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.
Ravi said:
9 years ago
But current carrying capacity is different thing compared to current decrease, that's what I feel.
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.
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.
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.
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.
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).
Sampath said:
8 years ago
Twice is the correct option.
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?
Rajashekar balya said:
7 years ago
R=(4ρ*l)÷D^2.
SANTOSH KUMAR said:
7 years ago
Since, J=I/A, A= πD^2/4 therefore I=JπD^2/4.
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