In a potentiometer experiment,the null point | vedclass

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(B) The potential gradient $x$ of a potentiometer wire is given by $x = \frac{V}{L} = \frac{E R}{(R + R') L}$,where $E$ is the e.m.f. of the driver ce

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Increase resistance in main circuit

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(B) The potential gradient $x$ of a potentiometer wire is given by $x = \frac{V}{L} = \frac{E R}{(R + R') L}$,where $E$ is the e.m.f. of the driver cell,$R$ is the resistance of the potentiometer wire,$R'$ is the resistance in the main circuit,and $L$ is the total length of the wire.For a given cell of e.m.f. $\epsilon$,the balancing length $l$ is given by $\epsilon = x \cdot l$,which implies $l = \frac{\epsilon}{x}$.To shift the null point from the $7^{	ext{th}}$ wire to the $9^{	ext{th}}$ wire,the balancing length $l$ must increase.Since $l = \frac{\epsilon}{x}$,increasing $l$ requires a decrease in the potential gradient $x$.Looking at the expression $x = \frac{E R}{(R + R') L}$,to decrease $x$,we must increase the denominator term $(R + R')$.Therefore,we must increase the resistance $R'$ in the main circuit.

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?

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