A 5V battery with internal resistance $2\Omega$  and  2V battery with internal resistance  $1\Omega$ are connected to a  $10\Omega$ resistor as shown in figure. The current in $10\Omega$  resistor, is

Applying Kirchhoff's second law for closed loop $A{P}_2{P}_{1}DA,$ we get $10I_1+2I-5=02I+10I_1=5\mathrm{......}\left(i\right)$

Applying Kirchhoff's second law for closed loop $P_{2}BC{P}_1{P}_2,$ we get $-2+1(I-I_1)-10I_1=0$

$I-11I_1=2\mathrm{.........}\left(ii\right) or 2I-22I_1=4\mathrm{.........}\left(iii\right)$

Subtracting eq. (iii) from eq. (i), we get  $32I_1=1or{I}_1=\frac{1}{32}A\approx 0.03A$ from $P_2 to P_1$