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Find the e.m.f(E(0)) end internal resis...

Find the `e.m.f(E_(0)) ` end internal resistance `(r_(0))` of a simple battery which is equivalent to a perallel combination of two battery of `e.m.fE_(1) and E_(2)` and internal resistance `v_(1) and v_(2)` respectively with polarities

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Let us connect the resistance `R` between points A and B
According to kirhhoff's first law at junction A
`I = I_(1) - I_(2)` …(i)
Appliying kiichoff's second law to closed circuit `BE_(2)ARB`
`- I_(2)v_(2) +IR = - E_(2)`
or `I R = - E_(2) + I_(2)v_(2)`...(iii)
from (i) and (ii)
`IR= -E_(2) + (I_(1) - I)r_(2)`
`I(R+ v_(2)) = -E_(2)+I_(2)r_(2)`
multipling (i) by `r_(2)` and (iv) by `r_(1)` end adding we get `IRr_(2) + 1(IR + r_(2))r_(1) = E_(1)r_(2) - E_(2)r_(1))`
or `I= (E_(1) r_(2) -E_(2)r_(1))/(R(r_(1) + r_(2)) + r_(1)r_(2)) = E_(1)r_(2) - E_(2r_(2))I(r_(1) + r_(2))/(R + (r_(1)r_(1))/(r_(1) + r_(2))) = (E_(0))/(R + r_(0))`
where `E_(0) = (E_(1) r_(2) - E_(2)r_(1))/((r_(1) + r_(2)) and r_(0) = (r_(1)r_(2))/(r_(1) + r_(2))`
Here `E_(0) = E.M.F` of bettery required
and `r_(0) =` initial resitence of battery
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