The figure shows a 2.0 ; V potentiometer used | vedclass

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(A) Let the internal resistance of the cell be $r$.The balance point of the cell in an open circuit is given by $l_1 = 76.3 \; cm$.When an external re

Vedclass · Accepted Answer

$1.68$

Vedclass · Answer

(A) Let the internal resistance of the cell be $r$.The balance point of the cell in an open circuit is given by $l_1 = 76.3 \; cm$.When an external resistance $R = 9.5 \; \Omega$ is connected in parallel to the cell,the new balance point is $l_2 = 64.8 \; cm$.The formula for the internal resistance $r$ of a cell using a potentiometer is given by:$r = \left( \frac{l_1 - l_2}{l_2} ight) R$Substituting the given values:$r = \left( \frac{76.3 - 64.8}{64.8} ight) 	imes 9.5$$r = \left( \frac{11.5}{64.8} ight) 	imes 9.5$$r \approx 0.17747 	imes 9.5 \approx 1.686 \; \Omega$.Rounding to two decimal places,the internal resistance of the cell is $1.68 \; \Omega$.

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