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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?
In a potentiometer experiment, a cell of emf 1.25 V gives a balance point at 35 cm mark of the wire. If this cell is replaced by another cell, the balance point is at 63 cm mark. Calculate the emf of the second cell.
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Solution
Emf of the cell, E1 = 1.25 V
Balance point of the potentiometer, l1 = 35 cm
The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm
The balance condition is given by the relation,
`E_1/E_2 = I_1/I_2`
`E_2 = E_1 xx I_2/I_1`
= `1.25 xx 63/35`
= 2.25 V
Therefore, the emf of the second cell is 2.25 V.
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