Factors that affect resistance of a wire?
... First, the total length of the wires will effect the amount of resistance. The longer the wire, the more resistance that there will be. There is a direct relationship between the amount of resistance encountered by charge and the length of wire it must traverse. After all, if resistance occurs as the result of collisions between charge carriers and the atoms of the wire, then there is likely to be more collisions in a longer wire. More collisions means more resistance. Second, the cross-sectional area of the wires will effect the amount of resistance. Wider wires have a greater cross-sectional area. Water will flow through a wider pipe at a higher rate than it will flow through a narrow pipe; this can be attributed to the lower amount of resistance which is present in the wider pipe. In the same manner, the wider the wire, the less resistance that there will be to the flow of electric charge. When all other variables are the same, charge will flow at higher rates through wider wires with greater cross-sectional areas than through thinner wires. A third variable which is known to effect the resistance to charge flow is the material that a wire is made of. Not all materials are created equal in terms of their conductive ability. Some materials are better conductors than others and offer less resistance to the flow of charge. Silver is one of the best conductors, but is never used in wires of household circuits due to its cost. Copper and aluminum are among the least expensive materials with suitable conducting ability to permit their use in wires of household circuits. The conducting ability of a material is often indicated by its resistivity. The resistivity of a material is dependent upon the material's electronic structure and its temperature. For most (but not all) materials, resistivity increases with increasing temperature. The table below lists resistivity values for various materials at temperatures of 20 degrees Celsius. ...
If a wire is stretched to double its length, what will its new resistance be?
What happens to the resistancece of the wire when the length is doubled?Since the length of the wire is doubled, its area of cross-section is halved so as to keep the volume of the wire constant.Now the resistance of the wire is proportional to length of the wire, since the length of the wire is doubled its resistance also doubles due to doubling of the length. But resistance is also inversely proportional to the area of cross-section, and as area of cross-section is halved due to doubling of length, the resistance is once more doubled due to the halfing of the area of cross-section. So overall the resistance of the wire becomes four times the original resistance.
If length is increased by 20%, what is the resistance?
if, length= l and resistance (R) =pl/A=p(l^2)/VThen , increase in length = l*20/100l'= l+(l/5)l'=6l/5Final resistance = p( l'^2)/VR'=p *36(l^2)/25VR'=(36/25)RIncrease in resistance (%) ={( R'-R) /R}*100=11R*100/25R %=44%
If a given wire is stretched 3 times its length, how will its resistance change?
The change in resistance can be easily calculated by applying the following formula:Resistance = (Specific Resistivity Constant x Length of wire) / Area of cross section of wireWhen the wire is stretched to 3 times its original length, its cross section is also reduced to one third of what it was before. By using the above relation, we find the new resistance to be 3x3= 9 times of what it was before!Hope this helps.
How do you calculate the percentage increase in resistance of a cylindrical wire that is stretched to increase its length by 10%?
Resistance of a cylindrical wire is:R=(rho) l/A =(rho)l^2/(Al)=(rho)l^2/V………(1)(rho) is resistivity of the material of the wire.l is length of the wire.A is area of normal cross section of the wireV is volume of the wire which remains constant on stretching the wire.New resistance after stretching,R'=[(rho)/V](1.1 l)^2=[(rho)/V](1.21)l^2=(1.21)R.Therefore,(R'-R)/R=(1.21 -1)=0.21OR[(R'-R)/R]x100=0.21x100=21%
An electric wire is stretched to increase its length by 25%. By what % will the resistance be increased and what will be increase in its resistivity?
