Discussion Forum

 Two cells having unknown emf E₁ and E₂ (E₁  > E₂)  are connected in a potentiometer circuit,to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E₂ is connected, so as to oppose cell E₁, the null point is obtained at 90 cm from the same end, the ratio of the EMFs of two cells $\left(\frac{E_{1}}{E_{2}}\right)$ is 

Answer:

Explanation:

  The emf of a cell in a potentiometer is 

                   $E=\frac{V}{L}l$

 where, l= length of wire at  null point,

 V= voltage  of source

 and L= total length  of  potentiometer wire

 $\therefore$                $E\propto l$

 When two cell  of emf  E₁ and E₂   (E₁ > E₂) are connected to assist each other , then

 E₁+E₂=490   ........(i)

 when the cell of emf E₂ is connected , so as to  oppose cell E₁ then

 E₁-E₂=90    ..............(ii)

 from ERqs. (i) and (ii) , we get

 E₁=290 V  and E₂= 200 V

$\therefore$    $\frac{E_{1}}{E_{2}}=\frac{290}{200}=\frac{29}{20}=1.45$