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When a resistor of 11 Omega is connected...

When a resistor of `11 Omega` is connected in series with an electric cell, the current flowing in it is `0.5 A`. Instead, when a resistor of `5 Omega` is connected to the same electric cell in series, the current increases by `0.4 A` The internal resistance of the cell is

A

`1.5 Omega`

B

`2 Omega`

C

`2.5 Omega`

D

`3.5 Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
`0.5(r + 11) = 0.9(r + 5)`
`rArr 5r + 55 = 9r + 45`
`rArr 4r =10 rArr r = 10/4 = 2.5 Omega`
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