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RECITATION-CH 26

# RECITATION-CH 26 - heater are related by 1250 W 10.9 A 115...

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RECITATION 6 11. Since the mass density of the material does not change, the volume remains the same. If 1 L is the original length, 2 L is the new length, 1 A is the original cross-sectional area, and 2 A is the new cross-sectional area, then 1 1 2 2 . L A L A = If 2 1 3 , L L = then 1 2 1 3 . A A = The new resistance of the wire is ( 29 1 2 1 2 1 2 1 1 3 9 9 , 3 L L L R R A A A ρ = = = = where 1 R is the original resistance. Thus, ( 29 2 9 6.0 54 . R = Ω = 19. (a) Electrical energy is converted to heat at a rate given by ( 29 2 , V P R = where V is the potential difference across the heater and R is the resistance of the heater. Thus, 2 3 (120 V) 1.0 10 W 1.0 kW. 14 P = = × = (b) The cost is given by (1.0 kW)(5.0 h)(5.0¢/kW h) 25¢. = 21. Using the relation ( 29 2 P V R = ∆ , the power dissipated in the second case is given by ( 29 ( 29 2 2 2 2 2 2 1 1 1.50 V 0.135 W. 3.00 V V R P P P V R = = = 23. (a) The current in the heater, the power dissipated, and the potential difference across the

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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. 26-5, 26-8, and 26-15, 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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RECITATION-CH 26 - heater are related by 1250 W 10.9 A 115...

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