A wite of given material having length land area of cross- section. A has a resistance of 4 π . What would be the resistance of another wire of the same material having length l 12 and area of cross- section 2 A?
Answers
Answer:
Wire of uniform cross-section has a resistance of R. Let the diameter be d, the length of the wire be l cross-sectional area be A.
Now the resistance of a wire can be expressed as:
R=ρ
A
l
Where resistivity is ρ.
Now a second wire is twice as long and of half cross section.
Let final length be l
1
, diameter be d
1
and
cross section Area be A
1
.
A
1
=
2
A
R
1
=ρ
A
1
l
1
Putting the value of l
1
=2l and A
1
=
2
A
We have, resistance of the new second wire, R
2
=ρ
A
4l
=4×R
1
=4×8=32Ω
Explanation:
Answer:
Explanation:
By the formula of resistance:
R=(σ) L/A
here σ is resistivity, L is length of wire and A is area of cross section
Given. σ1=σ2. ______eq 1
for first material R1=σ1 L/A.
we can write.. σ1=R1A/L
σ1=4A/L.
for second material σ2 = R2.4A/L
from eq1. 4A/L=R24A/L
R2=0.
hope you understand
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