A wire of resistance R is folded into two. Find the change in Resistance
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the resistance is now decreased by 2/x times the actual resistance
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Resistivity remains constant.
I m using € to denote resistivity
Case 1 , € = R 1 a/ l
Since the wire is folded , a becomes 2 times and length becomes 1 /2
Therefore
€ = R2 × 2a / l/2 = R2 ×4 /l
Therefore , since € is same
R1 a /l = R2 × 4 a /l
R1 / R2 = 4 /1 => R2 = R1/4
Therefore new resisitance is 1 4 th
a stands dor area of cross section
R is inversely proportional to area of cross section
Thank u★★★
#ckc
I m using € to denote resistivity
Case 1 , € = R 1 a/ l
Since the wire is folded , a becomes 2 times and length becomes 1 /2
Therefore
€ = R2 × 2a / l/2 = R2 ×4 /l
Therefore , since € is same
R1 a /l = R2 × 4 a /l
R1 / R2 = 4 /1 => R2 = R1/4
Therefore new resisitance is 1 4 th
a stands dor area of cross section
R is inversely proportional to area of cross section
Thank u★★★
#ckc
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