A 6 V lamp on a 9 V battery: designing the series resistor
A pupil is given a 9.0 V battery, a torch lamp marked '6.0 V, 3.0 W', an ammeter, a voltmeter, a rheostat and a box of resistors. A multimeter placed across the lamp while it is cold reads 1.8 Ω. The pupil first wires the lamp straight across the battery; the voltmeter across the lamp reads 8.1 V, the filament glows very brightly and the lamp fails after a few seconds. Testing the battery separately with a 10 Ω resistor gives a terminal p.d. of 8.0 V, from which the pupil works out that the battery behaves as a 9.0 V source with about 1.2 Ω internal resistance. He now has one identical spare lamp and wants it to run at its rated 6.0 V. The school resistor box contains carbon film resistors rated at 1 W and wirewound resistors rated at 5 W, in E12 values.
1. Use the potential-divider idea to explain the 8.1 V reading and show quantitatively why the lamp failed.
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At normal working temperature the lamp's resistance follows from its rating: R = V²/P = 6.0² / 3.0 = 12 Ω.
The battery and lamp form a series circuit with the internal resistance: I = 9.0 / (12 + 1.2) = 0.68 A V_lamp = 9.0 - 0.68 x 1.2 = 8.18 V, matching the observed 8.1 V. (Equivalent view: V_lamp = 9.0 x 12/13.2 = 8.18 V.)
The power actually dissipated is V²/R = 8.18²/12 = 5.6 W, nearly twice the 3.0 W rating. The filament therefore runs far hotter than design and evaporates; the lamp failed from over-power, not from a faulty battery.
The 1.8 Ω cold reading is not a contradiction: filament resistance rises steeply with temperature, roughly six-fold between room temperature and white heat. Using 1.8 Ω would predict 3.0 A and 27 W, which is the switch-on surge, not the steady state.
2. Select a resistor from the box that will run the spare lamp at about 6.0 V, and choose its power rating. Show the check calculation.
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Target current: at 6.0 V, I = P/V = 3.0/6.0 = 0.50 A. Total resistance needed: R_total = 9.0 / 0.50 = 18.0 Ω. Series resistor: 18.0 - 12.0 - 1.2 = 4.8 Ω, so choose the E12 value 4.7 Ω.
Check: R_total = 12.0 + 1.2 + 4.7 = 17.9 Ω, I = 9.0/17.9 = 0.503 A, V_lamp = 0.503 x 12.0 = 6.03 V - comfortably within a few per cent of the rating, and slightly on the safe side.
Power in the resistor: P = I²R = 0.503² x 4.7 = 1.19 W. A 1 W carbon film resistor would run at 120% of its rating and drift or burn, so specify the 5 W wirewound version. It costs nothing in performance: the resistance value is what sets the voltage, the power rating only decides whether the component survives.
3. Explain what happens during the first few milliseconds after switching on, and why this makes the 5 W wirewound the right choice rather than a 1 W part of the same value.
Read the worked answer
At the instant of switching on the filament is cold, about 1.8 Ω, while the resistor is at room temperature. Circuit resistance is then 1.2 + 1.8 + 4.7 = 7.7 Ω, so the initial current is 9.0/7.7 = 1.17 A, about 2.3 times the steady 0.50 A.
The instantaneous power in the resistor is 1.17² x 4.7 = 6.4 W, five times its steady dissipation. The surge lasts only until the filament reaches operating temperature (tens of milliseconds, because the filament has very small heat capacity), so the average extra energy is small, but a thin carbon film can still be damaged by the current density and the local hot spot. A wirewound 5 W part has enough thermal mass and surface area to absorb it repeatedly.
The same effect explains the lamp's own failure: the cold filament draws 3 A and dissipates tens of watts for a moment, which is why incandescent lamps almost always fail at the instant of switching on rather than while running.