AB is a potentiometer wire as shown in the fi | vedclass

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(B) In a potentiometer circuit,the potential difference across the wire $AB$ is given by $V_{AB} = I \cdot R_{AB}$,where $I$ is the current in the pri

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(B) In a potentiometer circuit,the potential difference across the wire $AB$ is given by $V_{AB} = I \cdot R_{AB}$,where $I$ is the current in the primary circuit and $R_{AB}$ is the resistance of the wire $AB$.The current $I$ in the primary circuit is given by $I = \frac{E}{R + R_{AB}}$,where $E$ is the $EMF$ of the primary cell and $R$ is the external resistance.When the value of $R$ is increased,the total resistance of the primary circuit increases,which causes the current $I$ to decrease.Since $V_{AB} = I \cdot R_{AB}$,a decrease in $I$ leads to a decrease in the potential difference $V_{AB}$ across the wire $AB$.The potential gradient $k$ is defined as $k = \frac{V_{AB}}{L}$,where $L$ is the length of the wire $AB$. Thus,the potential gradient $k$ decreases.The balance point $J$ is obtained when the $EMF$ of the secondary cell $\varepsilon$ equals the potential drop across the length $AJ$,i.e.,$\varepsilon = k \cdot AJ$.Therefore,$AJ = \frac{\varepsilon}{k}$. Since $k$ decreases,the length $AJ$ must increase to maintain the balance condition.As the length $AJ$ increases,the balance point $J$ shifts towards $B$.

$AB$ is a potentiometer wire as shown in the figure. If the value of $R$ is increased,in which direction will the balance point $J$ shift?

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