A uniform metallic wire having resistance $4\ \Omega$ is bent to form a square loop (ABCD) (see figure). A resistance of $2\ \Omega$ is connected between points B and D and a battery of $2$ V is connected across points A and C as shown in the figure. Now the value of current ($I$) is :

The total resistance of the uniform wire is $4\ \Omega$. Since it is bent to form a square loop ABCD, the resistance of each of its four sides is equal.

Resistance of each side $= \dfrac{4}{4} = 1\ \Omega$.
Thus, $R_{AB} = R_{BC} = R_{CD} = R_{DA} = 1\ \Omega$.

The given circuit can be redrawn as a Wheatstone bridge where the arms are AB, BC, AD, and DC, and the central arm is BD. The battery is connected across the opposite corners A and C.

Let's check the balance condition of the Wheatstone bridge:
$\dfrac{R_{AB}}{R_{AD}} = \dfrac{1}{1} = 1$
$\dfrac{R_{BC}}{R_{DC}} = \dfrac{1}{1} = 1$

Since $\dfrac{R_{AB}}{R_{AD}} = \dfrac{R_{BC}}{R_{DC}}$, the Wheatstone bridge is balanced. This means the potential at point B is equal to the potential at point D ($V_B = V_D$).

Therefore, no current will flow through the $2\ \Omega$ resistor connected between B and D, and it can be removed from the circuit for calculation.

Now, the circuit simplifies to two parallel branches across the battery:
1. Upper branch (A $\rightarrow$ B $\rightarrow$ C) with resistance $R_1 = R_{AB} + R_{BC} = 1 + 1 = 2\ \Omega$.
2. Lower branch (A $\rightarrow$ D $\rightarrow$ C) with resistance $R_2 = R_{AD} + R_{DC} = 1 + 1 = 2\ \Omega$.

The equivalent resistance $R_{eq}$ of the circuit is the parallel combination of $R_1$ and $R_2$:
$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{2} + \dfrac{1}{2} = 1\ \Omega^{-1}$
$\Rightarrow R_{eq} = 1\ \Omega$

The total current $I$ drawn from the battery is given by Ohm's law:
$I = \dfrac{V}{R_{eq}} = \dfrac{2\text{ V}}{1\ \Omega} = 2\text{ A}$.

Answer: $2$ A