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Electric Bulbs

The balancing...

Question

The balancing lengths of potentiometer wire are $l_{1}$ and $l_{2}$ when two cells of emf $E_{1}$ and $E_{2}$ are connected in the secondary circuit in series. first to support each other and next to oppose each other. Then the ratio of $\frac{E_{1}}{E_{2}}$ is equal to [if $E_{1} > E_{2}$ ]

A

\frac{l_{1} - l_{2}}{l_{1} + l_{2}}

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B

\frac{l_{1} + l_{2}}{l_{1} - l_{2}}

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C

\frac{l_{1}}{l_{2}}

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D

\frac{l_{2}}{l_{1}}

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Solution

The correct option is B $\frac{l_{1} + l_{2}}{l_{1} - l_{2}}$
For support each other :

$(E_{1} + E_{2}) \propto l_{1} \Rightarrow E_{1} + E_{2} = x l_{1} . . . . . . (1)$

For Oppose each other :

$(E_{1} - E_{2}) \propto l_{2} \Rightarrow E_{1} - E_{2} = x l_{2} . . . . . (2)$

On dividing (1) and (2),

$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{l_{1}}{l_{2}}$

By componendo and dividendo method.

$\frac{(E_{1} + E_{2}) + (E_{1} - E_{2})}{(E_{1} + E_{2}) - (E_{1} - E_{2})} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}$

$\frac{2 E_{1}}{2 E_{2}} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}$

$\frac{E_{1}}{E_{2}} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}$

 Hence, $(B)$ is the correct answer.

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