A uniform metallic wire having resistance $4 Ω$ is bent to form a square loop (ABCD) (see figure). A resistance of $2 Ω$ 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 :
NEET 2026 (Cancelled)

Question

TestHub · Accepted Answer

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

Resistance of each side $= \frac{4}{4} = 1 Ω$ .
Thus, $R_{A B} = R_{BC} = R_{C D} = R_{D A} = 1 Ω$ .

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:
$\frac{R _{A B}}{R _{A D}} = \frac{1}{1} = 1$
$\frac{R _{BC}}{R _{D C}} = \frac{1}{1} = 1$

Since $\frac{R _{A B}}{R _{A D}} = \frac{R _{BC}}{R _{D C}}$ , 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 $2 Ω$ 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 $\to$ B $\to$ C) with resistance $R_{1} = R_{A B} + R_{BC} = 1 + 1 = 2 Ω$ .
2. Lower branch (A $\to$ D $\to$ C) with resistance $R_{2} = R_{A D} + R_{D C} = 1 + 1 = 2 Ω$ .

The equivalent resistance $R_{e q}$ of the circuit is the parallel combination of $R_{1}$ and $R_{2}$ :
$\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{1}{2} + \frac{1}{2} = 1 Ω^{- 1}$
$\Rightarrow R_{e q} = 1 Ω$

The total current $I$ drawn from the battery is given by Ohm's law:
$I = \frac{V}{R _{e q}} = \frac{2 V}{1 Ω} = 2 A$ .

Answer: $2$ A

Physics - Current Electricity Question with Solution | TestHub

Options:

Doubts & Discussion