A uniform metallic wire having resistance 4\ is bent to form a square loop (ABCD) (see figure). A resistance of 2\ 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\ . Since it is bent to form a square loop ABCD, the resistance of each of its four sides is equal. Resistance of each side = 4 4 = 1\ . Thus, R_ AB = R_ BC = R_ CD = R_ DA = 1\ . 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: R_ AB R_ AD = 1 1 = 1 R_ BC R_ DC = 1 1 = 1 Since R_ AB R_ AD = 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\ 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 B C) with resistance R_1 = R_ AB + R_ BC = 1 + 1 = 2\ . 2. Lower branch (A D C) with resistance R_2 = R_ AD + R_ DC = 1 + 1 = 2\ . The equivalent resistance R_ eq of the circuit is the parallel combination of R_1 and R_2: 1 R_ eq = 1 R_1 + 1 R_2 = 1 2 + 1 2 = 1\ ^ -1 R_ eq = 1\ The total current I drawn from the battery is given by Ohm's law: I = V R_ eq = 2 V 1\ = 2 A . Answer: 2 A
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