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
one-fourth
half
twice
four times
Answer: Option
Explanation:

Current carrying capacity depends on area of cross-section.

Discussion:
23 comments Page 2 of 3.

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.

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.

Pankaj said:   1 decade ago
Yes, it is right.

Because R = (resitivity*Lenth)/Area.

Area = pi(r1*r1).

Ravi said:   9 years ago
But current carrying capacity is different thing compared to current decrease, that's what I feel.

Ravi said:   9 years ago
If Wire diameter increases, resistance increases, current decreases.

Harsha said:   10 years ago
r = pl/a.
i = v/r.

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.

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.

Ken said:   10 years ago
Can anyone explain the answer? I really need the solution.

Ken said:   10 years ago
The answer should be A, then why is that D is the answer?

Can anyone explain this?


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