Two cells having unknown e.m.f.s E_{1} and E_ | vedclass

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(D) In a potentiometer circuit,when two cells are connected to assist each other,the balancing length $\ell_{1}$ is proportional to $(E_{1} + E_{2})$.

Vedclass · Accepted Answer

$1.45$

Vedclass · Answer

(D) In a potentiometer circuit,when two cells are connected to assist each other,the balancing length $\ell_{1}$ is proportional to $(E_{1} + E_{2})$. Thus,$E_{1} + E_{2} = k \ell_{1}$,where $k$ is the potential gradient.When they are connected to oppose each other,the balancing length $\ell_{2}$ is proportional to $(E_{1} - E_{2})$. Thus,$E_{1} - E_{2} = k \ell_{2}$.Dividing the two equations,we get: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{\ell_{1}}{\ell_{2}}$.Using the given values $\ell_{1} = 490 \ cm$ and $\ell_{2} = 90 \ cm$:$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{490}{90} = \frac{49}{9}$.Applying componendo and dividendo:$\frac{E_{1}}{E_{2}} = \frac{49 + 9}{49 - 9} = \frac{58}{40} = 1.45$.

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