Use the worked example above to help you solve this problem. (a) Calculate the r

Question

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.

(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 RN as a multiple of the old resistance RO.

What is the resistance of a 5.7 m length of nichrome wire that has a radius of 0.321 mm? How much current does it carry when connected to a 120 V source?

EXAMPLE 17.3 The Resistance of Nichrome Wire 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 Rv 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 fy and Ay, 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 A) Calculate the resistance per unit length Find the cross-sectional area of the wire: A = 2 = (3.210 x 10-4 m)2 = 3.24 x 10-7 m2 Obtain the resistivity of Nichrome, solve for R/f, and substitute: 6 1.5 x 10- ·m = 4.6 /m A 3.24x10-7m2 (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: 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 n old area Ao, using the fact the volume doesn't change and fN = 210 Substitute to find the new resistance: ew area Ay in terms of the AN = Ao(€ 0121 o) = A012 Av (Ao/2) Ao LEARN MORE

Explanation / Answer

Resistance of Wire is Given by -
R = *L/A

Where,
R is Resistance
is Resistivity of Nichrome = 1.5*10^-6 ohm-m
L is Length of Wire
A is Area of wire = 3.14*(0.642*10^-3)^2 m^2

Resistance per unit length -
R/L = /A
R/L = (1.5*10^-6 )/( 3.14*(0.642*10^-3)^2) ohm/m
R/L = 1.16 ohm/m

(b)
If Potential Difference = 10.5 v
Resitance of 1m Length of Wire , R = 1.16 ohm

We know, I = V/R
I = 10.5/1.16
I = 9.1 Amp

Current in the wire, I = 9.1 Amp

(c)
Already The Answer is Correct,

Exercise Answers -

Resistance of Wire is Given by -
R = *L/A

Where,
R is Resistance
is Resistivity of Nichrome = 1.5*10^-6 ohm-m
L is Length of Wire = 5.7 m
A is Area of wire = 3.14*(0.321*10^-3)^2 m^2

Resistance per unit length -
R = *L/A
R = (1.5*10^-6 * 5.7 )/( 3.14*(0.321*10^-3)^2) ohm/m
R = 26.5 ohm

If Potential Difference = 10.5 v
We know, I = V/R
I = 120/26.5
I = 4.53 Amp

Current in the wire, I = 4.53 Amp

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