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 `epsilon_(eq)` of the two cells is between `epsilon_(1)` and `epsilon_(2), i.e., epsilon_(1)ltepsilon_(eq)ltepsilon_(2)`
B. The quivalent 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)` is independent of internal resistances `r_(1)` and `r_(2)`

Answer

A

Explanation

Refer Fig. 2(EP).5 the equivalent internal resistance of two cells between A and B is
`(1)/(r_(eq))=(1)/(r_(1))+(1)/(r_(2))=(r_(1)+r_(2))/(r_(1)r_(2))` or `r_(eq)=(r_(1)r_(2))/(r_(1)+r_(2))` ....(i)
If `epsilon_(eq)` is the equivalent emf of the two cells in parallel between A and B, then
`(espilon_(eq))/(r_(eq))=(epsilon_(1))/(r_(1))+(epsilon_(2))/(r_(2))=(epsilon_(1)r_(2)+epsilon_(2)r_(1))/(r_(1)r_(2))` [Refel Art. 2(a),26(b)]
`therefore epsilon_(eq)=(epsilon_(1)r_(2)+epsilon_(2)r_(1))/(r_(1)r_(2))xxr_(eq)=((epsilon_(1)r_(2)+epsilon_(2)r_(1))/(r_(1)r_(2))xx(r_(1)r_(2))/((r_(1)+r_(2))=(epsilon_(1)r_(2)+epsilon_(2)+epsilon_(2)r_(1))/((r_(1)+r_(2))`
This shows that whatever may be the values of `r_(1)` and `r_(2)`, the value of `epsilon_(eq)` is in between `epsilon_(1)` and `epsilon_(2)`. As `epsilon_(2)gtepsilon_(1)`, so `epsilon_(1)ltepsilon_(eq)ltepsilon_(2)`.