I-V Characteristics of a Filament Lamp and Resistor: Non-Ohmic Behaviour
A student investigates the current-voltage characteristics of a filament lamp and a resistor. The lamp is rated 12 V, 24 W. The resistor has a resistance of 10 Ω. They connect each component in turn in series with a variable power supply, an ammeter, and a voltmeter. For the lamp, they record: at 6.0 V, current = 1.2 A; at 12.0 V, current = 2.0 A. For the resistor, the current is directly proportional to voltage, with a gradient of 0.10 A/V. The student plots graphs of current against voltage. They calculate the resistance of the lamp at 6.0 V and 12.0 V and find they are different. They also calculate the power dissipated by the lamp at 12.0 V and compare with the rated power. They notice that the lamp's resistance increases with voltage. They are asked to explain this in terms of the filament's temperature and to evaluate whether the lamp obeys Ohm's law.
1. Calculate the resistance of the lamp at 6.0 V and 12.0 V, and the power dissipated at 12.0 V. Compare with the rated power.
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At 6.0 V, the current is 1.2 A, so R = V / I = 6.0 / 1.2 = 5.0 Ω. At 12.0 V, the current is 2.0 A, so R = 12.0 / 2.0 = 6.0 Ω. The power dissipated at 12.0 V is P = V I = 12.0 × 2.0 = 24 W. This exactly matches the rated power of 24 W, so the lamp is operating at its rated power when connected to 12.0 V. The resistance has increased from 5.0 Ω to 6.0 Ω as the voltage (and current) increased.
2. Explain why the resistance of the filament lamp changes with voltage, using the concept of temperature coefficient of resistance.
Read the worked answer
The filament of the lamp is made of tungsten, which has a positive temperature coefficient of resistance. As the voltage across the lamp increases, the current increases, and more electrical power is dissipated as heat (P = I²R or P = V²/R). This raises the temperature of the filament significantly. Since the resistance of tungsten increases with temperature according to R = R₀[1 + α(T − T₀)], the resistance of the filament increases as it gets hotter. At 12.0 V, the filament is hotter than at 6.0 V, so its resistance is higher (6.0 Ω compared with 5.0 Ω). This means the lamp does not obey Ohm's law, which requires resistance to be constant. The I-V graph for the lamp is a curve with a decreasing slope, characteristic of a non-ohmic component.
3. The student connects the lamp and the resistor in series across a 12.0 V supply. Calculate the current in the circuit and the potential difference across each component.
Read the worked answer
In series, the same current I flows through both the lamp and the resistor. The resistor is ohmic, so V_R = I R = 10 I. The lamp is non-ohmic; we can interpolate from the given data. At I = 1.2 A, V_lamp = 6.0 V; at I = 2.0 A, V_lamp = 12.0 V. A reasonable approximation is to model the lamp's resistance as increasing linearly with current: from 5.0 Ω at 1.2 A to 6.0 Ω at 2.0 A, giving R_lamp(I) = 5.0 + [(6.0 − 5.0)/(2.0 − 1.2)](I − 1.2) = 5.0 + 1.25(I − 1.2) = 3.5 + 1.25 I. Then V_lamp = I R_lamp = 3.5 I + 1.25 I².
The total voltage is V_total = V_lamp + V_R = (3.5 I + 1.25 I²) + 10 I = 1.25 I² + 13.5 I. Setting this equal to 12.0 V gives 1.25 I² + 13.5 I − 12.0 = 0. Solving the quadratic: I = [−13.5 + √(13.5² + 4 × 1.25 × 12.0)] / (2 × 1.25) = [−13.5 + √(182.25 + 60)] / 2.5 = [−13.5 + √242.25] / 2.5 = (−13.5 + 15.565) / 2.5 = 0.826 A. Thus I ≈ 0.83 A.
Then V_R = 10 × 0.826 = 8.26 V, and V_lamp = 12.0 − 8.26 = 3.74 V. Checking with the lamp model: V_lamp = 3.5(0.826) + 1.25(0.826)² = 2.891 + 0.853 = 3.744 V, consistent. So the current is about 0.83 A, the resistor takes about 8.3 V, and the lamp takes about 3.7 V.