Q. Five cells each of internal resistances $0.2\Omega$ and emf 2 V are connected in series with a resistance of 4$\Omega.$ The current through the external resistance is
A) 4 A
B) 2 A
C) 1 A
D) 0.5 A
Explanation
(b) Current through external resistance,
$i = \frac{nE}{nr + R} = \frac{5 \times 2}{5 \times 0.2 + 4} = 2A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.