Two cells of same emf E but internal resistance `r_(1) and r_(2)` are connected in series to an external resistor R(figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero?

Explanation

Applying Ohm's law
Effective resistance `=R+r_(1)+r_(2)` and effective emf of two cells `=E+E=2E`, so the electric current is given by
`I=(E+E)/(E+r_(1)+r_(2))`
The potential difference across the terminals of the first cell and putting it equal to zero.
`V_(1)=E-Ir_(1)=E-(2E)/(r_(1)+r_(2)+R)r_(1)=0`
or `E=(2Er_(1))/(r_(1)+r_(2)+R)implies1=(2r_(1))/(r_(1)+r_(2)+R)`
`r_(1)+r_(2)+R=2r_(1) impliesR=r_(1)-r_(2)`
this is the required relation.