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A cell of e.mf. E and internal resistanc...

A cell of e.mf. E and internal resistance r is connected in series with an external resistance nr. Then, the ratio of the terminal potential difference to E.M.F.is

A

`((1)/(n))`

B

`(1)/((n+1))`

C

`(n)/((n+1))`

D

`((n+1))/(n)`

Text Solution

Verified by Experts

The correct Answer is:
C
`E=1(n r +r)`
`=1r(n+1)`
But potential difference =1nr
`rArr ("Potential difference")/("Emf")=(n)/(n+1)`
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