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Question
In potentiometer experiment, if l1 is the balancing length for e.m.f. of cell of internal resistance r and l2 is the balancing length for its terminal potential difference when shunted with resistance R then :
Options
`l_1=l_2((R+r)/R)`
`l_1=l_2((R)/(R+r))`
`l_1=l_2((R)/(R-r))`
`l_1=l_2((R-r)/R)`
Solution
`l_1=l_2((R+r)/R)`
The internal resistance of a cell is given as
`r= ((l_1-l_2)/l_2)R`
→`r/R=(l_1-l_2)/l_2`
∴ `rl_2=Rl_1-Rl_2`
∴`l_2(R+r)=Rl_1`
∴`l_1=l_2(R+r)/R`
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