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The emf of a daniel cell is 1.08 V. When...

The emf of a daniel cell is 1.08 V. When the terminals of the cells are connected to a ressitance of `3Omega`, the potential difference across the terminals is found to be 0.6 V. Then the internal resistance of the cell is

A

`1.8Omega`

B

`2.4Omega`

C

`3.24Omega`

D

`0.2Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`r=((E-V)/(V))R`
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