The electric resisitance of a certain wire of iron is R. If its length and radius are both doubled. Then what will happen to the equivalent resistance?
Answers
Answered by
2
hello friend...!!
we know that R = Φ L 1 / A 1 .
where Φ represents the resistivity, L1 = length, A1 = area. ( π r^2 )
according to the given question, when area is doubled implies the new area will be
A2 = π (2r)^2 = 4πr^2.
similarly length is also doubled implies
L2 = 2L1.
therefore to calculate the new resistance,
R1/R2 = ΦL1/A1 x A2/ΦL2.
implies, R1/R2 = ΦL1/πr^2 x 4πr^2 /2ΦL1.
implies, R1 = 2 R2
therefore , the new resistance R2 = R1/2 .
implies that the new resistance obtained is reduced by half.
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hope it is useful..!!
we know that R = Φ L 1 / A 1 .
where Φ represents the resistivity, L1 = length, A1 = area. ( π r^2 )
according to the given question, when area is doubled implies the new area will be
A2 = π (2r)^2 = 4πr^2.
similarly length is also doubled implies
L2 = 2L1.
therefore to calculate the new resistance,
R1/R2 = ΦL1/A1 x A2/ΦL2.
implies, R1/R2 = ΦL1/πr^2 x 4πr^2 /2ΦL1.
implies, R1 = 2 R2
therefore , the new resistance R2 = R1/2 .
implies that the new resistance obtained is reduced by half.
---------------------------------------------
hope it is useful..!!
Answered by
0
The resistance will be halved and the specific resistance will remain unchanged
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