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Unformatted text preview: what we shall use to compute the power P . E VALUATE Substituting the values given in the problem, we find the power consumption to be pg 2 (125 A)(11 V) 1.38 kW P IV = = = 47. I NTERPRET This problem compares resistances of wires made up of different materials. Ohm’s law is involved. D EVELOP The resistance of a uniform wire of Ohmic material is given by Equation 24.6: / . R L A ρ = Therefore, the resistance per unit length is 2 /4 R L A d π = = E VALUATE Equal values of / R L for copper and aluminum wires imply that 8 Cu Al Al Al 2 2 8 Cu Cu Cu Al 2.65 10 m = 1.26 1.68 10 m d d d d× Ω⋅ = → = = × Ω⋅ where we have used Table 24.1 for resistivities. A SSESS Since 1/ 2 , d z the higher the resistivity, the greater the diameter of the wire....
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 Summer '08
 Hoston,Amahd,Bakanowski
 Physics, Current, Magnetic Field, Resistor, ohm, Electrical resistance

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