In a potentiometer arrangement- a cell of emf 1-25 V gives a balance point 35-0 cm length of the wire- If the cell is replaced by another cell and the balance point shifts to 63-0 cm- what is the emf of the second cell-

Question

Toppr Admin · Accepted Answer

This can be solved by the formula V1l1-V2l2-x2234-1-2535-V263-x2234-V2-2-25V

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

This can be solved by the formula $\frac{V_{1}}{l_{1}} = \frac{V_{2}}{l_{2}}$
$∴ \frac{1.25}{35} = \frac{V_{2}}{63}$
$∴ V_{2} = 2.25 V$

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

This can be solved by the formula V1l1=V2l2∴1.2535=V263∴V2=2.25V

This can be solved by the formula $\frac{V_{1}}{l_{1}} = \frac{V_{2}}{l_{2}}$
$∴ \frac{1.25}{35} = \frac{V_{2}}{63}$
$∴ V_{2} = 2.25 V$