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8.11 Electric power and energy  (Page 3/4)

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Making connections: energy, power, and time

The relationship E = Pt size 12{E = ital "Pt"} {} is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiation dose of an X-ray image is related to the power and time of exposure.

Calculating the cost effectiveness of compact fluorescent lights (cfl)

If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be?

Strategy

To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour.

Solution for (a)

The energy used in kilowatt-hours is found by entering the power and time into the expression for energy:

E = Pt = ( 60 W ) ( 1000 h ) = 60,000 W ⋅ h. size 12{E = ital "Pt" = \( "60 W" \) \( "1000 h" \) =" 60,000 W " cdot " h."} {}

In kilowatt-hours, this is

E = 60 . 0 kW ⋅ h. size 12{E =" 60" "." "0 kW " cdot " h."} {}

Now the electricity cost is

cost = ( 60.0 kW ⋅ h ) ( $0.12 /kW ⋅ h ) = $ 7.20.

The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day).

Solution for (b)

Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95 for 1000 hours.

Discussion

Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business must include for replacing the incandescent bulbs more often has not been figured in here.

Making connections: take-home experiment—electrical energy use inventory

1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use P = IV size 12{P = ital "IV"} {} . 2) Check out the total wattage used in the rest rooms of your school’s floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends?

Section summary

  • Electric power P size 12{P} {} is the rate (in watts) that energy is supplied by a source or dissipated by a device.
  • Three expressions for electrical power are
    P = IV, size 12{P = ital "IV,"} {}
    P = V 2 R , size 12{P = { {V rSup { size 8{2} } } over {R} } ","} {}

    and

    P = I 2 R . size 12{P = I rSup { size 8{2} } R"."} {}
  • The energy used by a device with a power P size 12{P} {} over a time t size 12{t} {} is E = Pt size 12{E = ital "Pt"} {} .

Conceptual questions

Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break?

The power dissipated in a resistor is given by P = V 2 / R size 12{P = V rSup { size 8{2} } /R} {} , which means power decreases if resistance increases. Yet this power is also given by P = I 2 R size 12{P = I rSup { size 8{2} } R} {} , which means power increases if resistance increases. Explain why there is no contradiction here.

Problem exercises

What is the power of a 1.00 × 10 2 MV lightning bolt having a current of 2.00 × 10 4 A ?

2 . 00 × 10 12 W size 12{2 "." "00"´"10" rSup { size 8{"12"} } " W"} {}

What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup?

A charge of 4.00 C of charge passes through a pocket calculator’s solar cells in 4.00 h. What is the power output, given the calculator’s voltage output is 3.00 V? (See [link] .)

Photograph of a small calculator having a strip of solar cells just above the keys.
The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons)

How many watts does a flashlight that has 6.00 × 10 2 C pass through it in 0.500 h use if its voltage is 3.00 V?

Find the power dissipated in each of these extension cords: (a) an extension cord having a 0 . 0600 - Ω size 12{0 "." "0600-" %OMEGA } {} resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of 0 . 300 Ω . size 12{0 "." "300" %OMEGA } {}

(a) 1.50 W

(b) 7.50 W

Verify that the units of a volt-ampere are watts, as implied by the equation P = IV size 12{P = ital "IV"} {} .

Show that the units 1 V 2 / Ω = 1 W size 12{1" V" rSup { size 8{2} } / %OMEGA =1" W"} {} , as implied by the equation P = V 2 / R size 12{P = V rSup { size 8{2} } /R} {} .

V 2 Ω = V 2 V/A = AV = C s J C = J s = 1 W

Show that the units 1 A 2 ⋅ Ω = 1 W size 12{1" V" rSup { size 8{2} } / %OMEGA =1" W"} {} , as implied by the equation P = I 2 R size 12{P = I rSup { size 8{2} } R} {} .

Verify the energy unit equivalence that 1 kW ⋅ h = 3 . 60 × 10 6 J size 12{1" kW" cdot "h = 3" "." "60"´"10" rSup { size 8{6} } " J"} {} .

1 kW ⋅ h= 1 × 10 3 J 1 s 1 h 3600 s 1 h = 3 . 60 × 10 6 J size 12{1" kW" cdot "h=" left ( { {1 times "10" rSup { size 8{3} } " J"} over {"1 s"} } right ) left (1" h" right ) left ( { {"3600"" s"} over {"1 h"} } right )=3 "." "60" times "10" rSup { size 8{6} } " J"} {}

Electrons in an X-ray tube are accelerated through 1.00 × 10 2 kV and directed toward a target to produce X-rays. Calculate the power of the electron beam in this tube if it has a current of 15.0 mA.

An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents /kW ⋅ h size 12{"12" "." 0" cents/kW" cdot h} {} ? See [link] .

Photograph of an electric hot water heater connected to the electric and water supply
On-demand electric hot water heater. Heat is supplied to water only when needed. (credit: aviddavid, Flickr)

$438/y

With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At 9.0 cents/kW · h , how much does this cost?

Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a car? (b) What current flows through it?

Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at 1 . 00 A ⋅ h size 12{1 "." "00"" A" cdot h} {} and 1.58 V keep a 1.00-W flashlight bulb burning?

1.58 h

A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path?

The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages 12 . 0 cents/kW ⋅ h size 12{"12" "." 0" cents/kW" cdot "h"} {} .

$3.94 billion/year
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Source:  OpenStax, Concepts of physics with linear momentum. OpenStax CNX. Aug 11, 2016 Download for free at http://legacy.cnx.org/content/col11960/1.9
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