Question

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.

A. two equivalent emf `epsi_(eq)` of the two cells is between `epsi_(1)` and `epsi_(2)`, i.e., `epsi_(1)ltepsi_(eq)ltepsi_(2)`
B. the equivalent emf `epsi_(eq)` is smaller than `epsi_(1)`
C. The `epsi_(eq)` is given by `epsi_(eq)=epsi_(1)+epsi_(2)` always
D. `epsi_(eq)` is independent of internal resistance `r_(1)` and `r_(2)`

Answer 1

The equivalent emf of this combination is given by
`epsi_(eq)=(epsi_(2)r_(1)+epsi_(1)r_(2))/(r_(1)+r_(2))`
This suggest that the equivalent emf `epsi_(eq)` of thhe two cells is given by
`epsi_(1)ltepsi_(eq)ltepsi_(2)`