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Ten identical batteries each of emf 2 V ...

Ten identical batteries each of emf 2 V are connected in series to a `8 Omega` resistor. If the current in the circuit is 2 A, then the internal resistance of each battery is

A

`0.2 Omega`

B

`0.3 Omega`

C

`0.4 Omega`

D

`0.5 Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
For this circuit,
`I= (E_("eff"))/( R_("eff"))`
So, `2 = (20)/( 8+ 10r)`
`16+ 20 r = 20`
`20 r = 20 - 16`
`r = (4)/(20)`
`r = (1)/(5) = 0.2 Omega`
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