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(A) The current $I$ in the potentiometer wire $AC$ is given by:$I = \frac{E}{R_{total}} = \frac{2}{10 + R}$The potential difference $V_{AC}$ across th

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$790 \, \Omega$

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(A) The current $I$ in the potentiometer wire $AC$ is given by:$I = \frac{E}{R_{total}} = \frac{2}{10 + R}$The potential difference $V_{AC}$ across the entire potentiometer wire is:$V_{AC} = I 	imes R_{wire} = \left( \frac{2}{10 + R} ight) 	imes 10$The potential gradient $k$ along the wire of length $L = 100 \, cm$ is:$k = \frac{V_{AC}}{L} = \left( \frac{2}{10 + R} ight) 	imes \frac{10}{100} = \frac{2}{10(10 + R)}$The source of $emf$ $E' = 10 \, mV = 10 	imes 10^{-3} \, V$ is balanced against a length $l = 40 \, cm$. The balancing condition is:$E' = k 	imes l$$10 	imes 10^{-3} = \left( \frac{2}{10(10 + R)} ight) 	imes 40$$10^{-2} = \frac{8}{10 + R}$$10 + R = \frac{8}{10^{-2}} = 800$$R = 800 - 10 = 790 \, \Omega$.

Similar Questions

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

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