Q. Two batteries of emf $varepsilon_{1}$ and $varepsilon_{2}left(varepsilon_{2}>varepsilon_{1}right)$i) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure

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Two batteries of emf $varepsilon_{1}$ and $varepsilon_{2}left(varepsilon_{2}>varepsilon_{1}right)$i) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure

Solution:

A. The equivalent emf $varepsilon_{ eq }$ of the two cells is between $varepsilon_{1}$ and $varepsilon_{2}$,

i.e., $varepsilon_{1}

The cells of emf $varepsilon_{1}$ and $varepsilon_{2}$ with internal resistance $r_{1}$ and $r_{2}$ respectively.

In parallel combination,

the relation between equivalent emf and equivalent resistance is,

$frac{varepsilon_{ eq }}{ r _{ eq }}=frac{varepsilon_{1}}{ r _{1}}+frac{varepsilon_{2}}{ r _{2}}$

The resistance in parallel combination is,

$r _{ eq }=frac{ r _{1} cdot r _{2}}{ r _{1}+ r _{2}}$

From the above equation,

$varepsilon_{ eq }=frac{varepsilon_{1} r _{2}+varepsilon_{2} r _{1}}{ r _{1}+ r _{2}}$

In question,

$varepsilon_{2}>varepsilon_{1}$

Then The equivalent emf $varepsilon_{ eq }$ of two cell is between $varepsilon_{1}$ and $varepsilon_{2}$.