Electronics and Communication Engineering - Materials and Components - Discussion
Discussion Forum : Materials and Components - Section 1 (Q.No. 12)
12.
If the diameter of a wire is doubled, its current carrying capacity becomes
Answer: Option
Explanation:
Current carrying capacity depends on area of cross-section.
Discussion:
23 comments Page 3 of 3.
Jing said:
6 years ago
From i=V/R and R=pL/A.
; i=(V*A)/pL.
i=V*{(r)^2*π)}/ pL if d is increased 3 times,
i=V*{(3*r)^2*π)}/pL; i is increased 9 times.
; i=(V*A)/pL.
i=V*{(r)^2*π)}/ pL if d is increased 3 times,
i=V*{(3*r)^2*π)}/pL; i is increased 9 times.
Gaurav said:
5 years ago
R = (resistivity*length)/Area
Area = π * Square of radius.
We know that radius = (diameter)/2.
so 2*radius = diameter.
take resistivity*length = k.
R' = k/π*(square of (2*radius))
R' = k/π*4*square of radius.
R' = R/4,
I = V/R,
I' = V/R',
I' = 4V/R,
I' = 4I.
Area = π * Square of radius.
We know that radius = (diameter)/2.
so 2*radius = diameter.
take resistivity*length = k.
R' = k/π*(square of (2*radius))
R' = k/π*4*square of radius.
R' = R/4,
I = V/R,
I' = V/R',
I' = 4V/R,
I' = 4I.
(1)
Malik said:
3 years ago
A = π*d.
If d = 2d.
A = π*2d,
A = 2A.
Area doubles resistance becomes half and current should be double or two times.
If d = 2d.
A = π*2d,
A = 2A.
Area doubles resistance becomes half and current should be double or two times.
(3)
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