A wire is bent in the form of a circle of radius 2 m. The resistance per unit length of the wire is 1/π Ω m⁻¹. A battery of 6 V is connected between points A and B of the circle, where ∠ AOB = 90° and O is the centre. Find the current through the battery.

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(1) 8 A. Given: radius 2 m, resistance per unit length 1/π Ω m⁻¹, ∠ AOB = 90°, battery 6 V. Idea: A and B split the ring into two arcs, and the two arcs are in parallel across the battery. Circumference = 2π×2 = 4π m, so the whole ring has 4π×1/π = 4 Ω. The 90° arc is a quarter of the ring, 1 Ω; the other arc is 3 Ω. R_AB = (1×3)/(1 + 3) = 0.75 Ω, so I = 6/(0.75) = 8 A. So the current through the battery is 8 A.

A wire is bent in the form of a circle of radius 2 m. The resistance per unit length of the wire is 1/π Ω m⁻¹. A battery of 6 V is connected between points A and B of the circle, where ∠ AOB = 90° and O is the centre. Find the current through the battery.

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