The balancing lengths of potentiometer wire are 800cm and 600cm when two cells of emf’s E1 and E2 are connected in the secondary circuit first in series and then terminals of one cell is reversed; E1E2 is equal to ______

ℓ1ℓ2=E1+E2E1-E2⇒E1E2=ℓ1+ℓ2ℓ1−ℓ2=800+600800−600=1400200=71

ℓ1ℓ2=E1+E2E1-E2⇒E1E2=ℓ1+ℓ2ℓ1−ℓ2=800+600800−600=1400200=71

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The balancing lengths of potentiometer wire are 800cm and 600cm when two cells of emf’s E₁ and E₂ are connected in the secondary circuit first in series and then terminals of one cell is reversed; $\frac{E_{1}}{E_{2}}$ is equal to ______

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$\frac{14}{11}$

$\frac{4}{3}$

$\frac{7}{1}$

$\frac{1}{11}$

answer is C.

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

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{ℓ_{1} + ℓ_{2}}{ℓ_{1} - ℓ_{2}} = \frac{800 + 600}{800 - 600} = \frac{1400}{200} = \frac{7}{1}$

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