Answer:

answer given below

Explanation:

The simplified circuit is attached here. We can analyze the circuit by making it two branches left and right.

The net resistance in the left branch

[tex]R_L=R_1+R_5||(R_3+R_4)=5\Omega+6\Omega||6\Omega\\{} \qquad\qquad\qquad\qquad \qquad \quad= 5\Omega+\frac{6\times 6}{6+6} \Omega\\{} \qquad\qquad\qquad\qquad \qquad \quad= 5\Omega+3\Omega\\{} \qquad\qquad\qquad\qquad \qquad \quad= 8\Omega\\and\\[/tex]

The net resistance in t he right branch is

[tex]R_R=R2+R_6||(R_7+R_8)=10\Omega+(6\Omega||6Omega)=10\Omega+3\Omega=13\Omega[/tex]

So the current in the left branch is

[tex]I_L=\frac{V}{R_L} =\frac{10}{8} A=\frac{5}{4} A[/tex]

The current in the right branch

[tex]I_R=\frac{V}{R_R} =\frac{10}{13} A[/tex]

We can see that the current [tex]I_L[/tex] divides into two parts. One of which flow through R5 and the other flows through R3 and R4.

The part of current flowing through the point A is

[tex]I_A=\frac{I_LR_5}{R5+R3+R_4} =\frac{(5/4)\times 6}{6+6} A=\frac{5}{8} A[/tex]

In finding this the current divider rule is used. Even with out making use of this, we can see that this current is half the incoming current, since it has two flow through two parallel 6 ohm resistors.

Similarly the current flowing through the point B is

[tex]I_B=\frac{10}{26} A[/tex]

due to the same explanation(  the incoming current has to flow through two 6 ohm resistors in parallel).

So,

The potential at the point A , [tex]V_A=I_A R_4=\frac{5}{8} \times 3=\frac{15}{8} \,V[/tex]

The potential at the poinbt B , [tex]V_B=I_B R_B=\frac{10}{26} \times 4=\frac{40}{26} \,V[/tex]

So the potential difference [tex]V_{AB}[/tex] is

[tex]V_{AB}=\frac{15}{8} V-\frac{40}{26} V=0.3365 \,V[/tex]