The current flowing through the two bulbs marked as 60 W, 240 V each when connected in series with 240 V source is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.125 A
The rating of a bulb (60 W, 240 V) fixes its resistance, since it draws 60 W when 240 V is applied across it.
\( \displaystyle R = \frac{V^{2}}{P} = \frac{(240)^{2}}{60} = 960\,\Omega \)
In series the resistances add, and the same current flows through both bulbs.
\( R_{\mathrm{s}} = 960 + 960 = 1920\,\Omega \)
\( \displaystyle I = \frac{V}{R_{\mathrm{s}}} = \frac{240\,\mathrm{V}}{1920\,\Omega} = 0.125\,\mathrm{A} \)