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 75.0 W bulbs can operate with this voltage source? (b) At

Question

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 [latex]20.0 \mathrm{~A}[/latex] at an operating voltage of [latex]1.20 imes 10^{2} \mathrm{~V}[/latex]. (a) How many [latex]75.0 \mathrm{~W}[/latex] bulbs can operate with this voltage source? (b) At [latex]\$ 0.120[/latex] per kilowatt-hour, how much does it cost to operate these bulbs for [latex]8.00[/latex] h?

STRATEGY Find the necessary power with [latex]P=I \Delta V[/latex] then divide by [latex]75.0 \mathrm{~W}[/latex] 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.

Accepted Answer

(a) Find the number of bulbs that can be lighted.

Substitute into Equation [latex]17.8[/latex]

[latex]P = IΔV[/latex] [17.8]

to get the total power:

[latex]P_{ ext {total }}=I \Delta V=(20.0 \mathrm{~A})\left(1.20 imes 10^{2} \mathrm{~V} ight)=2.40 imes 10^{3} \mathrm{~W} [/latex]

Divide the total power by the power per bulb to get the number of bulbs:

[latex] ext { Number of bulbs }=\frac{P_{ ext {total }}}{P_{ ext {bulb }}}=\frac{2.40 imes 10^{3} \mathrm{~W}}{75.0 \mathrm{~W}}=32.0 [/latex]

(b) Calculate the cost of this electricity for an 8.00-h day. Find the energy in kilowatt-hours:

[latex]\begin{aligned} ext { Energy } &=P t=\left(2.40 imes 10^{3} \mathrm{~W} ight)\left(\frac{1.00 \mathrm{~kW}}{1.00 imes 10^{3} \mathrm{~W}} ight)(8.00 \mathrm{~h}) \&=19.2 \mathrm{kWh} \end{aligned}[/latex]

Multiply the energy by the cost per kilowatt-hour:

[latex] ext { Cost } =(19.2 \mathrm{kWh})(\$ 0.12 / \mathrm{kWh})=\$ 2.30[/latex]

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.

Question 17.5: THE COST OF LIGHTING UP YOUR LIFE GOAL Apply the electric po...

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