Two Bulbs in Series: Who Was Right About the Brightness?
A physics student is given two filament bulbs marked "60 W, 220 V" and "100 W, 220 V". She connects them in series across the 220 V mains and predicts the 100 W bulb will glow more brightly because it is the more powerful bulb. Her lab partner predicts the opposite. The supply holds a steady 220 V, the filament resistances are taken as constant at their rated values, and the bench socket is protected by a 1 A fuse. The student has a second 100 W bulb available and a multimeter, but the fuse must not be replaced during the practical.
1. Compute the resistances and the power actually dissipated in each bulb, and decide who was right.
Read the worked answer
Step 1 - resistances from the ratings. The rating gives power at 220 V, so R = V²/P.
- R₆₀ = 220²/60 = 48,400/60 = 806.7 Ω
- R₁₀₀ = 220²/100 = 48,400/100 = 484 Ω
The "more powerful" bulb has the lower resistance - that is the whole issue.
Step 2 - series current. R_total = 806.7 + 484 = 1,290.7 Ω I = 220/1,290.7 = 0.170 A
Step 3 - power per bulb, P = I²R (same current in both).
- P₆₀ = (0.170)² × 806.7 = 0.0290 × 806.7 ≈ 23.4 W
- P₁₀₀ = 0.0290 × 484 ≈ 14.0 W
- Check: total 37.4 W; supply gives VI = 220 × 0.170 = 37.4 W ✓
Verdict: the partner is right. The 60 W bulb glows more brightly in series. In series the current is shared, so power splits in proportion to resistance, and the lower-rated bulb has the higher resistance.
The student's error is applying a parallel-circuit intuition (higher rating → brighter) to a series circuit. Also worth noting: both bulbs run well below their ratings (23.4 < 60, 14.0 < 100), so neither is at risk.
2. Is the 1 A fuse adequate for the series arrangement and for a parallel arrangement of the same two bulbs?
Read the worked answer
Series: I = 0.170 A. Comfortably under 1 A ✓. The fuse would not blow, but a 1 A fuse gives almost no discrimination here - a fault adding 0.83 A would pass unnoticed.
Parallel across the same 220 V supply: each bulb sees the full mains voltage, so each draws its rated current.
- I₆₀ = 220/806.7 = 0.273 A
- I₁₀₀ = 220/484 = 0.455 A
- Total = 0.727 A < 1 A ✓
So the fuse holds in both arrangements, and the experiment can be run either way on this bench.
The judgement to flag: the fuse is adequate, but only just, in parallel - 0.727 A against a 1 A rating is 73% loading. If the student later adds a third bulb in parallel the total passes 1 A and the fuse blows mid-practical. Since the fuse may not be replaced during the session, the safer plan is to run the parallel test last, or to use the series arrangement for any measurement involving added components.
3. The student swaps the 60 W bulb for the spare 100 W bulb, keeping the series connection. What happens to the untouched 100 W bulb's brightness, and to the current?
Read the worked answer
New circuit: two identical 484 Ω filaments in series.
- R_total = 968 Ω
- I = 220/968 = 0.227 A (up from 0.170 A - the total resistance fell, so current rose)
- P per bulb = I²R = (0.227)² × 484 = 0.0516 × 484 ≈ 25 W in each
- Check: 25 + 25 = 50 W; VI = 220 × 0.227 = 50 W ✓
The untouched 100 W bulb goes from about 14.0 W to about 25 W - it becomes noticeably brighter. Because the two bulbs are now identical, they share the 220 V equally (110 V each) and dissipate equal power.
Why this matters conceptually: the second bulb's brightness did not change because of anything done to it. It changed because removing the high-resistance 60 W filament raised the current through the whole series loop. Brightness in a series circuit is a property of the whole circuit, not of the individual bulb's rating.
Fuse check: 0.227 A is still far below 1 A, so the swap is safe. This third case is also the clean demonstration for the write-up: same bulb, three different brightnesses, purely from what it is connected in series with.