29. Let rbe the resistance of each of the narrow wires. Since they are in parallel the resistance Rof the composite is given by 19Rr=, or R = r/9. Now rd=42ρA/πand RD=42A/π, where is the resistivity of copper. A = πd 2/4 was used for the cross-sectional area of a single wire, and a similar expression
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.