Combine the concept of resistivity with Ohm\'s law. (a) Calculate the resistance

Question

Combine the concept of resistivity with Ohm's law. (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 R_N as a multiple of the old resistance R_O. 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 and l_N the A_N, 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. (A) Calculate the resistance per unit length. Find the cross-sectional area of the wire: A = pi r^2 = pui (3.210 times 10^-4 m)^2 = 3.24 times 10^-7 m62 Obtain the resistivity of Nichrome, solve for R/l, and substitute: R/l = rho/A = 1.5 times 10^-6 ohm middot m/3.24 times 10^-7 m^2 = 4.6 ohm/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: I = delta V/R = 10.0 V/4.6 ohm = 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 A_N in terms of the V_N = V_O rightarrow A_N l_O rightarrow A_N = A_O = A_O(l_O/l_n) 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. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error, ohm/m (b) If a potential difference of 12.0 V is maintained across a 1.00 m length of wire, what is the current in the wire? Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. A (c) The wire is melted down and recast with triple its original length. Find the new resistance R_N as a multiple of the old resistance R_O. Your response differs from the correct answer by more than 10%. Double check your calculations. R_O What is the resistance of a 6.9 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? R = ohm I = A

Explanation / Answer

Practice it

(a)

Resistance per unit length=resistivity/area=1.15 ohm/meter

(b) current=voltage/resistance=10.35 A

(c)now length will become thrice

since volume will be constant so area will decrease by 3 times

new resistance=resistivity*new length/new area=resistivity*3l/A/3=9R

so 9 times

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