Over-running a filament bulb from a 9 V battery
A student must run a 6.0 V, 3.0 W filament bulb from a 9.0 V battery whose internal resistance is 1.5 Ω. Connected directly across the battery, the bulb draws 0.62 A, glows brighter than normal, and the glass begins to darken after a few minutes. A multimeter set to resistance mode across a cold, unused bulb of the same type reads 1.4 Ω; the manufacturer states that the filament's resistance at normal operating temperature is about twelve times its cold value. The teacher asks for a 20-minute demonstration in which the bulb is not driven above its rated values, and provides a box of resistors, a rheostat, an ammeter and a voltmeter. A second identical bulb is available for comparison. The student proposes adding a 4.7 Ω resistor in series and claims the bulb will then be 'slightly under-run, which is fine for a demonstration'.
1. From the rating, find the current and resistance the filament should have at operating temperature. Use the measured 0.62 A to find the actual filament resistance and power, and explain what the 1.4 Ω cold reading means at the instant of switching on.
Read the worked answer
Rated values: I = P/V = 3.0/6.0 = 0.50 A and R = V/I = 6.0/0.50 = 12 Ω.
Direct connection at 0.62 A: total resistance = 9.0/0.62 = 14.5 Ω, so the filament is 14.5 − 1.5 = 13.0 Ω. Bulb voltage = 0.62 × 13.0 = 8.1 V and power = 0.62² × 13.0 = 5.0 W, about 1.7 times the rating. The resistance is above 12 Ω because the filament is hotter than its rated temperature - resistance and temperature rise together, so the two measurements are self-consistent. The darkening is tungsten evaporating from the overheated filament and plating onto the cooler glass; life falls steeply as over-run increases.
Cold reading: 1.4 Ω is the filament at room temperature, where the multimeter's tiny test current cannot heat it. If the filament were still 1.4 Ω when the switch closed, the current would be 9.0/(1.5 + 1.4) = 3.1 A, more than six times the rated current, with a momentary power of 3.1² × 1.4 = 13 W. The filament heats to operating temperature within milliseconds, so this surge is brief - but it produces the largest thermal and magnetic stresses, which is why bulbs usually fail at switch-on rather than while running.
2. What is the minimum series resistance needed for rated operation? Evaluate the student's 4.7 Ω choice numerically, and comment on whether the 20-minute demonstration is safe.
Read the worked answer
At the rated point the bulb takes 0.50 A at 6.0 V, so the rest of the circuit must drop 9.0 − 6.0 = 3.0 V at 0.50 A, a total of 6.0 Ω. Of that, 1.5 Ω is inside the battery, so the minimum series resistance is 4.5 Ω.
With 4.7 Ω and the filament at 12 Ω: total = 1.5 + 4.7 + 12 = 18.2 Ω, I = 9.0/18.2 = 0.494 A, V_bulb = 5.93 V, P = 2.93 W - about 2% below rated. The student's description is fair, with one correction: at the slightly lower current the filament runs a little cooler, so its resistance is slightly below 12 Ω and the true current will be a little above 0.494 A, probably 0.50–0.51 A. That should be checked on the ammeter and trimmed with the rheostat rather than assumed.
Safety: electrically this is fine, but the resistor must be rated for I²R = 0.494² × 4.7 = 1.15 W, so a 2 W or larger wirewound resistor is needed; a small 0.25 W carbon resistor would overheat and smell or fail during 20 minutes. The bulb itself stays at or just below its rating, so the demonstration is safe provided the current is monitored and the resistor is adequately rated.
3. A classmate concludes from the 1.4 Ω and 13 Ω readings that 'the bulb does not obey Ohm's law'. Assess this statement precisely.
Read the worked answer
The statement is loose rather than simply wrong. Ohm's law in the form R = V/I is the definition of resistance and always holds at any instant; what fails is the assumption that R is constant for a given component. For the bulb, R depends strongly on temperature: tungsten's resistivity rises roughly tenfold between room temperature and about 2700 °C, which is exactly what the 1.4 Ω and 13 Ω figures show.
The bulb is therefore a non-ohmic component: its I–V graph is a curve whose slope (conductance) falls as V and I increase, not a straight line through the origin. It behaves ohmically only over ranges where the temperature does not change appreciably - which is why the cold multimeter reading is reproducible for a given meter but is not the resistance during operation, and why the meter's own test current matters.
A better statement: the bulb obeys V = IR at every instant, but its resistance is not constant because the current changes its temperature. Plotting I against V and observing the decreasing slope would demonstrate this directly.