Two Bulbs, One Ammeter, and Ohm's Law
Two students are given a 6.0 V battery of negligible internal resistance, a fixed 8.0 ohm resistor, a small filament bulb, an ammeter and a voltmeter. In Circuit 1 the resistor and the bulb are connected in series across the battery, and the ammeter in the main branch reads 0.30 A. They then connect a second, identical bulb in parallel with the first, giving Circuit 2. The first bulb visibly dims, the ammeter in the main branch now reads 0.50 A, and a voltmeter connected across the filament of the first bulb reads 2.0 V. Their textbook states that the resistance of a metallic conductor is constant at constant temperature, and one student concludes that at least one of the readings must be faulty. The teacher asks them to work out the resistance of a bulb in each circuit, to explain why the bulb dimmed, and to decide whether the data are self-consistent.
1. For Circuit 1, find the voltage across the resistor, the voltage across the bulb, and the resistance of the bulb.
Read the worked answer
Resistor first. The same current flows through every part of a series circuit, so V_R = IR = 0.30 x 8.0 = 2.4 V.
Then the bulb. The two components share the battery voltage: V_bulb = 6.0 - 2.4 = 3.6 V. Hence R_bulb = V/I = 3.6 / 0.30 = 12 ohm, and the bulb is dissipating 3.6 x 0.30 = 1.08 W.
Check. Total resistance 8.0 + 12 = 20 ohm, and 6.0 / 20 = 0.30 A, which matches the ammeter.
2. For Circuit 2, use the readings to find the resistance of each bulb and show that the two bulbs behave identically.
Read the worked answer
Voltage across the parallel pair. The whole 0.50 A passes through the 8.0 ohm resistor, so it takes 0.50 x 8.0 = 4.0 V and leaves 6.0 - 4.0 = 2.0 V for the parallel section. This agrees with the voltmeter reading of 2.0 V, so the readings are internally consistent.
Each bulb. Both bulbs have 2.0 V across them and are identical, so the current splits equally: 0.25 A each. R = 2.0 / 0.25 = 8.0 ohm per bulb.
Check. 8.0 ohm in parallel with 8.0 ohm is 4.0 ohm; total 8.0 + 4.0 = 12 ohm; 6.0 / 12 = 0.50 A, matching the ammeter. Identical behaviour is confirmed: the data are exactly what two equal bulbs give.
3. The student says a reading must be wrong because the bulb's resistance changed. Judge this claim and explain the dimming using the two circuits.
Read the worked answer
Judgement. No reading is faulty. The assumption that fails is 'constant temperature', not Ohm's law. In Circuit 1 the filament carries 1.08 W and sits at 12 ohm; in Circuit 2 each filament carries only 0.5 x 2.0 = 0.50 W, is cooler, and sits at 8.0 ohm. For a metal, resistance rises with temperature, so a hot filament having the higher resistance is exactly what should happen. V = IR still holds at every instant; it is R that is not a constant for a filament.
Why it dims. Adding the second bulb in parallel drops the resistance of that section from 12 ohm to 4 ohm, so the total falls from 20 ohm to 12 ohm and the current rises from 0.30 A to 0.50 A. But the resistor now takes 4.0 V instead of 2.4 V, leaving only 2.0 V across the bulbs instead of 3.6 V. Each bulb therefore gets less power than before, so both are dimmer - the extra current goes into the fixed resistor, not into the bulbs.