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

The emf of a daniel cell is `1.08`V. When the terminals of the cell are connected to resistance 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

`0.2Omega`

D

`0.24Omega`

Text Solution

Verified by Experts

The correct Answer is:
B

Shunt resistance ,
`S = (i_gG)/(i-i_g)= (0.01)(500)/(5-0.01) = 0.1Omega`
therefore , `0.1Omega` resistor is connected in parallel .
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