Q. A house, served by 220 V supply line, is protected by a 9 A fuse. The maximum number of 60 W bulbs in parallel that can be turned on is
A) 11
B) 22
C) 33
D) 44
Explanation
(c) Resistance of each bulb $= \frac{(220)^{2}}{60}$
If the number of bulbs is $n$ then effective resistance
$= \frac{(220)^{2}}{60n}$
$\therefore i = 9 = \frac{V}{R_{eff}} = \frac{220 \times 60n}{220 \times 220}\ \Rightarrow \ \ n = 33$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.