Two students X and Y perform a potentiometer | vedclass

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(D) The potential drop across the potentiometer wire $AB$ is given by $V_{AB} = E \cdot \frac{R_{AB}}{R + R_{AB} + r}$,where $E$ is the $EMF$ of the d

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would shift towards point $B$,would shift towards point $A$

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(D) The potential drop across the potentiometer wire $AB$ is given by $V_{AB} = E \cdot \frac{R_{AB}}{R + R_{AB} + r}$,where $E$ is the $EMF$ of the driver cell,$R$ is the external resistance,$R_{AB}$ is the resistance of the wire,and $r$ is the internal resistance of the driver cell.$(i)$ When student $X$ increases $R$,the total resistance of the primary circuit increases,so the current $I = \frac{E}{R + R_{AB} + r}$ decreases. Consequently,the potential gradient $k = \frac{V_{AB}}{L}$ decreases. Since the balancing condition is $E_1 = k \cdot l$,where $l$ is the balancing length,if $k$ decreases,$l$ must increase to maintain the same $E_1$. Thus,the null point shifts towards $B$.(ii) When student $Y$ decreases $S$,the terminal potential difference across the cell $E_1$ is $V = E_1 - I_1 r_1$,where $I_1 = \frac{E_1}{S + r_1}$. Decreasing $S$ increases the current $I_1$ drawn from the cell $E_1$,which increases the voltage drop $I_1 r_1$ across the internal resistance $r_1$. Therefore,the terminal voltage $V$ across the cell $E_1$ decreases. Since $V = k \cdot l$,a decrease in $V$ requires a smaller balancing length $l$. Thus,the null point shifts towards $A$.

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