19_InstSolManual_PDF_Part8

# 19_InstSolManual_PDF_Part8 - 19-8 Chapter 19 V2 R Solve(a...

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19.43. Set Up: A kWh is power of 1 kW for a time of 1 h. Solve: (a) In 3.0 yr the bulbs are on for compact bulb The energy used is The cost of this energy is One bulb will last longer than this. The bulb cost is \$11.00, so the total cost is \$19.08. incandescent The energy used is The cost of this energy is Six bulbs will be used during this time and the bulb cost will be \$4.50. The total cost will be \$39.54. (b) The compact bulb will save (c) Reflect: The initial cost of the bulb is much greater for the compact fluorescent bulb but the savings soon repay the cost of the bulb. The compact bulb should last for over six years, so over a 6 year period the savings per year will be even greater. 19.44. Set Up: The total resistance is the resistance of the person plus the internal resistance of the power supply. Solve: (a) (b) (c) The resistance of the power supply would need to be 19.45. Set Up: Solve: (a) (b) (c) 19.46. Set Up: For resistors in parallel, For resistors in series, These rules may have to be applied in several ways.
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• Spring '07
• Shoberg
• Energy, Resistor, Trigraph, Electrical resistance, Series and parallel circuits, Bulb

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