P28_025 - area of the thick wire. Since the single thick...

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25. Let r be the resistance of each of the narrow wires. Since they are in parallel the resistance R of the composite is given by 1 R = 9 r , or R = r/ 9. Now r =4 ρ`/πd 2 and R =4 ρ`/πD 2 ,where ρ is the resistivity of copper. A = πd 2 / 4was used for the cross-sectional area of a single wire, and a similar expression was used for the cross-sectional
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Unformatted text preview: area of the thick wire. Since the single thick wire is to have the same resistance as the composite, 4 ` D 2 = 4 ` 9 d 2 = D = 3 d ....
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