Filament Lamp I–V Curve: Meter Placement and the Ohmic Claim
A student obtains the I–V characteristic of a 6.0 V, 0.50 A filament lamp. The circuit has a 0–12 V supply, a rheostat, the lamp, and an ammeter in series; a digital voltmeter is connected across the lamp and the ammeter together. The ammeter has resistance 0.40 Ω and the voltmeter has resistance 10 MΩ. At low current I = 0.100 A the voltmeter reads 0.290 V. At I = 0.500 A it reads 6.00 V. The student calculates R = V/I and reports that the lamp resistance rises from 2.90 Ω to 12.0 Ω. They conclude that the lamp is ohmic at low current and non-ohmic at high current because 'the resistance is constant at first and then increases'. A second student says the first student's low-current value is wrong by about 15%.
1. Calculate the corrected lamp resistance at I = 0.100 A and at I = 0.500 A. Hence determine whether the second student's 'about 15%' claim is justified at low current.
Read the worked answer
At I = 0.100 A the voltmeter reads the p.d. across lamp + ammeter: V_meas = V_lamp + I R_A. So V_lamp = 0.290 − 0.100 × 0.40 = 0.250 V, giving R_lamp = 0.250/0.100 = 2.50 Ω. The student's value is 2.90 Ω. The error is (2.90 − 2.50)/2.50 = 0.40/2.50 = 0.16, i.e. 16%, so the second student's 'about 15%' is justified. At I = 0.500 A, V_lamp = 6.00 − 0.500 × 0.40 = 5.80 V, so R_lamp = 5.80/0.500 = 11.6 Ω. The student's 12.0 Ω is only 0.4/11.6 = 3.4% high. The ammeter resistance therefore matters much more at low current because the ammeter p.d. is a larger fraction of the total when the lamp resistance is small. The voltmeter's 10 MΩ resistance is negligible in parallel with 2.5–12 Ω, so it does not explain the discrepancy.
2. Evaluate the first student's conclusion that the lamp is ohmic at low current and non-ohmic at high current. Your answer should consider the circuit and the number of data points.
Read the worked answer
The conclusion is not supported. Two readings cannot establish ohmic behaviour over a range; 'ohmic' means the resistance is constant for a given temperature, and the filament temperature changes as current changes. Even at 0.100 A the filament is slightly warmer than room temperature, so its resistance is not exactly constant. The apparent low-current constancy is partly an artefact of the circuit: the measured p.d. includes the ammeter, and the ammeter resistance is a significant fraction of the lamp resistance at low current. Correcting the readings gives 2.50 Ω at 0.100 A and 11.6 Ω at 0.500 A, so the resistance is already changing. To test ohmic behaviour properly, the student should take many readings, correct for meter resistance, and plot V against I; a straight line through the origin over a range would be evidence of ohmic behaviour at constant temperature, which a filament lamp does not show because of heating. The high-current increase is real, but the explanation should refer to increased lattice vibrations and electron scattering with temperature, not simply 'non-ohmic at high current'.
3. Suggest a modified circuit and a measurement procedure that would reduce the systematic error at low currents, and explain why it is less suitable at high currents.
Read the worked answer
Connect the voltmeter directly across the lamp, with the ammeter outside the voltmeter branch (ammeter in series with the lamp, voltmeter in parallel with the lamp only). Then the voltmeter reads the true lamp p.d., not the lamp plus ammeter. At low current, the lamp resistance is about 2.5 Ω, while the voltmeter is 10 MΩ, so the current drawn by the voltmeter is about 0.25 µA, negligible compared with 0.100 A; the loading error is about 0.00025%. At high current the lamp resistance is larger and the voltmeter current is still negligible compared with 0.500 A. The modified circuit is therefore suitable at both ends. One limitation is that the ammeter now measures the lamp current plus the small voltmeter current; this is negligible here but would matter if the voltmeter had a low resistance or the lamp current were comparable to the voltmeter current. Take readings quickly and at the same temperature, or use a pulse method, to reduce heating effects.