Bulb dimming on a loaded battery
A student is given a 12 V battery of internal resistance 0.5 Ω, a bulb marked '12 V, 24 W', a heater coil of resistance 3 Ω, a switch and connecting wires of negligible resistance. First the bulb alone is connected across the battery and it glows with a certain brightness. Then the heater is connected in parallel with the bulb using the same battery. The student observes that the bulb becomes noticeably dimmer and the battery warms up. A friend claims the battery is faulty because its terminal voltage should stay at 12 V. The laboratory has a 6 V battery of the same internal resistance and a variable resistor, but no other bulbs. The student must decide whether the dimming is expected behaviour and explain it using calculations, and also suggest one change that would make the bulb glow closer to its rated brightness while the heater is on.
1. Calculate the resistance of the bulb and the current, terminal voltage and power of the bulb when it is connected alone.
Read the worked answer
Bulb resistance: R = V²/P = 12²/24 = 6 Ω. With the bulb alone, total resistance = 6 + 0.5 = 6.5 Ω, so I = 12/6.5 = 1.85 A. Terminal voltage V_t = ε − Ir = 12 − (1.85)(0.5) = 11.08 V. Bulb power = V_t²/R = (11.08)²/6 ≈ 20.5 W, somewhat below the rated 24 W. The bulb is not faulty; even alone it runs slightly below its rating because of the internal resistance.
2. With the heater connected in parallel, calculate the new current drawn, the terminal voltage and the bulb power. Is the friend right that the battery is faulty?
Read the worked answer
6 Ω in parallel with 3 Ω gives R_p = (6×3)/(6+3) = 2 Ω. Total resistance = 2 + 0.5 = 2.5 Ω; I = 12/2.5 = 4.8 A. Terminal voltage = 12 − (4.8)(0.5) = 9.6 V. Bulb current = 9.6/6 = 1.6 A; bulb power = 9.6 × 1.6 = 15.36 W. This is about 25% less than 20.5 W, so the dimming is real and expected. The friend is wrong: 12 V is the emf, not the terminal voltage. The terminal voltage falls as the current rises because the internal resistance drops Ir volts inside the battery, which also explains the warming.
3. Suggest one change, using only the equipment available, that would make the bulb glow closer to its rated brightness while the heater is on, and justify it.
Read the worked answer
Put the variable resistor in series with the heater (not the bulb). This raises the resistance of the heater branch, lowers the total current drawn from the battery, and therefore reduces the internal voltage drop Ir. The terminal voltage rises towards 12 V and the bulb becomes brighter, though the heater produces less heat. Using a battery of lower internal resistance, or a separate battery for the heater, would be better if available, but with the given equipment the series resistor is the workable option.