GOAL Combine the concept of resistivity with Ohm\'s law. PROBLEM (a) Calculate t
ID: 1407720 • Letter: G
Question
GOAL Combine the concept of resistivity with Ohm's law.
PROBLEM (a) Calculate the resistance per unit length of a 22-gauge Nichrome wire of radius 0.321 mm. (b) If a potential difference of 10.0 V is maintained across a 1.00-m length of the Nichrome wire, what is the current in the wire? (c) The wire is melted down and recast with twice its original length. Find the new resistance RN as a multiple of the old resistance RO.
STRATEGY Part (a) requires substitution into Equation 17.5 after calculating the cross-sectional area, whereas part (b) is a matter of substitution into Ohm's law. Part (c) requires some algebra. The idea is to take the expression for the new resistance and substitute expressions for ?N and AN, the new length and cross-sectional area, in terms of the old length and cross-section. For the area substitution, remember that the volumes of the old and new wires are the same. GOAL Combine the concept of resistivity with Ohm's law.
PROBLEM (a) Calculate the resistance per unit length of a 22-gauge Nichrome wire of radius 0.321 mm. (b) If a potential difference of 10.0 V is maintained across a 1.00-m length of the Nichrome wire, what is the current in the wire? (c) The wire is melted down and recast with twice its original length. Find the new resistance RN as a multiple of the old resistance RO.
STRATEGY Part (a) requires substitution into Equation 17.5 after calculating the cross-sectional area, whereas part (b) is a matter of substitution into Ohm's law. Part (c) requires some algebra. The idea is to take the expression for the new resistance and substitute expressions for ?N and AN, the new length and cross-sectional area, in terms of the old length and cross-section. For the area substitution, remember that the volumes of the old and new wires are the same. SOLUTION Find the cross-sectional area of the wire: Obtain the resistivity of Nichrome, solve for R/?, and substitute: (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: I = ?V = 10.0 V = 2.2 A R 4.6 ? Substitute given values into Ohm's law: I = ?V = 10.0 V = 2.2 A R 4.6 ? Substitute given values into Ohm's law: I = ?V = 10.0 V = 2.2 A R 4.6 ? I = ?V = 10.0 V = 2.2 A R 4.6 ? (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 AN in terms of the old area AO, using the fact the volume doesn't change and ?N = 2?O. VN = VO ? AN?N = AO?O ? AN = AO(?O/?N) AN = AO(?O/2?O) = AO/2 Substitute to find the new resistance: RN = ??N = ?(2?O) = 4 ??O = 4RO AN (AO/2) AO Find the new area AN in terms of the old area AO, using the fact the volume doesn't change and ?N = 2?O. VN = VO ? AN?N = AO?O ? AN = AO(?O/?N) AN = AO(?O/2?O) = AO/2 Find the new area AN in terms of the old area AO, using the fact the volume doesn't change and ?N = 2?O. VN = VO ? AN?N = AO?O ? AN = AO(?O/?N) AN = AO(?O/2?O) = AO/2 VN = VO ? AN?N = AO?O ? AN = AO(?O/?N) AN = AO(?O/2?O) = AO/2 Substitute to find the new resistance: RN = ??N = ?(2?O) = 4 ??O = 4RO AN (AO/2) AO Substitute to find the new resistance: RN = ??N = ?(2?O) = 4 ??O = 4RO AN (AO/2) AO RN = ??N = ?(2?O) = 4 ??O = 4RO AN (AO/2) AO 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 an 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?
lower currenthigher current the same current 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 an 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?
lower currenthigher current the same current lower currenthigher current the same current PRACTICE IT EXERCISEHINTS: GETTING STARTED | I'M STUCK!
R = ? = 1.5 10-6 ? · m = 4.6 ?/m ? A 3.24 10-7 m2
Explanation / Answer
Exercise:
rho = 1.5*10^-6 ohm m
L = 6.3 m
radius, r = 0.321 mm = 3.21*10^-4 m
Area, A =pi*r^2 = pi * (3.21*10^-4)^2 = 3.24*10^-7 m^2
R = rho * L/ A
= (1.5*10^-6) * 6.3 / (3.24*10^-7 )
= 0.735 ohm
I = V/R
= 120/0.735
= 163.3 A
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