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?

Open in App

Solution

Given,
Emf of the cell, $E_{1} = 1.25 V$
Balance point of the potentiometer, $l_{1} = 35 c m$
The cell is replaced by another cell of emf $E_{2} .$
New balance point of the potentiometer, $l_{2} = 63 c m$
The balance condition is given by the relation,
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$E_{2} = E_{1} \times \frac{l_{2}}{l_{1}}$

$\Rightarrow E_{2} = 1.25 \times \frac{63}{35} = 2.25 V$

Therefore, end of the second cell is $2.25 V .$

Suggest Corrections

Similar questions

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?

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

Q. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell, the balance point shifts to 63 cm, then emf of the second cell is

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

Q. State the working principle of Potentiometer. Explain with the help of circuit diagram, how the potentiometer is used to determine the internal resistance of the given Primary cell.
In a potentiometer arrangement, a cell of emf