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 2 of 3.
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.
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.
Ravi said:
9 years ago
But current carrying capacity is different thing compared to current decrease, that's what I feel.
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).
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?
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).
Ravi said:
9 years ago
If Wire diameter increases, resistance increases, current decreases.
Ken said:
10 years ago
Can anyone explain the answer? I really need the solution.
SANTOSH KUMAR said:
7 years ago
Since, J=I/A, A= πD^2/4 therefore I=JπD^2/4.
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