Need help with question and practice it please (A) Calculate the resistance per

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Need help with question and practice it please

(A) Calculate the resistance per unit length Find the cross-sectional area of the A2(3.210 x 104m)2 3.24 x 107 m2 Obtain the resistivity of Nichrome, solve for R/t, and substitute 1.5x10-ba-m l A 3.24 x 10 m 4.6 /m (B) Find the current in a 1.00-m segment of the wire if the potential difference across it is 10.0 V: Substitute given values into Ohm's law: AV 10.0 V R 4.6 2.2 A (C) If the wire is melted down and recast with twice its original length, find the new resistance as a multiple of the old Find the new area Av in terms of the old area Ao, using the fact the volume doesn't change and2o Substitute to find the new resistance AN (Ao/2)A,o LEARN MORE REMARKS The resistivity of Nichrome is about 100 times that of copper, a typical good conductor. Therefore, a copper wire of the same radius would have a resistance per unit length of only 0.052 /m, and a 1.00-m length of copper wire of the same radius would carry the same current (2.2 A) with ar applied voltage of only 0.115 V. Because of its resistance to oxidation, Nichrome is often used for heating elements in toasters, irons and electric heaters QUESTION Would replacing the Nichrome with copper wire of the same size result in a higher current or lower current? the same current higher current lower current PRACTICE IT Use the worked example above to help you solve this problem (a) Calculate the resistance per unit length of a 44 gauge nichrome wire of radius 0.642 mm. 2/m (b) If a potential difference of 10.5 V is maintained across a 1.00 m length of wire, what is the current in the wire? (c) The wire is melted down and recast with triple its original length. Find the new resistance Rv as a multiple of the old resistance Ro-

Explanation / Answer

For Learn More problem:

Resistance is given by:

R = rho*L/A

So we can see that Resistance is directly proportional to resistivity of wire, Since resistivity of copper wire is lower than nichrome wire, So resistance of copper wire will be lower than nihrome wire.

Now Using ohm's law:

i = dV/R

So we can see that current is inversely proportional to resistance of wire and since resistance of copper wire is lower than nichrome wire, So current will be higher in copper wire.

Higher Current (Ans)

For PRACTICE IT problem:

Part A.

Area = pi*r^2 = pi*(0.642*10^-3)^2 = 1.29*10^-6 m^2

Now resistance per unit length will be

R/L = rho/A = 1.5*10^-6/(1.29*10^-6) = 1.16 Ohm/m

Part B.

Using Ohm's law:

i = dV/R

i = 10.5/1.16

i = 9.05 Amp

Part C.

Volume of both wire will remain same, So

Vn = Vo

An*Ln = Ao*Lo

An = Ao*(Lo/Ln)

Given that Ln = 3*Lo

An = Ao*(Lo/(3*Lo)) = Ao/3

So, Now

Rn = rho*Ln/An = rho*(3*Lo)/(Ao/3) = 9*(rho*Lo/Ao)

Rn = 9*Ro

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