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  • Can someone help me with this, - Current Electricity - JEE Main-2

In a potentiometer experiment the balancing with  a cell is at length 240 cm. On shunting the cell with a resistance of  2\Omega the balancing length becomes 120 cm The internal resistance of the cell is

  • Option 1)

    4\Omega

  • Option 2)

    2\Omega

  • Option 3)

    1\Omega

  • Option 4)

    0.5\Omega

 

Answers (2)

As we learnt in

Determine the internal resistance -

r=(\frac{l_{1}-l_{2}}{l_{2}})R

r=(\frac{E}{V}-1)R

\frac{E}{V}=\frac{l_{1}}{l_{2}}

- wherein

 

 

 

The internal resistance of a cell is given  by

r= R\left ( \frac{l_{1}}{l_{2}} -1\right )= R\left ( \frac{l_{1}-l_{2}}{l_{2}}\right )

\therefore \: \: \: \: \: \: r= 2\left [ \frac{240-120}{120} \right ]= 2\Omega


Option 1)

4\Omega

This is incorrect.

Option 2)

2\Omega

This is correct.

Option 3)

1\Omega

This is incorrect.

Option 4)

0.5\Omega

This is incorrect.

Posted by

Vakul

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