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(D) The total resistance of the wire is $4 \ \Omega$. Since it is bent into a square, the resistance of each side is $4 \ \Omega / 4 = 1 \ \Omega$.The

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(D) The total resistance of the wire is $4 \ \Omega$. Since it is bent into a square, the resistance of each side is $4 \ \Omega / 4 = 1 \ \Omega$.The circuit consists of two parallel branches: branch $ABC$ and branch $ADC$, connected across the battery terminals $A$ and $C$.Branch $ABC$ consists of two $1 \ \Omega$ resistors in series, so its total resistance is $1 \ \Omega + 1 \ \Omega = 2 \ \Omega$.Branch $ADC$ also consists of two $1 \ \Omega$ resistors in series, so its total resistance is $1 \ \Omega + 1 \ \Omega = 2 \ \Omega$.$A$ resistor of $2 \ \Omega$ is connected between points $B$ and $D$. However, due to the symmetry of the circuit, the potential at point $B$ is equal to the potential at point $D$ $(V_B = V_D)$.Since there is no potential difference across the $2 \ \Omega$ resistor, no current flows through it. Thus, it can be ignored in the calculation of the equivalent resistance.The equivalent resistance $R_{eq}$ of the two parallel branches of $2 \ \Omega$ each is given by:$1/R_{eq} = 1/2 + 1/2 = 1 \ \Omega^{-1} \implies R_{eq} = 1 \ \Omega$.The total current $I$ drawn from the $2 \ V$ battery is:$I = V / R_{eq} = 2 \ V / 1 \ \Omega = 2 \ A$.

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