Potential difference between $C$ and $A=V_{C}-V_{A}$

$=\frac{V}{8} \times 4=\frac{V}{2}\, \mathrm{volt}$

Current through $C B D$,   $I^{\prime}=\frac{V}{4}\, amp$

Potential difference between $C$ and $B=V_{C}-V_{B}$

$=\frac{V}{4} \times 1=\frac{V}{4} \,\text { volt. }$

Potential between $A$ and $B=V_{A}-V_{B}$

$\therefore \quad V_{A}-V_{B}=V_{C}-V_{B}-\left(V_{C}-V_{A}\right)=\frac{V}{4}-\frac{V}{2}=\frac{V}{4}$

$ \Rightarrow {V_A} - {V_B} < 0\quad or,\quad {V_A} < {V_B}$

As ${V_A} < {V_B}$, so direction of current will be from $B$ to $A$.