The above circuit potential difference between point A and B $(V_A-V_B)=$

$\frac{2}{3}V$

The correct answer option (2) : $\frac{2}{3}V$

$R_{eff}=4+\frac{3\times 6}{9}=6Ω$

$I=\frac{V}{R_{eff}}$

$=\frac{12}{6}=2\, A$

K.V.L in loop

$12-4\times 2-l_1\times 3=0$

$l_1=\frac{4}{3}A$

K.V.L in loop ......... (1)

$12-4\times 2 - 6l_2=0$

$l_2=\frac{2}{3}A$

K.V. L in loop ........ (2)

$12-4\times 2-6l_2=0$

$l_2=\frac{2}{3}A$

$V_0-V_A=1\times \frac{4}{3}$

$V_0-V_A=\frac{4}{3}$ .......... (i)

$V_0-V_B=\frac{2}{3}\times 3 $

$V_0-V_b=2V$ ............ (ii)

(ii) - (i)

$V_A-V_B=2-\frac{4}{3}=\frac{2}{3}V$