Two electric bulbs rated 25 W, 220 V and 100 W, 220 V are connected in series to a 440 V supply. Which of the bulbs will fuse?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The 25 W bulb
Each bulb has a fixed resistance \( R=V^2/P \) found from its rating. In series the same current flows through both, so the 440 V supply divides in the ratio of the resistances. A bulb fuses when the voltage across it is more than its rated 220 V.
The resistances of the two bulbs are:
\( \begin{aligned} R_{25} &= \frac{(220\,\mathrm{V})^2}{25\,\mathrm{W}} \\ &= 1936\,\Omega \end{aligned} \)
\( \begin{aligned} R_{100} &= \frac{(220\,\mathrm{V})^2}{100\,\mathrm{W}} \\ &= 484\,\Omega \end{aligned} \)
The current in the series circuit is:
\( \begin{aligned} I &= \frac{V}{R_{25}+R_{100}} \\ &= \frac{440\,\mathrm{V}}{(1936+484)\,\Omega} \\ &= \frac{2}{11}\,\mathrm{A} \end{aligned} \)
The voltage across each bulb is:
\( \begin{aligned} V_{25} &= IR_{25} \\ &= \frac{2}{11}\,\mathrm{A}\times1936\,\Omega \\ &= 352\,\mathrm{V} \end{aligned} \)
\( \begin{aligned} V_{100} &= IR_{100} \\ &= \frac{2}{11}\,\mathrm{A}\times484\,\Omega \\ &= 88\,\mathrm{V} \end{aligned} \)
From the above, the 25 W bulb gets 352 V, far more than its rated 220 V, so its filament burns out. The 100 W bulb gets only 88 V. Hence, the 25 W bulb will fuse.