If the potential difference across the internal resistance `r_(1)` is equal to the emf E of the battery, then

Options

A. `R=r_(1)+r_(2)`
B. `R=r_(1)//r_(2)`
C. `R=r_(1)-r_(2)`
D. `R=r_(2)//r_(1)`

Answer

c

Explanation

From circuit, `V_(1)+V_(2)=iR`
`E-ir+E-ir_(2)=iR`   `Alos ir_(1)=E=r_(1)E(give)`
`ir_(1)-ir_(1)+ir_(1)-ir_(2)=RimpliesR=r_(1)-r_(2)E-ir, + E-ir2 = iR`. Also ir - E (given) . ir -ir + ir - irR R= -12
Alternative :`i=(2E)/(r_(1)+r_(2)+R)=(E)/(r_(1))`
`Asir_(1)=E(give)i=E//r_(1)`
`2r_(1)=r_(1)+r_(2)+RimpliesR=r_(1)-r_(2)`