The emf of a cell is `Ev`, and its its internal resistance is `1Omega`. A resistance of `4Omega` is joined to battery in parallel. This is connected in secondary circuit of potentiometer. The balancing length is 160 cm. If 1 V cell balances for 100 cm of potentiometer wire, the emf of cell E is
A. 1 V
B. 3 V
C. 2 V
D. 4 V


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Correct Answer - C
`r=R[(l_(1)-l_(2))/(l_(2))]`

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