Question

Please answer to this question as soon as possible.​

Answer 1

Answer:

Rnet=(2π−θ)θR/4π²

Explanation:

Let  

R=ρLA

If a wire of length L is bend to form a loop, then radius of the loop:

r=L/2π

Length of the wire segment l=rθ

Resistance of the first wire segment:  

R1=ρrθ/A

Resistance of the second wire segment:  

R2=ρr(2π−θ)/A

These two segments are in parallel. Therefore net resistance:

1Rnet=Aρrθ+Aρr(2π−θ)

On solving:

Rnet=(2π−θ)θρr/2πA

Substitute r=L/2π on the above equation:

Rnet=(2π−θ)θR/4π²

Hope it helps!