The Dark Bulb in a Series Circuit
Year 9 practical. A teacher sets up a 12 V DC power pack, a switch and two bulbs, each marked '6 V, 0.5 A', connected in series with an ammeter. When the switch is closed, bulb B glows very brightly while bulb A stays completely dark. The ammeter reads 0.50 A. Students measure potential differences with a digital voltmeter: 12.0 V across the power pack, 0.0 V across bulb A and 11.6 V across bulb B. Two students offer explanations. Amara says bulb A has 'blown' and must be broken inside. Joseph says the current 'cannot get through' bulb A, so it takes the easiest path. A third student, Layla, notices the bulbs are getting hot and smells a faint burning smell from bulb B. The class has 25 minutes left, one spare bulb of the same rating, a multimeter and a box of leads, but no replacement power pack.
1. Which explanation fits the readings? State what an open filament and a short circuit would each predict, then give the next measurement you would take.
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The predictions differ, so the readings can discriminate between them.
- Open circuit ('blown' filament): the series path is broken. Current would be 0 A, the full 12 V would appear across the break, and both bulbs would be dark.
- Short circuit across bulb A: a near-zero-resistance path bypasses the filament. The potential difference across A collapses to about 0 V, nearly the whole supply appears across B, and the current is set by B alone. A 6 V, 0.5 A bulb has a hot resistance of about 12 Ω, so at 12 V the current would settle around 0.5–1.0 A depending on filament temperature.
The readings (0.50 A, 0.0 V across A, 11.6 V across B) match the short and refute the open circuit: a blown filament would give no current at all. Amara's explanation is therefore inconsistent with the ammeter. Joseph's wording is also wrong - current does not 'choose a path'. The short simply provides a parallel route of far lower resistance, so almost all charge carriers pass through it and almost none through the filament.
Next measurement: switch off, remove bulb A and measure its cold resistance with the multimeter. A shorted bulb reads close to 0 Ω; a blown one reads open circuit; a healthy bulb of this rating reads roughly 12 Ω. Compare with the spare.
2. A student argues: 'The ammeter reads exactly 0.50 A, which is the rated current, so nothing is wrong.' Assess this argument and quantify the risk to bulb B.
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The argument confuses current with power. A rating is a pair of values: 6 V and 0.5 A, which means a rated power of P = VI = 3.0 W.
Bulb B is now dissipating 11.6 × 0.50 ≈ 5.8 W, nearly double its rating. The current is right but the energy transferred per coulomb is doubled, so the filament runs much hotter than design. Tungsten evaporates faster, the envelope blackens, and the filament is likely to fail - and if it fails open, the whole series circuit goes dead and the lesson loses its apparatus.
The 'correct current' therefore gives false reassurance, which is the teaching point: an ammeter alone cannot reveal an over-power fault, and a rating must be respected as a pair. Meter readings only test a hypothesis when you know what the hypothesis predicts for each instrument.
Action: switch off at once (the hot smell is a warning, not a detail), allow cooling, replace bulb A with the spare, or substitute a 12 Ω resistor, before re-energising. If a 12 V bulb is available it is the safer substitute because it will not be overrun.
3. With 25 minutes left, one spare bulb, and a smell of burning, what should the teacher do, and what question would best check whether students have understood the diagnosis?
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Immediate action (safety before pedagogy): switch off and isolate the supply; do not leave the circuit energised while discussing it. Let bulb B cool, ventilate the room, and check the bulb base and holder for scorching. Replace bulb A with the spare, re-energise briefly and confirm that both bulbs now glow dimly (each should have about 6 V across it and the current should be near 0.5 A). If the spare is also suspect, use the 12 Ω resistor instead and abandon the demonstration.
Use the remaining time for the reasoning, not for more apparatus. Put the four readings on the board and ask pairs to write the prediction each hypothesis makes before revealing the diagnosis.
Best checking question: 'If I swapped bulb A for one that has definitely blown, what would the ammeter and the two voltmeters read, and why?' A student who has understood will say 0 A, 12 V across the gap, 0 V across B, because the series path is broken - and will distinguish this from the short, where the current survives. A student who only repeats 'it was short-circuited' without generating the contrasting prediction has memorised the outcome without the model.