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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?

3.12 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?

Views

Given,

$\dpi{100} E_1$ = 1.25 V ,  $\dpi{100} l_1$ = 35 cm

And, $\dpi{100} l_2$ = 63 cm

Let  $\dpi{100} E_2$ be the voltage in the second case.

Now, the balance condition is given by :

$\dpi{100} \frac{E_1}{E_2} = \frac{l_1}{l_2}$

$\dpi{100} \implies E_2 = \frac{l_2}{l_1}\times E_1 = \frac{63}{35}\times1.25$

$\dpi{100} \implies E_2 = 2.25 V$

Therefore, the emf of the second cell = 2.25 V

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