A 4ohm wire is doubled on it . calculate the new resistance.
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As we know that , Resistance is directly proportional to length and inversely proportional to the area of cross section.
The wire is doubled on it, that means the area is doubled and the length gets halved.
R1/R2 =[ (L1)×(A2)]/[(L2)×(A1)]
R1 = 4 ohm
L2 = (1/2) (L1)
⇒L1 = 2(L2) [∵length is halved]
and A2 = 2 (A1) [∵area is doubled]
∴4/R2 = [2(L2)×2(A1)]/[(L2)×(A1)] = 4
⇒R2 =1 = 1 ohm
The wire is doubled on it, that means the area is doubled and the length gets halved.
R1/R2 =[ (L1)×(A2)]/[(L2)×(A1)]
R1 = 4 ohm
L2 = (1/2) (L1)
⇒L1 = 2(L2) [∵length is halved]
and A2 = 2 (A1) [∵area is doubled]
∴4/R2 = [2(L2)×2(A1)]/[(L2)×(A1)] = 4
⇒R2 =1 = 1 ohm
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The answer is 1 ohm .....
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