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Home » 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 thesecond cell?
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?
CBSE | Class 12 | Current Electricity | NCERT | Physics
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?

Answer –

According to the question statement –

Emf of the cell, E 1 = 1.25 V

Balance point of potentiometer, given by l 1 = 35 cm

In the question, the cell is replaced by another cell of emf E 2.

New balance point of the potentiometer is given by l 2 = 63 cm

We know that the balance condition is given by the relation –

    \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{I}_{1}}}{{{I}_{2}}}\]

    \[{{E}_{2}}={{E}_{1}}\frac{{{I}_{1}}}{{{I}_{2}}}=1.25\times \frac{63}{35}\]

    \[{{E}_{2}}=2.25V\]

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