The current I flowing in the circuit is:

The correct answer is Option (2) → 2.4 A

Redraw the circuit with its nodes. The top wire joins the two upper corners, so it is one node, T. The left-hand lower point is the node X (the diagonal from there to the top-right corner starts on the left side), and the right-hand lower point is the node Y. The battery is connected between X and Y. The two diagonals cross without touching, so they are not joined at the centre.

Between X and T there are the 4 Ω resistor and the 12 Ω resistor (on the diagonal) in parallel. Between T and Y there are the 3 Ω resistor and the 6 Ω resistor (on the other diagonal) in parallel. These two groups are in series across the battery.

    \( \displaystyle R_{XT} = \frac{4\times12}{4+12}\,\Omega = 3\,\Omega \)

    \( \displaystyle R_{TY} = \frac{3\times6}{3+6}\,\Omega = 2\,\Omega \)

    \( R = R_{XT} + R_{TY} = 5\,\Omega \)

    \( \displaystyle I = \frac{V}{R} = \frac{12\,\mathrm{V}}{5\,\Omega} = 2.4\,\mathrm{A} \)