A piece of wire of resistance 20 is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.
guys please help me
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Step-by-step explanation:
When length of wire is doubled i.e. 2l
Also it's area of cross section becomes half i.e
2
A
Resistance R=
A
ρl
=20Ω
New Resistance ,
R
′
=
A
′
ρl
′
=
A
ρ(2l)×2
=4
A
ρl
=4(
A
ρl
)
=4×20
=80Ω
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