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The balancing lengths of potentiometer wire are $l_{1} a n d l_{2}$ when two cells of emf $E_{1} a n d E_{2}$ are connected in the secondary circuit in series first to help each other and next to oppose each other $\frac{E_{1}}{E_{2}}$ is equal to $(E_{1} > E_{2})$ .

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

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

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

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

answer is C.

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Detailed Solution

Step 1: When $E_{1}$ and $E_{2}$ are connected to help each other

$K = \frac{E_{1} + E_{2}}{L_{1}} \Rightarrow E_{1} + E_{2} = K L_{1}$

Step 2: When $E_{1}$ and $E_{2}$ are connected to oppose each other

$K = \frac{E_{1} - E_{2}}{L_{2}} \Rightarrow E_{1} - E_{2} = K L_{2}$

$E_{1} = \frac{K (L_{1} + L_{2})}{2}$ , $E_{2} = \frac{K (L_{1} - L_{2})}{2}$

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

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