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In the experiment to find internal resistance of Daniel cell using potentiometer with E as Emt of the cell 'L' the balancing length of the potentiometer wire when potentiometer circuit is closed. (when no current flows through cell. When $R= 5 \Omega $ is introduced the new balancing length is l.) Then the internal resistance of the cell is

$(a)\;r= \large\frac{l}{(L-l)} \times R \\(b)\;r= \large\frac{L-l}{e \times R} \\(c)\;r= \large\frac{l}{(L)} \times R \\(d)\;r= \large\frac{L}{l} \times R$
Can you answer this question?
 
 

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When the cell is short circuited with resistance R the emf falls to $\large\frac{ER}{R+r}$
$\therefore \large\frac{r+R}{R}=\frac{L}{l}$
$\therefore r= \large\frac{R(L-l)}{l}$$ \;Ohms$
Hence a is the correct answer.
answered Mar 19, 2014 by meena.p
 

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