near zero. The supply may also be overloaded, and its terminal voltage may sag because of internal resistance. The meter could be damaged if the current exceeds its range. Those data do not represent the bulb's I–V characteristic, so they must be discarded. The ammeter must be in series with the bulb, and the voltmeter in parallel with the bulb.
3. Design a cleaner I–V measurement for the bulb and explain how to correct for the finite resistance of the voltmeter and the internal resistance of the supply.
Read the worked answer
Use a four-terminal or compensated measurement. Put a low-value shunt resistor in series with the bulb and measure the voltage across the shunt with a high-impedance voltmeter to get the current; measure the voltage across the bulb directly with the same meter. Use a DC supply with a known current limit, and measure the supply terminal voltage as well as the bulb voltage so that the internal drop I·r_supply can be accounted for. If the voltmeter is across the bulb only, its finite resistance draws a small current I_V = V/R_V; the true bulb current is the ammeter current minus I_V. If the voltmeter is across both the bulb and ammeter, subtract the ammeter's burden voltage I·R_A from the reading. Keep leads short and use the same contact points. Take readings quickly or use pulsed current to reduce heating during each point, then wait for thermal equilibrium for the steady-state curve. Plot R versus I or versus estimated filament temperature, and state the measurement uncertainty.