Bulb \( B_1 \) is a 50 W bulb working on 100 V and Bulb \( B_2 \) is a 200 W bulb also working on 100 V. The ratio of resistances of bulbs \( B_1 \) to \( B_2 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4 : 1
For a bulb rated at power \( P \) on voltage \( V \), the power rating gives the resistance through \( P = V^{2}/R \).
\( \displaystyle R = \frac{V^{2}}{P} \)
\( \displaystyle R_1 = \frac{(100)^{2}}{50} = 200\,\Omega \)
\( \displaystyle R_2 = \frac{(100)^{2}}{200} = 50\,\Omega \)
\( \displaystyle \frac{R_1}{R_2} = \frac{200}{50} = 4 : 1 \)
At the same voltage, \( R \propto 1/P \): the lower-power bulb has the higher resistance.