Two cells of voltage `10V`and `2 V` and internal resistance `10 Omega` and `5Omega` respectively , are connected in parallel with the pesition and of `10V` battery connected to negative pole of `2V` battery Find the effected voltage and effected resistance of the combination

Explanation

According to kirchboff's first rule , at junction A
`I_(1) = l + l_(2)` or `l = l_(1) - l_(2)` ….(i)
According to kirchboff's second , rule in closed loop `ABCDA`: we have
`IR - I_(2) xx 5 = -2` or `I_(2) = (IR + 2)/(5)`….(ii)
in closed by `BCDEFAB, IR + I_(1) xx 10 = 10 or I_(1) = (10 - IR)/(10)` ....(iii)
putting value in (i) we get `I = (10 - IR)/(10) - (10 + IR)/(5) = (10 - IR 2IR - 4)/(10) = (6 - 3IR)/(10)`
or `10 I = 6 - 3IR` or `I(10+3R) = 6` or `I = (6)/(10 +3R) = ((6//3)/((10//3) +R)`....(iv)
if the `epsilon_("eff")` and `r_("eff")` are the effective voltage and effective internal of the combination , then from
`I = epsilon_("eff"))/(r_("eff") + R)` ...(v)
comparing (iv) and (v) we get , `epsilon_("eff") = 6//3 = 2V : r_("eff") = 10//3 Omega`