Find the value of current in the circuit.

The correct answer is Option (4) → 1 A

The five \( 2\,\Omega \) resistors form a Wheatstone bridge between the two terminals of the cell. From the left terminal one resistor goes to the first inner junction and one (the lower resistor) goes to the second inner junction; from these junctions one resistor each (the upper resistor and the right one) goes to the right terminal. The \( 2\,\Omega \) resistor on the main line between the two inner junctions is the middle arm.

All four arms are equal, so the bridge is balanced:

    \( \displaystyle \frac{2\,\Omega}{2\,\Omega} = \frac{2\,\Omega}{2\,\Omega} \)

The two inner junctions are at the same potential, no current flows in the middle arm, and it can be removed. Two branches of \( 2 + 2 = 4\,\Omega \) are then in parallel:

    \( \displaystyle R_{\mathrm{eq}} = \frac{4 \times 4}{4 + 4} = 2\,\Omega \)

    \( \displaystyle i = \frac{V}{R_{\mathrm{eq}}} = \frac{2\,\mathrm{V}}{2\,\Omega} = 1\,\mathrm{A} \)