Two batteries of emf `epsilon_(1)` and `epsilon_(2) (epsilon_(2)gtepsilon_(1)` and internal resistances `r_(1)` and `r_(2)` respectively are connected in parallel as shown in Fig. 2 (EP).1.

Options

A. The equivalent emf `e_(eq)` of the two cells is between `epsilon_(1)` i.e.`epsilon_1 lt epsilon_(eq) lt epsilon_2`. .
B. The equivalent emf `epsilon_(eq)` is smaller than `epsilon_1` .
C. The `epsilon_(eq)` is given by `epsilon_(eq) = epsilon_(1) + epsilon_(2)` always.
D. `epsilon_(eq)` independent of internal resistances `r_1 and r_2` .

Answer

A

Explanation

`sum_(eq) = ((epsilon_1//r_1 + epsilon_2//r_2)/(1//r_1 + 1//r_2))`.