Q. A torch bulb rated as $4.5\ W,\ 1.5\ V$ is connected as shown in the figure. The $e.m.f.$ of the cell needed to make the bulb glow at full intensity is
A) $4.5\ V$
B) $1.5\ V$
C) $2.67\ V$
D) $13.5\ V$
Explanation
(d) Current in the bulb $= \frac{P}{V} = \frac{4.5}{1.5} = 3A$
Current in $1\Omega$ resistance $= \frac{1.5}{1} = 1.5A$
Hence total current from the cell $i = 3 + 1.5 = 4.5A$
By using $E = V + ir \Rightarrow E = 1.5 + 4.5 \times (2.67) = 13.5V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.