An electric bulb rated 220 V, 60 W is connected in series with another bulb rated 220 V, 40 W, and the combination is connected across a 220 V source. If \( P_1' \) and \( P_2' \) are the powers drawn by the 60 W bulb and the 40 W bulb respectively, then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( P_1'\lt P_2' \)
The rating of a bulb gives its resistance through \( P=V^{2}/R \), so the bulb of lower rated power has the higher resistance. In a series combination the same current \( I \) flows through both bulbs. The power drawn by each is \( P=I^{2}R \), so it is larger for the larger resistance.
The resistances from the ratings are:
\( \begin{aligned} R_1 &= \frac{V^{2}}{P_1} \\ &= \frac{(220\,\mathrm{V})^{2}}{60\,\mathrm{W}} \\ &\approx 807\,\Omega \end{aligned} \)
\( \begin{aligned} R_2 &= \frac{V^{2}}{P_2} \\ &= \frac{(220\,\mathrm{V})^{2}}{40\,\mathrm{W}} \\ &= 1210\,\Omega \end{aligned} \)
With the same current through both bulbs:
\( \begin{aligned} \frac{P_1'}{P_2'} &= \frac{I^{2}R_1}{I^{2}R_2} \\ &= \frac{807\,\Omega}{1210\,\Omega} \\ &\approx 0.67 \end{aligned} \)
From the above, \( P_1'\lt P_2' \). Hence, in series the 40 W bulb draws more power and glows brighter than the 60 W bulb.