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A series combination of two resistors `1 Omega` each is connected to a 12 V battery of internal resistance `0.4 Omega`. The current flowing through it will be

A

`12A`

B

`6A`

C

`5A`

D

`3.2A`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=(E)/(R_(1)+R_(2)+r)`
`I=(12)/(1+1+0.4)=(12)/(2.4)=5A`
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