for loop $AFDBEA:$ $E_{F}-E_{E}=i_{1}\left(r_{F}+r_{E}+R\right)-R i_{2}$

or $1-2=i_{1}(1+2+2)-2 i_{2} \Rightarrow 5 i_{1}-2 i_{2}=-1 \ldots(1)$

for loop $CGDBHC:$ $E_{H}-E_{G}=i_{2}\left(r_{H}+r_{G}+R\right)-R i_{1}$

or $1-3=i_{2}(1+3+2)-2 i_{1} \Rightarrow-2 i_{1}+6 i_{2}=-2 \ldots(2)$

$(1)$ $\times 3 \Rightarrow 15 i_{1}-6 i_{2}=-3 \ldots(3)$

$(2)+(3), 13 i_{1}=-5 \Rightarrow i_{1}=-\frac{5}{13} A$

putting the value of $I_{1}$ in $(1),$ we get $I_{2}=-\frac{6}{13}$

$V_{D}-V_{B}=$ potential drop across $R=2 \Omega$ is $R\left(i_{2}-i_{1}\right)=2\left[-\frac{6}{13}+\frac{5}{13}\right]=-\frac{2}{13} V$

Thus, $V_{G}=E_{G}+i_{2} r_{G}=3+3\left(-\frac{6}{13}\right)=\frac{21}{13} V$ (as current

drawn from cell and $i_{2}$ are opposite direction)

and $V_{H}=E_{H}-i_{2} r_{H}=1-\left(-\frac{6}{13}\right) 1=1+\frac{6}{13}=\frac{19}{13} V$ (as

current drawn from cell and $i_{2}$ are same direction.