Skip to content

Rheostat as a brightness control

Curricula: KS3, GCSE, Sek I

Open the full lab ↗

Aim

Smoothly control the current and the brightness of a lamp with a rheostat.

The bench

A 6 V source, an ammeter, a 100 Ω rheostat and a 6 V / 3 W lamp in series. The part of the rheostat from the left end to the slider is in the circuit; the slider is at 30% (30 Ω).

Procedure

  1. Write down the ammeter reading.
  2. Move the slider to the left end and write down the current again.
  3. Move the slider to the right end.
  4. Find the position at which the lamp glows at half brightness.
Expected result

At 30 Ω the current is about 165 mA. At the left end the rheostat's resistance is close to zero: the current is about 0.5 A and the lamp lights at full brightness. At the right end (100 Ω) the current falls to about 55–60 mA and the lamp does not glow at all, even though current is flowing: there is less than 0.2 V across it, the filament heats to only 900 K, and visible glow begins at about 1000 K.

Questions

  1. Why does the current change smoothly rather than in jumps?
  2. What part of the rheostat must be included in the circuit to halve the current?
Answers
  1. Moving the slider changes the length of wire in the circuit, and with it the resistance, by as little as the slider moves, so the current \(I = U / R\) changes gradually too. On a real wire-wound rheostat the slider steps from one turn of wire to the next, and the resistance changes in very small steps of one turn; in the simulator it changes continuously.
  2. With a lamp of constant resistance the rheostat would have to add as much as the lamp's own 12 Ω (6 V / 0.5 A). The filament cools as the current falls and its resistance drops, so more is needed: at half the current, about 0.24 A, the lamp has only about 1.8 V across it, and the rheostat must take the rest with about 17 Ω, a sixth of its length.