AB is a potentiometer wire of length 100, cm | vedclass

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(D) $1$. First,calculate the current $I$ flowing through the potentiometer wire $AB$. The total resistance of the circuit is $R_{total} = R_{wire} + R

Vedclass · Accepted Answer

$0.16$

Vedclass · Answer

(D) $1$. First,calculate the current $I$ flowing through the potentiometer wire $AB$. The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} = 10\,\Omega + 40\,\Omega = 50\,\Omega$.$2$. The current $I$ in the circuit is given by $I = \frac{V}{R_{total}} = \frac{2\,V}{50\,\Omega} = 0.04\,A$.$3$. The potential drop across the potentiometer wire $AB$ is $V_{AB} = I 	imes R_{wire} = 0.04\,A 	imes 10\,\Omega = 0.4\,V$.$4$. The potential gradient $k$ (potential drop per unit length) is $k = \frac{V_{AB}}{L} = \frac{0.4\,V}{100\,cm} = 0.004\,V/cm$.$5$. The unknown $e.m.f.$ $E$ is balanced by a length $l = 40\,cm$. Therefore,$E = k 	imes l = 0.004\,V/cm 	imes 40\,cm = 0.16\,V$.

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