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A galvanometer connected with an unknown...

A galvanometer connected with an unknown resistor and two identical cells in series each of emf 2 V shows a current of 1 A. If the cells are connected in parallel, it shows 0.8 A. Then the internal resistance of the cell is

A

`1 Omega`

B

`0.5 Omega`

C

`0.25 Omega`

D

`0.33 Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
Series,
`I = (2E)/(2r + R)`
`rArr 2r + R = 4`…….. (i)
Parallel,
`I = 2/(R + r//2)`

`rArr 0.8 = 2/(R + r//2)`
`rArr 0.8R + 0.4R = 2`
`R + r/2 = 2/0.8 = 5/2`…… (ii)
Subtract (i) from (ii)
`-(3r)/2 =-3/2 rArr r = 1 Omega`
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