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.
M.A.Salam said:
9 years ago
We know that Resistance R = (resistivity *L)/A.
Where A is cross-sectional area & L is the length of wire.
Area = pi * r^2.
=pi * d^2/4.
d is diameter of wire.
But Current = voltage/Resistance.
I = V/R.
It can be written as the I proportional to d^2.
That means current will be four times.
Where A is cross-sectional area & L is the length of wire.
Area = pi * r^2.
=pi * d^2/4.
d is diameter of wire.
But Current = voltage/Resistance.
I = V/R.
It can be written as the I proportional to d^2.
That means current will be four times.
Nitin said:
10 years ago
Conductivity = 1/restivity.
Conductivity = l/RA.
There is length of conductor is l.
Cross section area of conductor is A.
A = 3.14*r^2.
Conductivity = l/R*3.14*r^2.
Then diameter is double then r = 2r.
Then Conductivity = l/R*3.14*(2r)^2.
= l/R*3-14*4r^2.
= 1/4(l/R*3-14r^2).
Then answer is 1/4.
Conductivity = l/RA.
There is length of conductor is l.
Cross section area of conductor is A.
A = 3.14*r^2.
Conductivity = l/R*3.14*r^2.
Then diameter is double then r = 2r.
Then Conductivity = l/R*3.14*(2r)^2.
= l/R*3-14*4r^2.
= 1/4(l/R*3-14r^2).
Then answer is 1/4.
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)
Banti said:
10 years ago
Its so simple in practically if we increase the core diameter of wire then current carrying capacity in increase like example in your home you have used big core wire compare to any low voltage and current electronics devices.
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.
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?
Mariappan said:
9 years ago
R= pl/a if a = φ d^2/ 4 sub value if R= pl*4/d^2 if d=2 cm assume if d = 4cm double sub value R = pl * 4/16.
R =V/I.
V/I = pl/4.
I = 4V/pl.
R =V/I.
V/I = pl/4.
I = 4V/pl.
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.
Jai said:
10 years ago
R = (Resistivity*Length)/Area.
In both case resistivity = 1 & l = 1.
R = 1/A.
V/I = 1/A if V = 1.
R = 2r.
I = 4*r^2.
In both case resistivity = 1 & l = 1.
R = 1/A.
V/I = 1/A if V = 1.
R = 2r.
I = 4*r^2.
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)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers