Lamp Brightness when a Parallel Branch Is Removed
A student is given a battery of emf 12.0 V and internal resistance 2.0 Ω, a fixed resistor R1 = 5.0 Ω, a resistor R2 = 20 Ω and a lamp marked 6.0 V, 3.0 W. The lamp is connected in parallel with R2, and this combination is in series with R1 and the battery. The student measures the current through the lamp as 0.52 A, close to the predicted value, and then removes R2 from the circuit, expecting the lamp's brightness to stay the same 'because the battery emf has not changed'. The lamp is assumed to have a constant resistance equal to its rated value. A voltmeter connected across the battery terminals reads 10.3 V before R2 is removed. The student also notes that the voltmeter reading changes after the removal and calls this a 'faulty battery'.
1. Calculate the lamp's resistance, the lamp current before R2 is removed, and the terminal p.d. Compare with the measured 0.52 A and 10.3 V.
Read the worked answer
Rated lamp resistance: R_L = V²/P = 6.0²/3.0 = 12 Ω.
With R2 in parallel: R_p = (20 × 12)/(20 + 12) = 7.5 Ω. Total resistance = 5.0 + 7.5 + 2.0 = 14.5 Ω.
Battery current I = ε/R_total = 12.0/14.5 = 0.828 A.
p.d. across the parallel section = I R_p = 0.828 × 7.5 = 6.21 V, so I_lamp = 6.21/12 = 0.518 A.
Terminal p.d. = ε − I r = 12.0 − 0.828 × 2.0 = 10.34 V.
Both predictions (0.518 A, 10.34 V) agree with the measurements (0.52 A, 10.3 V) to about 1%, so the model is consistent.
2. After R2 is removed, calculate the new lamp current and the new terminal p.d. Is the lamp brighter or dimmer? Explain using the internal resistance.
Read the worked answer
Removing R2 leaves 5.0 Ω in series with the 12 Ω lamp and the 2.0 Ω internal resistance: R_total = 19.0 Ω.
I = 12.0/19.0 = 0.632 A, and this is now the lamp current because the lamp is in series with the rest.
Terminal p.d. = 12.0 − 0.632 × 2.0 = 10.74 V; p.d. across the lamp = 0.632 × 12 = 7.58 V.
The lamp current rises from 0.518 A to 0.632 A, so the power (I²R) rises from about 3.2 W to 4.8 W and the lamp is brighter. The emf is unchanged, but the terminal p.d. rises because the smaller current reduces the lost volts Ir. Brightness depends on the p.d. across the lamp, not on the emf directly.
3. The student calls the changed voltmeter reading a 'faulty battery'. Evaluate this, and state one assumption in your calculation that would fail for a real lamp.
Read the worked answer
The changing terminal p.d. is not a fault: it is the normal signature of internal resistance. A real battery has ε = terminal p.d. + Ir, so any change in current changes the terminal reading. The voltmeter reading should change when R2 is removed, and the direction (rising from 10.3 V to about 10.7 V) is exactly what the model predicts.
The calculation assumes the lamp resistance stays at 12 Ω. A real filament lamp heats up as current increases, and its resistance rises with temperature. Therefore the real current after removing R2 will be somewhat less than 0.632 A, and the real p.d. across the lamp will be higher than 7.58 V. The lamp is also rated at 6.0 V; at about 7.6 V it would be overrun, which could shorten its life or burn it out.