a piece of wire of resistance 20 ohm is drawn out so that its length is increased to twice its original length. Calculate the resistance of wire in a new situation.
Answers
Answered by
462
When length of wire is doubled i.e. 2l (say)
Also it's area of cross section becomes half i.e A/2
Resistance R = ρl/A= 20
Resistance ,
R' = ρl/A
=ρ(2l)*2/A
=4ρl/A= 4(pl/A)
= 4*20
= 80ohm
Also it's area of cross section becomes half i.e A/2
Resistance R = ρl/A= 20
Resistance ,
R' = ρl/A
=ρ(2l)*2/A
=4ρl/A= 4(pl/A)
= 4*20
= 80ohm
Answered by
187
Answer:
The resistance of wire in a new situation is 80 ohm
Explanation:
Given
Resistance = R1 = 20 ohm
Let length of the wire be L
According to question,
New length of the wire = 2L
As we know,
When the length is doubled, area of the wire becomes half.
Therefore,
New resistance = R = 2L/ (A/2) =4L/A
R = 4 * R1
Substituitng value of R1
R = 4 x 20
R = 80 ohm.
Hence,
The resistance of wire in a new situation is 80 ohm
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