Q. A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by :
A) $\frac{LE_0r}{(r+r_1)l}$
B) $\frac{LE_0r}{lr_1}$
C) $\frac{E_0r}{(r+r_1)} \cdot \frac{l}{L}$
D) $\frac{E_0l}{L}$
Explanation
[Solution] Potential gradient = $\frac{E_0}{r+r_1} \cdot \frac{r}{L}$, so $E = \frac{E_0r}{r+r_1} \cdot \frac{l}{L}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.