Two Lamps, One Supply, One Fuse: Is the Series Circuit Actually Safe?
A first-year student is given a 12 V bench supply, two filament lamps rated 6 V/3 W and 6 V/2 W, a box of fuses marked 0.1 A, 0.3 A, 0.5 A, 1 A and 2 A, and connecting wire. She must run both lamps from the 12 V supply and protect the circuit with exactly one fuse.
She first proposes a parallel connection with the 1 A fuse: 'Each lamp is rated 6 V and the supply is 12 V, so in parallel each one gets the full 12 V; the currents add to 0.5 + 0.33 = 0.83 A, which is under 1 A, so the fuse holds.'
The demonstrator stops her. She rewires the lamps in series and keeps the 0.5 A fuse. The demonstrator says the series circuit is 'safe enough to switch on, but one lamp will age quickly'.
Assume the lamp resistances stay constant at their rated values and the supply is ideal (zero internal resistance).
1. Calculate the rated resistance and rated current of each lamp, then show why the parallel arrangement with the 1 A fuse is wrong.
Read the worked answer
Rated values. For a lamp rated at V and P: R = V²/P, I = P/V.
- 6 V/3 W lamp: R₁ = 36/3 = 12 Ω; I₁ = 3/6 = 0.50 A.
- 6 V/2 W lamp: R₂ = 36/2 = 18 Ω; I₂ = 2/6 = 0.33 A.
What parallel across 12 V actually does. Each lamp receives the full 12 V, twice its rating. Currents become:
- I₁ = 12/12 = 1.0 A (twice rated)
- I₂ = 12/18 = 0.67 A (twice rated)
- Total = 1.67 A, not 0.83 A.
Powers: 12²/12 = 12 W in the first lamp and 12²/18 = 8 W in the second - four times their ratings. Both filaments would be destroyed almost immediately, and the 1 A fuse would blow while the lamps burned.
Where her arithmetic went wrong. She used the rated currents 0.5 A and 0.33 A, which are the currents at 6 V. In parallel the lamps do not sit at 6 V; they sit at the supply voltage, 12 V. Substituting rated current into a circuit at twice the rated voltage is the error. The correct tool is I = V/R with the actual voltage across each lamp.
2. The lamps are now in series with the 0.5 A fuse. Find the current and the voltage across each lamp, and decide whether the circuit is safe, and for which lamp it is not.
Read the worked answer
Series resistance: R_total = 12 + 18 = 30 Ω. The same current flows through both.
Current: I = 12/30 = 0.40 A. This is below each lamp's rated current (0.50 A and 0.33 A? - note it exceeds the second lamp's rated 0.33 A, which is the first sign of trouble).
Voltages: V = IR
- 12 Ω lamp: V₁ = 0.40 × 12 = 4.80 V - below its 6 V rating.
- 18 Ω lamp: V₂ = 0.40 × 18 = 7.20 V - 20% above its 6 V rating.
Check: 4.80 + 7.20 = 12 V. ✓
Powers: P₁ = I²R = 0.16 × 12 = 1.92 W (rated 3 W, under-run); P₂ = 0.16 × 18 = 2.88 W (rated 2 W, over-run by 44%). Total 4.80 W, not 3 W + 2 W - the rated powers apply only at 6 V each, which the series circuit never gives them.
Verdict. The circuit will work, and it is safe in the narrow sense that nothing draws a fault current: the 0.5 A fuse carries 0.40 A and does not blow. But the 2 W lamp is the one that ages quickly - it runs at 7.2 V instead of 6 V, so its filament runs hotter and evaporates faster. The demonstrator's remark identifies that lamp, not the 3 W one. A correct fix is a series resistor or a 6 V supply with the lamps in parallel.
3. Is the 0.5 A fuse the correct choice? Compare the alternatives and explain what the fuse does and does not protect against here.
Read the worked answer
Operating current is 0.40 A. The fuse must pass that continuously, so any rating below 0.40 A is disqualified: the 0.1 A and 0.3 A fuses would blow during normal running, making the circuit unusable. That removes two of the five options immediately.
Choosing among 0.5 A, 1 A and 2 A. The rule for fuse selection is the smallest standard rating comfortably above the normal current. 0.5 A sits just above 0.40 A (a 25% margin), so it is the correct choice - it carries the working current and interrupts the smallest possible fault current. The 1 A fuse would allow two and a half times the working current before opening, and the 2 A fuse five times; both would let a faulty lamp or a short draw damaging current through the wiring first.
What the fuse does not do. A fuse protects the wiring and supply against overcurrent; it is not a voltage regulator. Here the danger to the 2 W lamp is over-voltage (7.2 V against a 6 V rating) at a perfectly normal current of 0.40 A. No fuse in the box can detect that condition, which is exactly why the demonstrator calls the circuit 'safe enough to switch on' while still warning about lamp life. Choosing a smaller fuse would not protect the lamp - it would simply blow, and choosing a larger one would protect nothing at all. The lamp's over-voltage has to be fixed in the circuit design, not in the fuse rating.