(a) Calculate the resistance per unit length of a 44 gauge nichrome wire of radi

Question

(a) Calculate the resistance per unit length of a 44 gauge nichrome wire of radius 0.642 mm. 0.85 X Did you accidentally divide or take the inverse in your calculation? (b) If a potential difference or 10.1 V is maintained across a 1.00 m length or wire, what is the current in the wire? Your response differs from the correct answer by more than 10%. Double check your calculations. (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 R0. X Your response differs from the correct answer by more than 10%. Double check your calculations R0

Explanation / Answer

something is wrong with question

first of all

radius for 44 gauge wire is = 0.081 mm (not 0.642 mm)

second, u havn't given the length of nichrmoe wire

well, resistivity of nicrome wire is = 100 *10^(-8) ohm*m

Resistance per unit length = ressitivity / (pi*r^2) = 100*10^(-8) / (pi*(8.1*10^(-5))^2) = 0.48515 ohm/m

(b):

current in wire = V / (resistance per unit length * length of wire )

current in wire = 10.1 / (0.48515*1) = 20.8183 A

(C):

volume remains constant therefore

pi*new radius^2 * 3*L = pi*r^2*L

new radius = r/sqrt(3) = 0.081/sqrt(3) = 0.0468 mm

new R = resistivity*3*L / (pi*r^2) = 100 *10^(-8) * 3 / (pi* (4.68*10^(-5)) ^2) = 436 ohm

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