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_1gtR_1).` if the potential difference across the source having internal resistance `R_2` is zero, then

Options

A. `R = (R_(2) xx (R_(1) + R_(2)))/((R_(1) - R_(1))`
B. `R = R_(1) - R_(2)`
C. `R = (R_(1) R_(2))/((R_(1) + R_(2)))`
D. `R = (R_(1) R_(2))/((R_(1) - R_(2)))`

Answer

B

Explanation

(b) `R_(eq) = R_(1) + R_(2) + R`
`:. I = (2 E)/(R_(1) + R_(2) + R)`
According to the question,
`- (V_(A) - V_(B)) = e - IR_(2)`
`0 = E - IR_(2)`
`E = IR_(2)`
`E = (2E)/(R_(1) + R_(2) + R) R_(2)`
`R_(1) + R_(2) + R = 2 R_(2)`
`:. R = R_(2) - R_(1)`