a wire has a resistance of 10omh.it is comprised such that it's length becomes one third of its original length.calculate the resistance of the new wire also find the percentage change in its resistance
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Answer:
A wire has a resistance of 10 ohms. It is stretched to 3 times its original length. What will be the new resistance and new resistivity?
Resistivity is the property of the material and it will not change with changing the length of the wire. Now, we will focus on the effect of resistance due to change in length of the wire.
We know that,
R = (rho)*length/Cross Sectional Area
Since, the length of the wire becomes 3 times its original length; hence there will definitely be some change in cross sectional area.
Now we know that, volume = Mass/Density and since neither mass nor density has changed due to change in length, thus volume must remain constant.
We further know that,
Volume = Length * Cross Sectional Area
Assuming, Initial Volume = V1 & Final Volume = V2
Initial Cross Section Area = A1 & Final Cross Sectional Area = A2
Initial Length of Wire = L1 & Final Length of Wire = L2
Thus we have, L2 = 3 L1
Now, V2 = L2A2 & V1 = L1A1, we know that V1 = V2
i.e. L1A1 = L2A2
i.e. A2 = (L1/L2)A1
i.e. A2 = 1/3 A1
Putting the values of A2 & L2 into the following
R2/R1 = (L2/L1)*(A1/A2)
We get,
R2/R1 = 3*3
R2 =9 * R1
Thus, R2 = 9 *10= 90 Ohm.
We can assume the wire to be a cylinder so since volume will be constant then we have
π*r^2*l=π*R^2*3*l
r=√3*R
R=(1/√3)*r
So resistance previously was R1=(pl)/(π*r^2)
Present resistance will be
R2=(p*3*l)/(π*R^2)
=(p*3*l)/((π*r^2)/3)
=9*(pl)/(π*r^2)
=9*R1
=9*10=90ohms
Resistivity will remain unchanged since it is the property of a material and it itself does not depend on length or cross sectional area
R=(ρ∗L)/A
If the wire is an isolated system (no extra metals/wires coming in or going out) with a specified length and cross-sectional area, then stretching it would increase the length 3 times and presumably reduce the cross-sectional area 3 times as well. Then,
R=(ρ∗3L)/(1/3∗A)
R=9(ρ∗L)/A)
You get 9 times more resistance you initially had, this makes sense as you bring the wire to a thinner cross-sectional area, the electrons have more material to go through.
EDIT: Rho is the resistivity of the metal; L being the total length; and A being the cross-sectional area of the wire.
Fixed a math mistake! Thanks, Brett!
A wire has a resistance of 10 ohms. It is stretched by 10% of its original length, what will be the new resistance?
What is the new resistance of a 10-ohm wire if its length is stretched doubly?
A wire has a resistance of 2 ohms. It has been stretched to the length 3 times that of original, what will be the resistance of wire?
resistance=pl/a where l is the length,p is resistivity and a is area of crossection.When length is increased by 3 times,area of crossection decreases by 3 times So new resistance is p*3l/(a/3)=9*p*l/a=9*original r=90 ohms
resistivity is a property of material and it does not change for given temperature conditions
When the wire is stretchedto 3 times radius will be reduced by √3. Cross sectional area will be reduced 3 times. Resistance will increase proportionally for the same length. is. 30 ohms for same original length and resistance for the 3 times the length will be 90 ohms.
Resistivity will become 3 times the resistance (90/10) and 3 times the length. ie 27.
Here wire is stretched, it means density is constant. So if the length is stretched to 3 times the original length then radius will decrease to root three times.
So, l’=3l, a’=a/3
We have,
R=p*l/a=90ohms where R is resistance, p is resistivity
p=R*a/l=10ohms
Just visualize that wire as being "stranded", as most lamp cords around the house are. Let it consist of 3 strands, each having, of course, a resistance of 30 ohms. Now reconnect them in series (to triple the length as specified), thus obtaining a total resistance of 90 ohms.
Resistivity remain same and resistance is 90 ohms
resistivty is independent of dimensons…it only depends on temp n material
R=resistivity*L/A …when L is streched to three times A also decrease to 3 time so R’=9R
R=90ohms