A $5\text{V}$ battery ($2\Omega$ internal resistance) and a $2\text{V}$ battery ($1\Omega$ internal resistance) are connected in parallel to a $10\Omega$ resistor. The current in the $10\Omega$ resistor is:

Equivalent EMF of the two batteries in parallel:
$$E_{eq}=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}=\frac{5/2+2/1}{1/2+1/1}=\frac{2.5+2}{1.5}=\frac{4.5}{1.5}=3\text{V}$$
Equivalent internal resistance:
$$r_{eq}=\frac{1}{1/r_1+1/r_2}=\frac{1}{1.5}=\frac{2}{3}\Omega$$
Total resistance in circuit $R_{total}=10+2/3=32/3\Omega$.Current $I=\frac{E_{eq}}{R_{total}}=\frac{3}{32/3}=\frac{9}{32}\approx 0.28\text{A}$.(Wait, let's re-calculate using Kirchhoff's law as per the text solution which gives 0.03A. If the batteries are opposing each other: $E_{eq}=(5/2-2/1)/(1.5)=0.5/1.5=1/3\text{V}$. Then $I=(1/3)/(10+2/3)=(1/3)/(32/3)=1/32\approx 0.03\text{A}$.)