A Bulb, an Ammeter and a Mystery Resistor
A student is given a 6.0 V battery of negligible internal resistance, a filament bulb rated 6.0 V, 3.0 W, a fixed resistor of unknown value, an ammeter, a voltmeter and connecting wires. She connects the bulb and the resistor in series across the battery. The ammeter reads 0.35 A and the voltmeter, connected across the resistor, reads 3.9 V. The bulb glows dimly. She is asked whether the bulb is operating at its rated power and what the resistance of the fixed resistor is. She is also warned that the bulb's resistance changes with temperature.
1. Calculate the resistance of the fixed resistor and the power dissipated in the bulb, and decide whether the bulb is at its rated power.
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The voltmeter reads 3.9 V across the fixed resistor, and the same current of 0.35 A flows through it because the components are in series. So:
R = V/I = 3.9/0.35 ≈ 11 Ω (11.1 Ω to three significant figures).
The battery is 6.0 V, so the voltage across the bulb is 6.0 - 3.9 = 2.1 V. The power in the bulb is P = VI = 2.1 × 0.35 ≈ 0.74 W.
The bulb is rated 3.0 W at 6.0 V, so it is operating at about 0.74/3.0 ≈ 25% of its rated power. It is not at its rated power; it is well below it, which is consistent with the dim glow.
As a check, the bulb's resistance at this operating point is R = V/I = 2.1/0.35 = 6.0 Ω. Its rated resistance would be V²/P = 36/3.0 = 12 Ω. The measured resistance is lower, which is expected because the filament is cooler than at full power and a cooler filament has lower resistance.
2. The student is asked to predict what would happen to the ammeter reading if the fixed resistor were replaced by one of half the resistance. Explain why a simple ratio prediction would be misleading.
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A simple prediction might say: total resistance falls, so current rises in proportion. But that assumes the bulb's resistance stays at 6.0 Ω, which is not true. If the current rises, the filament gets hotter and its resistance increases. So the current will rise, but by less than a simple ratio would suggest.
A careful estimate can be made. With the original resistor halved to about 5.6 Ω, suppose the current rises to I. The bulb voltage is 6.0 - 5.6I, and the bulb resistance at the new current is unknown but higher than 6.0 Ω. If we guess the bulb resistance rises to about 8 Ω, then total resistance is about 13.6 Ω and I ≈ 6.0/13.6 ≈ 0.44 A. This is a rough estimate, not an exact answer, because the bulb's resistance depends on temperature in a way that is not given.
The key judgement is that the bulb is a non-ohmic component. Its V-I graph curves, so the resistance cannot be treated as constant. The correct statement is qualitative: the current increases, the bulb becomes brighter, and the increase is smaller than a constant-resistance calculation would predict. A precise value would require the bulb's characteristic curve.