A potentiometer circuit shown in the figure i | vedclass

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Question

(A) In a potentiometer, the potential drop across the length $XP$ is given by $V_{XP} = I \cdot R_{XP}$, where $I$ is the current in the potentiometer

Vedclass · Accepted Answer

Decrease the resistance $R$

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(A) In a potentiometer, the potential drop across the length $XP$ is given by $V_{XP} = I \cdot R_{XP}$, where $I$ is the current in the potentiometer wire $XY$. For a balance point to exist, the potential drop across $XP$ must be equal to the $e.m.f.$ of the cell $E$ $(V_{XP} = E)$. Since the galvanometer deflection decreases as $P$ moves towards $Y$, it implies that the potential drop across $XP$ is still less than $E$ even at the end $Y$ $(V_{XY} To obtain a balance point between $X$ and $Y$, we need to increase the potential gradient along the wire $XY$. This can be achieved by increasing the current $I$ in the wire $XY$. Since $I = V / (R_{wire} + R)$, decreasing the external resistance $R$ will increase the current $I$, thereby increasing the potential drop across any segment $XP$ for a given position of $P$. Thus, decreasing $R$ allows the balance point to be reached within the length $XY$.

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