Q. Two batteries $A$ and $B$ each of e.m.f. $2\ V$ are connected in series to an external resistance $R = 1\ ohm.$ If the internal resistance of battery $A$ is $1.9\ ohm$ and that of $B$ is $0.9\ ohm,$ what is the potential difference between the terminals of battery $A$
A) $2\ V$
B) $3.8\ V$
C) Zero
D) None of the above
Explanation
(c) $i = \frac{2 + 2}{1 + 1.9 + 0.9} = \frac{4}{3.8}A$
For cell $A\ E = V + ir \Rightarrow V = 2 - \frac{4}{3.8} \times 1.9 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.