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Two cells of emfs $E_{1}$ and $E_{2}$ are connected as shown in figure $(E_{1} > E_{2})$ , when a potentiometer is connected between A and B, balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio $E_{1} / E_{2}$ is

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detailed solution

Correct option is C

E1E1−E2=300100⇒2E1=3E2⇒E1E2=32

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Two cells of emfs E1 and E2 are connected as shown in figure (E1>E2), when a potentiometer is connected between A and B, balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio E1/E2 is

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Ready to Test Your Skills?

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