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 :

(A) 2 A
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 \(=\frac{4}{4}=1 \Omega\).
Thus, \(R_{A B}=R_{B C}=R_{C D}=R_{D A}=1 \Omega\).
The given circuit can be redrawn as a Wheatstone bridge where the arms are \(A B, B C, A D\), and \(D C\), 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:
\(\frac{R_{A B}}{R_{A D}}=\frac{1}{1}=1\)
\(\frac{R_{B C}}{R_{D C}}=\frac{1}{1}=1\)
Since \(\frac{R_{A B}}{R_{A D}}=\frac{R_{B C}}{R_{D C}}\), the Wheatstone bridge is balanced. This means the potential at point B is equal to the potential at point \(D \left(V_B=V_D\right)\).
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 \(R_1=R_{A B}+R_{B C}=1+1=2 \Omega\).
2. Lower branch ( \(A \rightarrow D \rightarrow C\) ) with resistance \(R_2=R_{A D}+R_{D C}=1+1=2 \Omega\).
The equivalent resistance \(R_{e q}\) of the circuit is the parallel combination of \(R_1\) and \(R_2\) :
\(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2}+\frac{1}{2}=1 \Omega^{-1}\)
\(\Rightarrow R_{e q}=1 \Omega\)
The total current \(I\) drawn from the battery is given by Ohm's law:
\(I=\frac{V}{R_{e q}}=\frac{2 V}{1 \Omega}=2 A\)