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

Applying kirchhoff's junction rule, `I_(1)=I+I_(2)`
Applying kirchhoff's II law/loop rule applied in outer loop containing 10 V cell and resistance R, we have
`10=IR+10I_(1)`
Applying kirchhoff' II law/ loop rule applied in outer loop containing 2 V cell and resistance R, we have
`2=5I_(2)-RI=5(I_(1)-I)-RI`
or `4=10I_(1)-10I-2RI`
Solving Eqs. (i) and (ii), gives
`6=3RI+10I`
`2=I(R+(10)/(3))`
Also, the external resistance is R. the Ohm's law states that
`V=(R+R_(eff))`
On comparing, we have `V=2V` and effective internal resistance `(R_(eff))=((10)/(3))Omega`
Since, the effective internal resistance `(R_(eff))` of two cells is `((10)/(3))Omega`, being the parallel combination of `5Omega and 10Omega`. The equivalent circuit is given below.