question_answer

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. is balanced at a length \[I\] of the potentiometer wire. The e.m.f. E will be given by                                                                                                                             [NEET (Re) 2015]

A)  \[\frac{L{{E}_{0}}f}{l\,{{r}_{1}}}\]    

B)       \[\frac{{{E}_{0}}r}{(r+{{r}_{1}})}.\frac{l}{L}\]

C)  \[\frac{{{E}_{0}}l}{L}\]                      

D)       \[\frac{L{{E}_{0}}r}{(r+{{r}_{1}})l}\]

Correct Answer: B

Solution :

[b] Consider 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 \[{{r}_{1}}\] as shown in figure.

Current in wire \[AB=\frac{{{E}_{0}}}{{{r}_{1}}+r}\]

Potential gradient,

\[x=\frac{Ir}{L}\left[ \frac{{{E}_{0}}}{{{r}_{1}}+r} \right]\frac{r}{L}\]

e.m.f. produced across E will be given by

\[E=x.l=\left[ \frac{{{E}_{0}}r}{{{r}_{1}}+r} \right]\frac{\text{l}}{L}\]