The resistance of a very fine aluminum wire with a 10μm×10μm square cross section is role="math" localid="1649061344586" 1000Ω. A 1000Ω resistor is made by wrapping this wire in a spiral around a 3.0mmdiameter glass core. How many turns of wire are needed?

Short Answer

Expert verified

The number of turns of wire are needed is379turns.

Step by step solution

01

Given Information

Aluminum wire dimension=10μm×10μm

Resistancelocalid="1649061848996" =1000Ω

Resistor=1000Ω

Spiral wire diameter=3.0mm

02

Explanation

The total length of the wire will be such that

R=ρLA(1)

Express this length as,

L=RAρ=Ra2ρ(2)

where ais the thickness of the wire.

The length fulfills the following condition:

L=nP

Where P=πDis the perimeter of the glass core.

The number of turns is therefore,

n=LP=LπD

Substitute the second expression in the last, finding

localid="1649061681694" n=Ra2ρπD=Ra2πρD

Substitute the expression,

localid="1649062038246" n=1000Ω×0.00001m2π×2.8×10-8Ωm×0.003m

=379turns.

03

Final Answer 

Hence, the number of turns needed are379turns.

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