Q. 4 cells each of emf $2\ V$ and internal resistance of $1\Omega$ are connected in parallel to a load resistor of $2\Omega$. Then the current through the load resistor is
A) $2\ A$
B) $1.5\ A$
C) $1\ A$
D) $0.888\ A$
Explanation
(d) $I = \frac{E}{R + r/4} = \frac{2}{2 + 1/4} = \frac{2}{2.25} = 0.888A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.