Given that the potential at \( A \) and \( B \) is identical, the 30 \( \Omega \) resistor in the middle experiences no current flow.
The two parallel branches are configured as follows:
\[
R_1 = 10 \, \Omega + 20 \, \Omega = 30 \, \Omega
\]
\[
R_2 = 40 \, \Omega
\]
The equivalent resistance is calculated as:
\[
R_{{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(30)(40)}{30 + 40} = \frac{1200}{70} = 17.14 \, \Omega
\]
Applying Ohm's law, the total current is:
\[
I = \frac{40V}{17.14} = 2.33 \, A
\]