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17/07/2021 4:46 pm
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A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.
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17/07/2021 4:53 pm
R = ρ\(\frac{I}{A}\)
Now,
I' = 2I and A' = \(\frac{A}{2}\)
ρ' = ρ(since the material of the wires is the same)
So, R' = ρ'\(\frac{I'}{A'}\)
= ρ\(\frac{2I}{\frac{A}{2}}\)
= 4ρ\(\frac{I}{A}\) = 4R
R' = 4 x 20 = 80 Ω