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?

Solution

Given, Emf of cell, E₁ = 1.25 V
Balance point is obtained at, l₁ = 35 cm
When, the cell is replaced by another cell
New balance point is, l₂= 63 cm
Emf of the second cell can be found out using the relation,
$\frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}}$

Now, substituting values we get,

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

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Current Electricity

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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