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Work and power of the current

Curricula: KS3, GCSE, cycle 4, 1re (effet Joule), Sek I, AP2

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Aim

Measure the power the lamp consumes and the energy the current delivers over time; check that \(E = P \cdot t\).

The bench

A 6 V source, a wattmeter and a 6 V / 3 W lamp. The current circuit of the wattmeter (terminals "±" and "A") is connected in series with the lamp, the voltage circuit (terminals "±" and "V") across the lamp. The top line of the display is the power, the bottom line is the energy or the counting time: the keys E and t switch between them. Counting is stopped until the ▶■ key is pressed.

Procedure

  1. Record the power \(P\) on the top line of the display.
  2. Press ▶■ and after 10–20 s press it again to stop counting. Record the energy \(E\) (key E) and the time \(t\) (key t).
  3. Calculate \(P = E / t\) and compare it with the reading of the top line.
  4. Set the source to 3 V and record the power again.
Expected result

The lamp consumes about 2.75 W — slightly less than its rating: part of the voltage drops inside the source. For example, in 10.8 s the counter reaches about 29.7 J, and \(E / t\) = 2.75 W — the same as on the top line. At 3 V the power is about 0.89 W: the voltage has dropped by half, but the power only by a factor of 3.1, not 4. The filament is cooler, and its resistance is lower.

Questions

  1. What does the work done by the current in the lamp turn into?
  2. Why did the power of the lamp at half the voltage decrease not by a factor of four, as it would for a resistor?
  3. How many joules is 1 W · h?
Answers
  1. Into the internal energy of the hot filament, which the lamp gives off as heat and light. Of the 2.75 W most leaves as heat and infrared radiation; visible light is only a few percent.
  2. For a resistor \(P = U^2 / R\) with a constant \(R\), so half the voltage gives a quarter of the power. The resistance of a tungsten filament grows with its temperature: at 3 V the filament is cooler and its resistance lower, so the current falls by less than half, and the power falls only 3.1 times, from 2.75 W to 0.89 W.
  3. 1 W · h = 1 W · 3600 s = 3600 J. The lamp on this bench uses 2.75 W · 3600 s ≈ 9900 J in an hour.