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 :
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 \[E=x.l=\left[ \frac{{{E}_{0}}r}{{{r}_{1}}+r} \right]\frac{\text{l}}{L}\]You need to login to perform this action.
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