A wire had a resistance of 20Ω. It was then melted and reduced to one-fifth of its original length and its cross section area was doubled. Find its new resistance.
Answer fast please
Answers
Answered by
0
Answer:
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
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
please mark me as brainliest
Answered by
0
Answer:
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
Explanation: pls mark me brainlist
Similar questions