When cell of E.M.F. ' E 1 ' is connected to potentiometer wire, the balancing length is l 1 . Another cell of E.M.F. ' E 2 ' ( E 1 E 2 ) is connected so that two cells

Question

When cell of E.M.F. ' E 1 ​ ' is connected to potentiometer wire, the balancing length is l 1 ​ . Another cell of E.M.F. ' E 2 ​ ' ( E 1 ​ > E 2 ​ ) is connected so that two cells oppose each other, then the balancing length is l 2 ​ . The ratio E 1 ​ : E 2 ​ is

TestHub · Accepted Answer

Let the potential gradient of the potentiometer wire be $k$ .

When the cell of EMF $E_{1}$ is connected, the balancing length is $l_{1}$ .

$E_{1} = k l_{1}$

When the cell of EMF $E_{2}$ is connected in opposition to $E_{1}$ , the net EMF is $E_{1} - E_{2}$ . The balancing length is $l_{2}$ .

$E_{1} - E_{2} = k l_{2}$

Dividing the first equation by the second equation:

$\frac{E _{1}}{E _{1} - E _{2}} = \frac{l _{1}}{l _{2}}$

$E_{1} l_{2} = E_{1} l_{1} - E_{2} l_{1}$

$E_{2} l_{1} = E_{1} (l_{1} - l_{2})$

$\frac{E _{1}}{E _{2}} = \frac{l _{1}}{l _{1} - l _{2}}$

Answer: $\frac{l _{1}}{l _{1} - l _{2}}$

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