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The length of a wire of a potentiometer is 100cm and emf of its standard cell is E volt.It is employed to measure the emf of a battery whose internal resistance is 0.5 $\Omega$ . If the balance point is obtained at l=30cm from the positive end,emf of the battery is
Option: 1
$\frac{30E}{100}$

Option: 2
$\frac{30E}{100.5}$

Option: 3
$\frac{30E}{100-0.5}$

Option: 4
$\frac{30\left ( E-50i \right )}{100}$

Answers (1)

best_answer

As we learnt

Potential gradient -

Potential difference per unit length of wire

- wherein

$x=\frac{V}{L}$

Using the principle of potentiometer $v\alpha l$

$\frac{V}{E}= \frac{l}{L}\: \: or\: \: V=\frac{l}{L}E= \frac{30E}{100}$

Posted by

SANGALDEEP SINGH

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