Q. A torch battery consists of two cells of $1.45\ volt$ and an internal resistance $0.15\ \Omega.$ If each cell sends current through the filament of the lamps having resistance $1.5\ ohm,$ the value of current will be
A) $16.11\ amp$
B) $1.611\ amp$
C) $0.1611\ amp$
D) $2.6\ amp$
Explanation
(b) Here two cells are in series
Therefore total emf $= 2E$
Total resistance $= R + 2r$
$\therefore i = \frac{2E}{R + 2r} = \frac{2 \times 1.45}{1.5 + 2 \times 0.15} = \frac{2.9}{1.8} = \frac{29}{18} = 1.611\ amp$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.