An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which of the following statement is true?
Answer & explanation
Correct answer: option 2
The correct answer is option 2: $ P_1'< P_2'$
When two bulbs are connected in series,
→ Same current flows through both.
→ Power dissipated in each bulb P′∝R
Now,
From the bulb ratings:
R= $\frac{V^2}{P}$
So resistance is inversely proportional to rated power.
Compare Resistances
R(60W) :R (40W) = 1/ 60 : 1/ 40 = 40 : 60 = 2 : 3
Hence, R(40W) :R (60W)
Compare Powers in Series
Since P′∝R (same current in series),
P′ (60W) : P′ (40W) = R(60W) :R (40W) = 2:3
Therefore, P′ (60W) < P′ (40W)
P1′<P2′
The 40 W bulb (which has higher resistance) glows more brightly.