Question

A piece of silver has a resistance of 1 Ω . What will be the resistance of a constantan wire of one-third length and one-half diameter, if the specific resistance of constantan is 30 times that of silver ?

Answer 1

Explanation:

For silver wire R1 = l1/A1; For manganin wire, R = ρ2(l2/A2)

∴ R2/R1 = ρ2/ρ1 x l2/l1 x A1/A2

Now A1 = πd12/4 and A2 = πd22/4

∴ A1/A2 = d12/d22

∴ R2/R1 = ρ2/ρ1 x l2/l1 x (d1/d2)2

R1 = 1Ω; l2/l1 = 1/3, (d1/d2)2 = (3/1)2 = 9; ρ2/ρ1 = 30 ∴ R2 = 1 x 30 x (1/3) x 9 = 90Ω

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