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 bulbThe energy used isThe cost of this energy isOne bulb will last longer than this. The bulb cost is $11.00, so the total cost is $19.08.incandescentThe energy used is The cost of this energy isSix bulbs will be used during this time and the bulb cost will be $4.50. Thetotal 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 thecost of the bulb. The compact bulb should last for over six years, so over a 6 year period the savings per year will beeven greater.19.44.Set Up:The total resistance is the resistance of the person plus the internal resistance ofthe power supply.Solve: (a)(b)(c)The resistance of the power supply would need to be19.45.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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Resistor, Trigraph, Electrical resistance, Series and parallel circuits, Bulb