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.
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.
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.
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.
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.
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.
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?
Can anyone explain this?
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers