NEET 2026 — Physics solution

NEET2026Physics

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 :

A
4 A
B
4.5 A
C
8 A
D
2 A

Solution

Each side of the square has resistance $4/4 = 1 Ω$ . The two paths from A to C (A→B→C and A→D→C) each have resistance $1 + 1 = 2 Ω$ .

The 2 $Ω$ resistor between B and D acts as the galvanometer arm of a Wheatstone bridge. By symmetry the bridge is balanced and no current flows through this arm, so it can be ignored.

Net resistance: two 2 $Ω$ paths in parallel give $R_{n e t} = \frac{2 \times 2}{2 + 2} = 1 Ω$ .

$I = \frac{E}{R _{n e t}} = \frac{2}{1} = 2$ A.

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