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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

`1Omega`

B

`0.5Omega`

C

`0.25Omega`

D

`0.33Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
In series, the current
`i=(nE)/(R+nr)`
`1+(2xx2)/(R+2r)`
`R+2r=4….(i)`
Again, in parallel the current
`i=(E)/(R+(r )/(n))`
`0.8=(2)/(R+(r )/(2))`
`0.8=(2xx2)/(2R+r)`
From Eq. (i)
`0.8=(4)/(2(4-2r)+r)`
`0.8=(4)/(8-3r)`
`8-3r=5`
`r=1Omega`
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