Q. A current of two ampere is flowing through a cell of $e.m.f.\ 5\ volt$ and internal resistance $0.5\ ohm$ from negative to positive electrode. If the potential of negative electrode is $10V,$ the potential of positive electrode will be
A) $5\ V$
B) $14\ V$
C) $15\ V$
D) $16\ V$
Explanation
(b) $V_{2} - V_{1} = E - ir = 5 - 2 \times 0.5 = 4volt$
$\Rightarrow V_{2} = 4 + V_{1} = 4 + 10 = 14\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.