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Unformatted text preview: heater are related by 1250 W 10.9 A. 115 V P i V = = = ∆ (b) The resistance of the heater, the potential difference across the heater, and the power dissipated are related by ( 29 ( 29 2 2 115 V 10.6 . 1250 W V R P ∆ = = = Ω (c) The thermal energy E generated by the heater in time ∆ t = 1.0 h = 3600 s is 6 (1250 W)(3600 s) 4.5 10 J. E P t = ∆ = = × 33. The current density is related to the current by ( 29 2 , 2 i J d π = where d is the diameter of the wire. Therefore, ( 29 2 2 0.50 A 2 2 3.8 10 m. 440 A m i d J= = = × 39. From Eq. 265, 268, and 2615, we obtain the resistivity of the wire: ( 29 ( 29 4 4 2 115 V 8.2 10 m. 10 m 1.4 10 A m AR A V V L L i LJ ρ∆ ∆ = = = = = × Ω⋅ ×...
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 Spring '08
 Goksu
 Physics, Thermodynamics, Energy, Mass, Trigraph

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