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

Equivalent internal resistance r = `r_(1) + r_(2)`
equivalent emf E. = E + E = 2E
`therefore ` Current flowing in the circuit,
I =`(E.)/(R + r) = (2E)/(R + r_(1) + r_(2))`
`rArr ` Potential difference,
V = E - Ir
`therefore ` p.d. across terminals of first cell,
`V_(1) = E - Ir_(1)`
From data given,
0 =E - `(2Er_(1))/(R + r_(1) + r_(2))`
`therefore E = (2 Er_(1))/(R + r_(1) + r_(2))`
`therefore 1 = (2r_(1))/(R + r_(1) + r_(2))`
`therefore R + r_(1) + r_(2) = 2 r_(1)`
`therefore R = r_(1) - r_(2)`
Thus value of resistance should be `r_(1) - r_(2)`.