Q. When a resistor of $11\ \Omega$ is connected in series with an electric cell, the current flowing in it is $0.5\ A.$ Instead, when a resistor of $5\ \Omega$ is connected to the same electric cell in series, the current increases by $0.4\ A.$ The internal resistance of the cell is
A) $1.5\ \Omega$
B) $2\ \Omega$
C) $2.5\ \Omega$
D) $3.5\ \Omega$
Explanation
(c) By using $i = \frac{E}{R + r}$
$\Rightarrow 0.5 = \frac{E}{11 + r} \Rightarrow E = 5.5 + 0.5r$ …(i)
and $0.9 = \frac{E}{5 + r} \Rightarrow E = 4.5 + 0.9r$ …(ii)
On solving these equations, we have $r = 2.5\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.