Chapter 24 Practice Pr soln

Chapter 24 Practice Pr soln - what we shall use to compute...

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pg 1 Chapter 24 Practice Pr soln: 21. I NTERPRET This problem is about applying Ohm’s law to find the resistivity of a rod. D EVELOP If the rod has a uniform current density and obeys Ohm’s law (Equations 24.4a and 4b), then its resistivity is 2 ( /4) / E E E d J I A I π ρ = = = E VALUATE Substituting the values given in the problem, we find the resistivity of the rod to be 2 2 2 6 ( /4) (1.4 V/m) (10 m) /4 2.20 10 m 50 A E d I - - = = = × Ω⋅ 23. I NTERPRET This problem involves using Ohm’s law to calculate the resistance of a heating coil. D EVELOP The macroscopic form the Ohm’s law is probably applicable to the heating coil, which is typically a coil of wire. Equation 24.5 gives / . R V I = E VALUATE Substituting the values given in the problem statement, we find the resistance to be 120 V 25 4.8 A V R I = = = 29. I NTERPRET This problem is about electric power of a motor, given the current drawn and the voltage across the terminals. D EVELOP Equation 24.7, , P IV = provides the connection between electric current, voltage and electric power. This equation is
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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. Ohms law is involved. D EVELOP The resistance of a uniform wire of Ohmic material is given by Equation 24.6: / . R L A = Therefore, the resistance per unit length is 2 /4 R L A d = = 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-- = = = 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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Chapter 24 Practice Pr soln - what we shall use to compute...

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