how much resistance change if the diameter is doubled?
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Let the length of the wire is L, resistivity is ρ and diameter is d. The resistance of the wire is R=4ρL ∕ (πd²). Now if the diameter is doubled then the new resistance is R’=4ρL∕(π(2d)²) =4ρL∕(4πd²) =0.25R. Hence we see that the resistance of the wire becomes 25% of original resistance. This result is showing that the resistance decreases as we increase the cross-section area keeping the length constant.
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we know that
R=(density *length)/area
R is inversely proportional to area
let initial diameter be x
new diameter=2x
now according to question
R is inversely proportional to square of radius since π is constant
final resistance=1/4x^2 -1/8x^2
=1/8x^2
in percentage 2x^2*100/8x^2
25%
R=(density *length)/area
R is inversely proportional to area
let initial diameter be x
new diameter=2x
now according to question
R is inversely proportional to square of radius since π is constant
final resistance=1/4x^2 -1/8x^2
=1/8x^2
in percentage 2x^2*100/8x^2
25%
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