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?

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Solution

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

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In a potentiometer arrangement, a cell of emf

1.25 V

gives a balance point at

35.0 c m

length of the wire. If the cell is replaced by another cell and the balance point shifts to

63.0 c m

, what is the emf of the second cell?

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1.5 V

gives a balance point at

27 c m

length of wire. If the cell is replaced by another cell and balance point shifts to

54 c m

, the emf of the second cell is

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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 $\frac{V_{1}}{l_{1}} = \frac{V_{2}}{l_{2}}$
$∴ \frac{1.25}{35} = \frac{V_{2}}{63}$
$∴ V_{2} = 2.25 V$