Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are` R_(1) and R_(2) (R_(1) gt R_(1))`. If the potential difference across the source having internal resistance `R_(2)` is zero, then

Explanation

Let, us consider the current `i`, flowing through the circuit, as shown in the figure. Appling loop rule for the circuit, `Delta varphi = 0`
`2 xi + iR_(1) + iR_(2) + iR = 0`
or, `i(R_(1) + R_(2) + R) = 2 xi`
or, `i = (2xi)/(R + R_(1) + R_(2))`
Now, if `varphi_(1) - varpi_(2) = 0`
`xi + iR_(1) = 0`
or. `(2 xi R_(1))/(R + R_(1) + R_(2)) = xi` and `2R_(1) = R_(2) + R + R_(1)`
Thus, `varphi_(2) - varphi_(3) = -xi + iR_(2) = 0`
or, `(2 xi R_(2))/(R + R_(1) + R_(2)) = xi`
So, `R = R_(2) - R_(1)`, which is the required resistance.