Resistivity of a substance is and intrinsic property and cannot be changed by stretching or physical deformation. Hence the answer to your send part is No. The resistivity will remain constant.Resistance of a wire can be calculated by the formulaR = rho * L / A ………………………..eq(i)where rho is resistivity of the materialL is the length of the wireA is the cross sectional area of the wireSince no material is added in the elongation therefore the volume of wire is constant.Volume = L * A ……………………………..eq (ii)Hence if L is increased by 25% ie 5/4 times then the area shall decrease 4/5 times to keep volume constant.L’=L+(25% *L) = L(1+0.25) =L*5/4L’ * A’ = V = L * AL * 5/4 * A’ = V = L * A ……………………………….using eq(ii)5/4* A’ =Aie A’ =4/5 * A …………………………………….eq (iii)hence the new resistance is calculated asR’ = rho * L’ / A’ ……………………………………..using eq(i)R’ = (rho * 5/4 * L )/ (4/5 * A) ……………………using eq(iii)R’ = rho * L / A *{(5/4) / (4/5)}R’ = R * ( 25/16)R’ = R * (1.5625)Hence the resistance shall increase by 56.25%.I hope that this answers your question, for further queries please drop in a comment.further referenceElectrical resistivity and conductivity - Wikipedia
A kilometer of wire having a diameter of 11.7mm and a resistance of 0.031 ohm is drawn down so that its diamet?
resistivity r = RA/L Before drawn down: r = 0.031 * [pi * 11.7^2 /4] / L --- (1) After drawn down: r = R * [pi * 0.5^2 /4] / L1 ----(2) volume of metal remains unchanged, pi [11.7^2 /4 ] * L = pi [0.5^2 /4 ] * L1 [11.7^2 ] * L = [0.5^2 ] * L1 L/ L1 = [ 0.5 / 11.7]^2 = 0.0018262 (1) = (2) R * [pi * 0.5^2 /4] / L1 = 0.031 * [pi * 11.7^2 /4] / L R * 0.5^2 / L1 = 0.031 * 11.7^2 / L R = 0.031 * [11.7 / 0.5]^2 [ L1/L] R = 1.06089 * [1/ 0.0018262] R = 580.93 Ohm answer
Change in resistance in a conducting wire that's quadrupled in length and tripled in diameter...?
Quadrupling length will increase the resistance by 4 times: - (it's like having 4 equal resistors in series) Resistance is inversely proportional to the cross-sectional area (the larger the area the lower the resistance - the wider pipe analogy) Because cross-sectional area = pi r^2 resistance must be inversely proportional to r^2 With a tripled diameter the radius is tripled also, so the resistance is reduced by a factor of 3^2 i.e. the resistance is 9 times less. The overall effect is thus a change to 4 /9 ths of the original resistance value.
How does cross sectional area affect thermal resistance?
I'm working on a heat transfer problem but have been scratching my head where I am calculating thermal resistances. I'm looking at a black powder charge emitting heat which I am modeling to be conducted through an insulation then to some steel wire. This all takes place in a cylinder 4in in diameter. So what I have is Qin --> insulation --> wire. This all makes sense but then when I go to calculate the final temp, I'm getting ridiculous values using the equation T2=T1-(Qin*(sumR)). The equation for thermal resistance is R=x/kA and this has me confused. The value I calculated of R for the insulation makes sense but then the wire is where I'm lost. The cross sectional area of the wire is very small (5.6e-7 meters) which makes the thermal resistance go through the roof which is the cause of the error. It seems backwards to me that a decrease in area causes an increase in resistance. Am I looking at this all wrong? Any help is appreciated.
The length of a given cylindrical wire is increased by 100%. Due to consequent decrease in diameter the change in the resistance of the wire will be?
Resistance of a conductor is proportional to the ratio of it's length to it's cross sectional area.Just remember this and you'll be fine!So we can write the equation for resistance(R)R = p*(L/A)Here p is called the resistivity of conductor, which is the property of the material of conductor.Now coming back to your question, let's take initial resistance of conductor be R1R1' = p*(L/A)Now the length of the conductor is doubled.The mass of the conductor must remain unaffected and consequently the volume too( mass = d(volume) , d= density (here constant))So (initial volume) = (final volume)L*A =2*L*A(new)This yeilds A(new) = A/2So the area has been halved!!Now new resistance R2= p*(2*L/(A/2))R2'= 4p*(L/A)R2'= 4*R1So the new resistance is 4 times the original one!Hope this helps!Please do comment for any query.