Two Lamps in Series at the Exhibition Stall
An exhibition stall is lit by two incandescent lamps taken from a store: one rated 100 W, 220 V and the other 60 W, 220 V. Because only one socket is free, a student connects them in series across the 220 V mains, expecting both to give their rated brightness. Both glow dull, and the 60 W lamp is visibly brighter than the 100 W lamp, so the student concludes that the labels must have been interchanged. Before switching on, he had measured the 100 W lamp with a digital multimeter on the resistance range and read about 40 Ω instead of the 484 Ω he had calculated, which convinced him the lamp was defective. The mains voltage is steady at 220 V. A spare holder and a short piece of insulated wire are available, and he now proposes to short out the dimmer lamp so that the other one receives the full mains voltage. He asks whether the surviving lamp will burn out.
1. Calculate the resistance of each lamp from its rating, then the current, the power dissipated in each and the total power when they are in series on 220 V. Which lamp is brighter, and why does the total fall so far short of 160 W?
Read the worked answer
Resistances from ratings (valid at normal operating temperature):
- R₁₀₀ = V²/P = 220²/100 = 484 Ω
- R₆₀ = 220²/60 = 806.7 Ω
Series circuit. R_total = 484 + 806.7 = 1290.7 Ω; I = 220/1290.7 = 0.1704 A.
Power in each (P = I²R, same current in both):
- P₁₀₀ = (0.1704)² × 484 = 14.1 W
- P₆₀ = (0.1704)² × 806.7 = 23.4 W
- Total ≈ 37.5 W, against 160 W if both ran at rating.
Which is brighter. The 60 W lamp, because in series the current is common and P ∝ R; the 60 W lamp has the larger resistance. The labels are correct - they refer to operation at 220 V each, not in series.
Why the total collapses. Power at a given current is I²R; the series resistance is 2.7 times that of the 100 W lamp alone, so the current is only 0.17 A instead of 0.45 A. Power scales as the square of current, and the circuit now dissipates (0.17/0.45)² ≈ 14 % of the single-lamp power. Voltage sharing, not defect, explains the dimness.
2. Reconcile the measured 40 Ω with the calculated 484 Ω, and explain what this implies for the instant of switching on.
Read the worked answer
No contradiction. The ohmmeter passes a tiny current, so the filament is at room temperature (≈ 300 K). Tungsten's resistivity rises steeply with temperature; at the normal operating temperature of about 2700–2800 K it is roughly 12–14 times its room-temperature value. Here 484/40 ≈ 12, exactly the expected ratio - the lamp is healthy.
Implication at switch-on. The cold resistance is only about 40 Ω, so the initial current is I = 220/40 ≈ 5.5 A, about 12 times the steady 0.45 A. This inrush lasts only while the filament heats (tens of milliseconds), but it is the usual reason lamps fail at the moment of switching on, and it is why filament lamps are rated with a surge-tolerant switch and why dimmers use soft-start.
Practical caution. Because the resistance is not constant, the series calculation above gives only the final steady state; the transient is far worse. Any fuse or switch in the lamp circuit must survive ~5.5 A per lamp briefly, not just 0.45 A.
3. The student shorts out one lamp holder with the wire. Predict what happens to the remaining lamp in each of the two possible cases, quantify the current, and identify the real hazard in this arrangement.
Read the worked answer
Case A - the 60 W lamp is shorted, the 100 W lamp remains. The 100 W lamp now sees the full 220 V, draws 220/484 ≈ 0.45 A and runs at its rated 100 W. It does not burn out: it is designed for exactly this voltage.
Case B - the 100 W lamp is shorted, the 60 W lamp remains. The 60 W lamp sees 220 V, draws 220/806.7 ≈ 0.27 A and runs at its rated 60 W. Again normal.
So the prediction that the surviving lamp must burn out is wrong; shorting a holder simply removes one lamp from the circuit and restores rated operation of the other. The wire carries at most 0.45 A, well within its capacity.
The real hazard. The shorted holder is now directly connected across the mains: the two exposed terminals sit at 220 V with no lamp to drop the voltage. Touching them, or pushing a lamp into that holder later, gives a shock or a short circuit at the mains. The correct practice is to remove the holder from the circuit, insulate the terminals, and never leave an empty, energised lampholder in a string.