The wire of a potentiometer has a resistance of $4 \Omega$ and a length of $1 \ m$. It is connected to a cell of emf $2 \ V$ and internal resistance $1 \Omega$. The current flowing through the potentiometer wire is: (in $A$)

  • A
    $0.1$
  • B
    $0.2$
  • C
    $0.4$
  • D
    $0.8$

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Similar Questions

In a potentiometer arrangement,a cell gives a balancing point at $75 \, cm$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $3:2$,the difference in the balancing length of the potentiometer wire in above two cases will be ......... $cm$.

It is preferable to measure the $e.m.f.$ of a cell by a potentiometer rather than by a voltmeter because of the following possible reasons.
$(i)$ In the case of a potentiometer,no current flows through the cell.
$(ii)$ The length of the potentiometer wire allows for greater precision.
$(iii)$ Measurement by the potentiometer is quicker.
$(iv)$ The sensitivity of the galvanometer,when using a potentiometer,is not relevant.
Which of these reasons are correct?

In a potentiometer experiment,the balancing point with a cell is at a 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: (in $\Omega$)

In the experiment of a potentiometer,at balance,there is no current in the

In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

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