a wire of given material having length L and area of cross section has resistance of 4 what would be the resistance of another wire of same material having 1/2 length and area of cross section 2a
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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
have a nice day..
☺️✌️
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
have a nice day..
☺️✌️
Answered by
3
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_______________
We know that the
resistance is P(rho) l/a.
We can ignore P over here because the material is same in both cases.
Now,
R=l/a
4=l/a.
In the second case,
they are asking resistance if the area is double and length is halved.
So, R=l/2
whole divided by 2a.
We get R=l/4a.
But
we l/a=4.
So,
4/4=1
and
so the answer is 1 ohm.
good evening____________✌️✌️✌️✌️✌️
_____________________
__________________________________
_______________
We know that the
resistance is P(rho) l/a.
We can ignore P over here because the material is same in both cases.
Now,
R=l/a
4=l/a.
In the second case,
they are asking resistance if the area is double and length is halved.
So, R=l/2
whole divided by 2a.
We get R=l/4a.
But
we l/a=4.
So,
4/4=1
and
so the answer is 1 ohm.
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