Given, circuit diagram is shown in the figure, If $I_1$ and $I_2$ be the loop currents in loop (1) and loop (2), respectively then


By $K V L$ in loop (1),
$$
\begin{aligned}
& -5+10 I_1+10\left(I_1-I_2\right)=0 \\
& 20 I_1-10 I_2=5
\end{aligned}
$$

By $K V L$ in loop (2),
$$
\begin{aligned}
& 10 I_2+3+10\left(I_2-I_1\right)=0 \\
& -10 I_1+20 I_2=-3 \\
& -20 I_1+40 I_2=-6
\end{aligned}
$$

Solving Eqs. (i) and (ii), we get
$$
\begin{aligned}
& 30 I_2=-1 \\
& I_2=\frac{-1}{30} \mathrm{~A}
\end{aligned}
$$

From Eq. (i), $20 I_1-10\left(\frac{-1}{30}\right)=5 \Rightarrow I_1=\frac{7}{30} \mathrm{~A}$
$\therefore$ Current in branch $A B, I_{A B}=I_1-I_2=\frac{7}{30}-\left(\frac{-1}{30}\right)$
$$
I_{A B}=\frac{8}{30} \mathrm{~A}
$$
$\therefore$ Voltage across a terminal $A B$,
$$
V_{A B}=I_{A B} \times R_{A B}=\frac{8}{30} \times 10=\frac{8}{3} \mathrm{~V}
$$