Q.
A 100 W bulb $B_{1}$ and two 60 W bulbs $B_{2}\ and\ B_{3}$ are connected to a 250 V source as shown in figure. Now $W_{1},W_{2}\ and\ W_{3}$ are the output powers of the bulbs $B_{1},B_{2}\ and\ B_{3}$respectively, then
A) $W_{1} > W_{2} = \ W_{3}$
B) $W_{1} > W_{2}\ > \ W_{3}$
C) $W_{1} < W_{2} = W_{3}$
D) $W_{1} < W_{2}\ < \ W_{3}$
Explanation
(d) As resistance of a bulb $R = \frac{V^{2}}{P},$
Hence $R_{1}:R_{2}:R_{3} = \frac{1}{100}:\frac{1}{60}:\frac{1}{60}$
Now the combined potential difference across $B_{1}\ and\ B_{2}$is same as the potential difference across $B_{3}$. Hence, $W_{3}$ is more than $W_{1}\ and\ W_{2}$, being in series, carry same current and $R_{1} < R_{2}$, therefore $W_{1} < W_{2}$,
$\therefore\ W_{1} < W_{2} < W_{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.