Two cells of voltage $10 \ V$ and $2 \ V$ and internal resistances $10 \ \Omega$ and $5 \ \Omega$ respectively,are connected in parallel with the positive end of $10 \ V$ battery connected to the negative pole of $2 \ V$ battery (figure). Find the effective voltage and effective resistance of the combination.

(N/A) By using Kirchhoff's first law at point $A$,$I_{1} = I + I_{2}$ ... $(1)$
By using Kirchhoff's second law for loop $EDCFE$,$-IR - 10I_{1} + 10 = 0$
$\therefore IR + 10I_{1} = 10$ ... $(2)$
For loop $ADCBA$,$-IR + 5I_{2} - 2 = 0$
$-IR + 5(I_{1} - I) = 2$
$\therefore -IR + 5I_{1} - 5I = 2$
$\therefore -2IR + 10I_{1} - 10I = 4$ ... $(3)$
Subtracting equation $(3)$ from equation $(2)$:
$(IR + 10I_{1}) - (-2IR + 10I_{1} - 10I) = 10 - 4$
$3IR + 10I = 6$
$I(R + 10/3) = 2$ ... $(4)$
From Ohm's law,$I(R + R_{eq}) = V_{eq}$ ... $(5)$
Comparing equation $(4)$ and $(5)$,we get:
$R_{eq} = 10/3 \ \Omega$ and $V_{eq} = 2 \ V$.