Current Electricity: Conceptual Doubts Cleared (8)

Question

When two bulbs of $40\ \text{W}$ and $100\ \text{W}$ (both rated $220\ \text{V}$) are connected in series across a $220\ \text{V}$ supply, which one glows brighter? Most students answer “the $100\ \text{W}$ one”. Is that correct?

Solution — Step by Step

Final answer: The $40\ \text{W}$ bulb glows brighter in series.

Why This Works

The deep idea is that “wattage rating” is a conditional statement: “when fed $220\ \text{V}$, this bulb dissipates this much”. Once you change the voltage across the bulb, the rating no longer applies.

In series, the bulb with higher resistance gets a larger share of voltage (voltage divider) and dissipates more $I^2 R$. In parallel, both get full $220\ \text{V}$, so the higher-rated bulb wins as expected.

Alternative Method

Apply the voltage divider directly: $V_{40}/V_{100} = R_{40}/R_{100} = 1210/484 \approx 2.5$. The $40\ \text{W}$ bulb gets $2.5\times$ the voltage of the $100\ \text{W}$, so it dissipates about $2.5\times$ the power.

Common Mistake