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A new flashlight cell of e.m.f. 1.5 volt...

A new flashlight cell of e.m.f. 1.5 volts given a current of 15 amps. When connected directly to an ammeter of resistance `0.04 Omega`. The internal resistance of cell is

A

`0.06Omega`

B

`0.04Omega`

C

`0.10Omega`

D

`10Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
`E=l(R+r)`
`1.5=15(0.04+r)rArr(1.5)/(15)=0.04+r`
`rArr" "1/(10)=0.04+r`
`rArr" "r=0.1-0.04=0.06Omega`
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