Tow cells with the same emf E and different internal resistances `r_(1)` and `r_(2)` are connected in series to an external resistances R . The value of R so that the potential difference across the first cell be zero is

Options

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

Answer

C

Explanation

There are two batteries with emf E each and the internal resistances `r_(1)` and `r_(2)` respectively.
Hence, we have `l(R+r_(1)+r_(2))=2E`,
Thus, `l = (2E)/(R+r_(1)+r_(2))`

Now, the potential difference across the first cell would be equal to `V = E - lr_(1)`. From the question, V = 0
Hence, `E=lr_(1)=(2Er_(1))/(R+r_(1)+r_(2))`,
Thus, `R+r_(1)+r_(2)=2r_(1)`,
Hence, `R=r_(1)-r_(2)`.