Two Lamps in Series: Rating Versus Actual Power
In the SSC Science and Technology practical examination a group is given two incandescent lamps marked '100 W, 220 V' and '60 W, 220 V', a steady 220 V supply, holders, connecting wires and a 5 A fuse. The group connects the two lamps in series across the supply. Both glow dimly, and the 60 W lamp is clearly brighter than the 100 W lamp. A teammate insists that the 100 W lamp must be brighter 'because it is the more powerful one', and proposes replacing the 5 A fuse with a 15 A fuse so that the lamps receive 'full current'. The supervisor refuses to allow the change until the group justifies both the observation and the proposal, and asks what would happen if the lamps were connected in parallel instead. Lamp resistances may be treated as constant and the supply as ideal.
1. Calculate which lamp glows brighter in series and explain the reasoning, not just the answer.
Read the worked answer
The wattage printed on a lamp is the power it consumes at its rated voltage, not a fixed output. Resistance is found from P = V²/R, so R = V²/P.
- 100 W lamp: R = 220²/100 = 484 Ω
- 60 W lamp: R = 220²/60 = 806.7 Ω
In series the same current passes through both, so the lamps share the 220 V and cannot each receive 220 V.
- Total resistance = 484 + 806.7 = 1290.7 Ω
- Current I = 220/1290.7 = 0.170 A
- Power in 100 W lamp = I²R = (0.170)² × 484 = 14.1 W
- Power in 60 W lamp = (0.170)² × 806.7 = 23.4 W
Since the current is common, P = I²R means power is proportional to resistance. The lower-rated lamp has the higher resistance, so it dissipates more and glows brighter. The teammate has confused the rating with the actual operating condition.
2. Evaluate the proposal to fit a 15 A fuse. Is it physically effective and is it safe?
Read the worked answer
It is neither effective nor safe.
Physically ineffective: the current in the circuit is fixed by the supply voltage and the load resistance, I = V/R = 220/1290.7 = 0.17 A. A fuse does not drive current; it only melts when current exceeds its rating. Changing 5 A to 15 A therefore changes nothing about the brightness, because the current is already limited to 0.17 A by the lamps. The total power drawn is only 14.1 + 23.4 ≈ 37.5 W, roughly 0.17 A at 220 V.
Unsafe: a fuse is a protective device, chosen to be just above normal current but below the current that would overheat the wiring. A 5 A fuse is already far above the 0.17 A that flows, so it protects the connecting wires and the supply, not the lamps. Raising it to 15 A removes that protection: if a fault or short circuit occurred, the wiring could overheat and burn before the fuse acted.
The correct physics conclusion is that the fuse rating is not the cause of the dim glow - the series connection is.
3. What changes if the lamps are connected in parallel? Which lamp is brighter, and is the 5 A fuse adequate?
Read the worked answer
In parallel each lamp receives the full 220 V, so each now operates at its rated power.
- Current in 100 W lamp = P/V = 100/220 = 0.455 A
- Current in 60 W lamp = 60/220 = 0.273 A
- Total current drawn = 0.727 A
The 100 W lamp is now brighter, because it converts 100 W compared with 60 W. This confirms that the earlier result was a property of the series circuit, not of the lamps.
The 5 A fuse is comfortably adequate: the normal current of 0.73 A is about one-seventh of the rating. That is exactly how a fuse should be chosen - above the normal operating current but low enough to protect the wiring. The parallel connection also has a practical consequence worth stating: if one lamp fails, the other continues to glow, whereas in series a broken filament stops the current and both go out.