Question 17.5: The Cost of Lighting Up Your Life GOAL Apply the electric po...
The Cost of Lighting Up Your Life
GOAL Apply the electric power concept and calculate the cost of power usage using kilowatt-hours.
PROBLEM A circuit provides a maximum current of 20.0 A at an operating voltage of 1.20 × 10² V. (a) How many 75 W bulbs can operate with this voltage source? (b) At $0.120 per kilowatt-hour, how much does it cost to operate these bulbs for 8.00 h?
STRATEGY Find the necessary power with P = I ΔV then divide by 75.0 W per bulb to get the total number of bulbs. To find the cost, convert power to kilowatts and multiply by the number of hours, then multiply by the cost per -kilowatt-hour.
(a) Find the number of bulbs that can be lighted.
Substitute into Equation 17.8 to get the total power:
P_{\mathrm{total}}=I\Delta V=(20.0\,\mathrm{A})(1.20\times10^{2}\,\mathrm{V})=2.40\times10^{3}\,\mathrm{W}
Divide the total power by the power per bulb to get the number of bulbs:
{\mathrm{Number~of~bulbs}}\ ={\frac{P_{\mathrm{total}}}{P_{\mathrm{bulb}}}}={\frac{2.40~\times~10^{3}\,\mathrm{W}}{75.0\,\mathrm{W}}}= 32.0
(b) Calculate the cost of this electricity for an 8.00-h day.
Find the energy in kilowatt-hours:
{\mathrm{Energy}}=P t=(2.40\times10^{3}\,{\mathrm{W}}){\bigg(}{\frac{1.00\,{\mathrm{kW}}}{1.00~\times~10^{3}\,{\mathrm{W}}}}{\bigg)}(8.00\,{\mathrm{h)}}
= 19.2 kWh
Multiply the energy by the cost per kilowatt-hour:
Cost = (19.2 kWh)($0.12/kWh) = $2.30
REMARKS This amount of energy might correspond to what a small office uses in a working day, taking into account all power requirements (not just lighting). In general, resistive devices can have variable power output, depending on how the circuit is wired. Here, power outputs were specified, so such considerations were unnecessary.