Parallel Lamps: A Dim Bulb and a Surprising Voltmeter
A student connects a 12.0 V battery to two lamps in parallel. Lamp A is rated 12 V, 12 W, and lamp B is rated 12 V, 6 W. An ammeter in the main branch reads 1.40 A. A voltmeter across lamp A reads 11.8 V, but across lamp B it reads 8.0 V. Lamp B is noticeably dimmer than lamp A. The student has checked that both lamps work when connected alone to the battery. The battery's internal resistance is negligible for this question. The student's partner says the dim lamp proves that current is used up as it passes through lamp A, so less current reaches lamp B. The student suspects a bad connection instead and wants to use the readings to decide.
1. Calculate each lamp's resistance from its rating and the current each should draw at 12 V. Compare with the ammeter reading.
Read the worked answer
From P = V²/R:
- R_A = 12²/12 = 12 Ω
- R_B = 12²/6 = 24 Ω
At 12 V, the branch currents should be:
- I_A = 12/12 = 1.0 A
- I_B = 12/24 = 0.50 A
- Total = 1.50 A
The ammeter reads 1.40 A, about 7% low. That is consistent with a small extra resistance in the main branch or with non-ideal meter/battery behaviour. It does not support the partner's idea that current is used up in lamp A; in a parallel circuit the branch currents add.
2. Use the voltmeter readings to locate the fault and evaluate the partner's “current used up” claim.
Read the worked answer
Lamp A has 11.8 V across it, so its branch is nearly normal. Lamp B has only 8.0 V, so about 4.0 V is dropped somewhere in series with B. The current in B is I = V/R = 8.0/24 = 0.33 A, below its normal 0.50 A. Its power is V²/R = 8.0²/24 = 2.7 W, far below the rated 6 W, so it is dim.
Check the main current: I_A ≈ 11.8/12 = 0.98 A; I_B = 0.33 A; sum = 1.31 A, close to the 1.40 A reading within meter and rounding uncertainty. The partner's claim violates conservation of charge: current is not consumed by lamp A. The fault is a voltage drop across a bad connection in lamp B's branch.
3. Explain why the bad connection makes B dimmer even though the battery voltage is unchanged, and predict the voltmeter reading across the bad connection.
Read the worked answer
The bad connection acts as a resistor R_c in series with lamp B only. Branch B current is 0.33 A. If the battery provides 12.0 V and lamp B takes 8.0 V, the remaining 4.0 V is across R_c:
R_c ≈ 4.0 V / 0.33 A ≈ 12 Ω
The voltmeter across the bad connection should read about 4.0 V. The branch resistance rises from 24 Ω to about 36 Ω, so the current falls and the lamp's power falls. Lamp A is barely affected because it remains in parallel and the battery can supply the extra current needed by its branch.