In a potentiometer of 10 wires,the balance po | vedclass

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(A) The potential drop across the potentiometer wire is given by $V = I \cdot R_{wire} = \frac{E_{main}}{R_{main} + R_{wire}} \cdot R_{wire}$.To shift

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Increase resistance in the main circuit

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(A) The potential drop across the potentiometer wire is given by $V = I \cdot R_{wire} = \frac{E_{main}}{R_{main} + R_{wire}} \cdot R_{wire}$.To shift the balance point to a higher wire number (greater length),the potential gradient $(x = V/L)$ must be decreased.Since $x = \frac{V}{L} = \frac{I \cdot \rho}{A}$,decreasing the current $I$ in the primary circuit will decrease the potential gradient.By increasing the resistance in the main (primary) circuit,the current $I$ decreases,which reduces the potential gradient.Consequently,a greater length of the wire is required to balance the same emf of the secondary cell.Therefore,we should increase the resistance in the main circuit.

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should:

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