Parallel circuit with a filament lamp and fuse choice
A student builds a circuit from a 6.0 V battery of negligible internal resistance. A 4.0 Ω resistor is in series with a parallel pair made of a 12 Ω resistor and a 6.0 Ω resistor. An ammeter reads the total current, and a voltmeter reads the p.d. across the parallel pair. The student predicts that the 6.0 Ω resistor will carry twice the current of the 12 Ω resistor and will therefore become hotter. The 6.0 Ω resistor is then replaced by a filament lamp marked '6 V, 3 W'. The lamp glows dimly. The student says the lamp must be faulty because it is rated at 6 V and the battery is 6 V. A 1.0 A fuse protects the circuit. The teacher asks the class to check the student's prediction, to explain the dim lamp, and to judge whether the fuse is well chosen.
1. Check the student's prediction by calculating the equivalent resistance, total current, p.d. across the parallel pair, branch currents and power in each resistor.
Read the worked answer
Parallel pair: 1/R_p = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12, so R_p = 4.0 Ω.
Total resistance = 4.0 + 4.0 = 8.0 Ω, so I = V/R = 6.0/8.0 = 0.75 A.
p.d. across the pair = I × R_p = 0.75 × 4.0 = 3.0 V (the other 3.0 V is across the 4.0 Ω series resistor, so the battery p.d. splits equally).
Branch currents: 3.0/12 = 0.25 A and 3.0/6.0 = 0.50 A. They add to 0.75 A, which checks.
Power: 12 Ω resistor P = VI = 3.0 × 0.25 = 0.75 W; 6.0 Ω resistor P = 3.0 × 0.50 = 1.5 W.
The prediction is correct: the 6.0 Ω branch carries twice the current because it has half the resistance and both branches have the same p.d. It also dissipates twice the power, so it will run hotter, provided the two components are similar in size and cooling.
2. Calculate the lamp's behaviour when it replaces the 6.0 Ω resistor, and explain why it is dim without being faulty.
Read the worked answer
From its rating: operating resistance R = V²/P = 36/3 = 12 Ω, and rated current = P/V = 3/6.0 = 0.50 A.
The parallel pair is now 12 Ω with 12 Ω, so R_p = 6.0 Ω and total resistance = 4.0 + 6.0 = 10.0 Ω. Total current = 6.0/10.0 = 0.60 A, and the p.d. across the pair = 0.60 × 6.0 = 3.6 V.
Each branch takes 3.6/12 = 0.30 A, so the lamp's power = 3.6 × 0.30 = 1.08 W, roughly a third of its 3 W rating.
The lamp is not faulty. It only receives 3.6 V, not the 6.0 V it is rated for, because the series 4.0 Ω resistor takes the other 2.4 V. A lamp rated 6 V needs the full 6 V across it to run at full brightness; a dim glow is the expected result. The filament is also non-ohmic: its cold resistance is much lower than 12 Ω, so the current is briefly higher at switch-on, and the resistance rises as the filament heats. Treating it as a fixed 12 Ω is only an approximation.
3. Judge whether the 1.0 A fuse is well chosen for this circuit.
Read the worked answer
Normal operation: 0.75 A with the resistor, 0.60 A with the lamp. A 1.0 A fuse carries both without nuisance blowing, so it is safe in that sense.
Fault condition: the worst case is a short circuit across the parallel section. The current is then limited only by the 4.0 Ω series resistor: I = 6.0/4.0 = 1.5 A. A 1.0 A fuse at 1.5 times its rating will take a considerable time to melt, so it gives poor protection against a short circuit. The series resistor itself limits the maximum possible current to 1.5 A, which is why the circuit is not dangerous, but the fuse is not doing much.
Judgement: 1.0 A is adequate against a mild overload but too slow for a short circuit. A 0.75 A fuse would protect better, provided it does not blow on the switch-on surge from the cold filament, whose resistance is much lower than 12 Ω. The best choice balances protection against nuisance blowing; 1.0 A is a compromise, not an ideal selection.