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 1 of 3.
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)
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)
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.
SANTOSH KUMAR said:
7 years ago
Since, J=I/A, A= πD^2/4 therefore I=JπD^2/4.
Rajashekar balya said:
7 years ago
R=(4ρ*l)÷D^2.
Rohinipriya said:
7 years ago
If the diameter of the wire increased to three times compared with original value, then what will be the current carrying capacity of the wire?
Can anyone answer this?
Can anyone answer this?
Sampath said:
8 years ago
Twice is the correct option.
Partha said:
8 years ago
If the diameter of the wire is halved then what will be the answer?
Will it be one-fourth (1/4).
Will it be one-fourth (1/4).
Dharani said:
9 years ago
J = I/A.
A = π*r*r.
Diameter=r/2.
So, J = I/((π*r*r)/4).4J
= I/(π*r*r) Hence current increases by 4times.
A = π*r*r.
Diameter=r/2.
So, J = I/((π*r*r)/4).4J
= I/(π*r*r) Hence current increases by 4times.
Bhaskar said:
9 years ago
Current is directly proportional to area
I~a,
A = φR^2.
And diameter is proportional to radius then finally current carrying capacity increases by R^2 time I.e 4 times.
I~a,
A = φR^2.
And diameter is proportional to radius then finally current carrying capacity increases by R^2 time I.e 4 times.
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