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Question
The emf of the cell of internal resistance 1.275 Ω balances against a length of 217 cm of a potentiometer wire. Find the balancing length when the cell is shunted by a resistance of 15 Ω.
Numerical
Solution
The internal resistance of the cell is
`r = R(l/(l_1)-1)`
= `15(217/200 - 1)`
= `(15 xx 17)/200`
= `255/200`
= 1.275 Ω
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Potentiometer
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