Homework 6 Solutions

# 9 1018 j 43 the relation p v 2r implies p v 2

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Unformatted text preview: Electrical energy is converted to heat at a rate given by P = V 2 / R, where V is the potential difference across the heater and R is the resistance of the heater. Thus, 1050 CHAPTER 26 (120 V) 2 . . P= = 10 × 103 W = 10 kW. 14 Ω (b) The cost is given by (1.0kW)(5.0h)(5.0cents/kW ⋅ h) = US\$0.25. 42. (a) Referring to Fig. 26-32, the electric field would point down (toward the bottom of the page) in the strip, which means the current density vector would point down, too (by Eq. 26-11). This implies (since electrons are negatively charged) that the conduction electrons would be “drifting” upward in the strip. (b) Equation 24-6 immediately gives 12 eV, or (using e = 1.60 × 10−19 C) 1.9 × 10−18 J for the work done by the field (which equals, in magnitude, the potential energy change of the electron). (c) Since the electrons don’t (on average) gain kinetic energy as a result of this work done, it is generally dissipated as heat. The answer is as in part (b): 12 eV or 1.9 × 10−18 J. 43. The relation P = V 2/R implies P ∝ V 2. Consequently, the power dissipated in the second case is F 150 V IJ (0.540 W) = 0135 W. . . P=G H 3.00 V K 2 44. Since P = iV, the charge is q = it = Pt/V = (7.0 W) (5.0 h) (3600 s/h)/9.0 V = 1.4 × 104 C. 45. (a) The power dissipated, the current in the heater, and the potential difference across the heater are related by P = iV. Therefore, i= P 1250 W = = 10.9 A. 115 V V (b) Ohm’s law states V = iR, where R is the resistance of the heater. Thus, R= V 115 V = = 10.6 Ω. i 10.9 A (c) The thermal energy E generated by the heater in time t = 1.0 h = 3600 s is E = Pt = (1250W)(3600s) = 4.50 ×106 J. 46. (a) Using Table 26-1 and Eq. 26-10 (or Eq. 26-11), we have 1051 ⎛ ⎞ 2.00A | E | = ρ | J | = (1.69 ×10−8 Ω⋅ m ) ⎜ = 1.69 ×10−2 V/m. −6 2⎟ 2.00 ×10 m ⎠ ⎝ (b) Using L = 4.0 m, the resistance is found from Eq. 26-16: R = ρL/A = 0.0338 Ω. The rate of thermal energy generation is found from Eq. 26-27: P = i2 R = (2.00 A)2(0.0338...
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## This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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