A lamp gives (potential difference in V, current in A) readings (0, 0), (0.5, 0.18), (1, 0.35), (1.5, 0.49), (2, 0.6), (2.5, 0.68). (a) Calculate resistance at 0.50 V and 2.50 V. (b) Explain the change. (c) Find power at 2.0 V and energy in 5.0 min. (d) Describe a circuit and safe method for both signs of the characteristic.

Question

Accepted Answer

Resistances 2.78 Ω and 3.68 Ω; power 1.20 W; five-minute energy 360 J; heating raises filament resistance. At 0.50 V, R = V/I = 0.50/0.18 = 2.78 Ω. At 2.50 V, R = 2.50/0.68 = 3.68 Ω, so resistance is not constant. Greater current heats the filament. Stronger lattice vibrations increase charge-carrier collisions, causing the curved non-ohmic characteristic. At 2.0 V and 0.60 A, P = VI = 1.20 W. Five minutes is 300 s, so E = Pt = 1.20 × 300 = 360 J. Place ammeter and variable resistor in series with the lamp and voltmeter in parallel. Adjust a low-voltage supply gradually and stay within the lamp rating. Reverse supply connections for negative values, take readings quickly and repeat. A filament characteristic should be approximately symmetric, but should not be forced into a straight line.

Filament-lamp characteristic

What a complete response should do

Lamp current–voltage data

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Lamp current–voltage data — supplied values
Potential difference / V	Current / A
0	0
0.5	0.18
1	0.35
1.5	0.49
2	0.6
2.5	0.68