Let the emf of each source be E. When they are connected in series, the current in the circuit I = RωtEωt = r1+r2+RE+E = r1+r2+R2E ∴ potential drop across the cell of internal potential drop across the cell of internal r2 , (r1+r2+R2E)r2 Hence, E - r1+r2+R2Er2 = 0 ⇒ r1+r2+R = 2r2 ⇒ R = r2−r1