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.
Pankaj said:
1 decade ago
Yes, it is right.
Because R = (resitivity*Lenth)/Area.
Area = pi(r1*r1).
Because R = (resitivity*Lenth)/Area.
Area = pi(r1*r1).
Reddy said:
1 decade ago
Are is how related to current carrying capacity with area of cross section any one explain please clearly or shortly.
Ayas said:
10 years ago
But the resistance should be inversely proportional to current carrying capacity, hence the answer should be A.
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.
Ken said:
10 years ago
The answer should be A, then why is that D is the answer?
Can anyone explain this?
Can anyone explain this?
Ken said:
10 years ago
Can anyone explain the answer? I really need the solution.
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.
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.
Harsha said:
10 years ago
r = pl/a.
i = v/r.
i = v/r.
Ravi said:
9 years ago
If Wire diameter increases, resistance increases, current decreases.
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