Q. In the given figure, battery $E$ is balanced on $55\ cm$ length of potentiometer wire but when a resistance of $10\ \Omega$ is connected in parallel with the battery then it balances on $50\ cm$ length of the potentiometer wire then internal resistance $r$ of the battery is
A) $1\ \Omega$
B) $3\ \Omega$
C) $10\ \Omega$
D) $5\ \Omega$
Explanation
(a) $r = \left( \frac{l_{1} - l_{2}}{l_{2}} \right) \times R^{'} \Rightarrow r = \left( \frac{55 - 50}{50} \right) \times 10 = 1\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.