If You Know the Power Rating of an Appliance and the Voltage of the Line It Is Attached to Brainly

Learning Objectives

By the cease of this department, you volition be able to:

• Calculate the power dissipated by a resistor and power supplied by a ability supply.
• Calculate the cost of electricity under diverse circumstances.

Ability in Electric Circuits

Power is associated by many people with electricity. Knowing that power is the rate of energy use or free energy conversion, what is the expression for electrical power? Ability transmission lines might come to heed. We also recall of lightbulbs in terms of their ability ratings in watts. Let us compare a 25-West seedling with a threescore-W bulb. (See Figure 1(a).) Since both operate on the aforementioned voltage, the 60-Westward bulb must draw more than electric current to have a greater ability rating. Thus the threescore-W bulb'southward resistance must be lower than that of a 25-Due west bulb. If nosotros increase voltage, we also increment power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and and so burns out. Precisely how are voltage, current, and resistance related to electric power?

Figure 1. (a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W seedling (upper right), has the higher resistance? Which draws more current? Which uses the nearly free energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a unlike color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light every bit the sixty-W seedling, but at 1/4 to 1/10 the input ability. (credit: dbgg1979, Flickr)

Electric free energy depends on both the voltage involved and the accuse moved. This is expressed most simply equally PE = qV, where q is the accuse moved and V is the voltage (or more precisely, the potential difference the charge moves through). Ability is the rate at which energy is moved, and and so electric power is

[latex]P=\frac{PE}{t}=\frac{qV}{t}\\[/latex].

Recognizing that electric current is I=q/t (note that Δt=t here), the expression for power becomes

P = IV

Electric power (P) is merely the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅V= 1 W. For instance, cars often have ane or more auxiliary power outlets with which you lot can charge a jail cell phone or other electronic devices. These outlets may exist rated at 20 A, then that the circuit can deliver a maximum ability P = 4 = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed every bit volt-amperes or even kilovolt-amperes (one kA ⋅V = 1 kW). To encounter the relationship of ability to resistance, nosotros combine Ohm's police withP = IV. Substituting I = V/R gives P= (Five/R)V=5 ²/R. Similarly, substituting Five = IR gives P = I(IR) = I²R. Iii expressions for electric ability are listed together here for convenience:

[latex]P=\text{IV}\\[/latex]

[latex]P=\frac{{Five}^{2}}{R}\\[/latex]

[latex]P={I}^{2}R\\[/latex].

Note that the showtime equation is always valid, whereas the other two tin be used just for resistors. In a simple excursion, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more than complicated circuits, P can be the power dissipated by a single device and not the full power in the excursion.) Different insights tin can be gained from the 3 different expressions for electric power. For example, P=Five ²/R implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in P=V ²/R, the effect of applying a higher voltage is perchance greater than expected. Thus, when the voltage is doubled to a 25-West bulb, its power nearly quadruples to about 100 W, called-for it out. If the seedling's resistance remained abiding, its power would be exactly 100 W, simply at the higher temperature its resistance is higher, besides.

Case 1. Calculating Ability Dissipation and Current: Hot and Cold Power

(a) Consider the examples given in Ohm's Police force: Resistance and Simple Circuits and Resistance and Resistivity. Then find the power prodigal by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it depict when common cold?

Strategy for (a)

For the hot headlight, nosotros know voltage and electric current, so we can use P = IV to detect the power. For the cold headlight, we know the voltage and resistance, so we can use P=V ²/R to find the power.

Solution for (a)

Entering the known values of current and voltage for the hot headlight, we obtain

P = 4 = (2.l A)(12.0 V) = 30.0 W.

The cold resistance was 0.350 Ω, and then the power information technology uses when first switched on is

[latex]P=\frac{{V}^{2}}{R}=\frac{{\left({12.0}\text{ V}\right)}^{2}}{0.350\text{ }\Omega }=411\text{ W}\\[/latex].

Discussion for (a)

The 30 W dissipated by the hot headlight is typical. But the 411 West when cold is surprisingly college. The initial power speedily decreases as the bulb's temperature increases and its resistance increases.

Strategy and Solution for (b)

The current when the bulb is cold can be found several different means. We rearrange one of the power equations, P=I ² R, and enter known values, obtaining

[latex]I=\sqrt{\frac{P}{R}}=\sqrt{\frac{411\text{ West}}{{0.350}\text{ }\Omega }}=34.3\text{ A}\\[/latex].

Discussion for (b)

The cold electric current is remarkably college than the steady-state value of 2.50 A, but the current will quickly decline to that value every bit the bulb'southward temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly equally a device comes on. In some cases, such every bit with electric motors, the current remains loftier for several seconds, necessitating special "slow blow" fuses.

The Cost of Electricity

The more electrical appliances y'all use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. Yous pay for the free energy used. Since P=E/t, we see that

Eastward = Pt

is the energy used by a device using power P for a fourth dimension interval t. For example, the more lightbulbs burning, the greater P used; the longer they are on, the greater t is. The energy unit on electric bills is the kilowatt-hour (kW ⋅ h), consistent with the relationshipE = Pt. Information technology is easy to estimate the toll of operating electric appliances if you lot have some thought of their power consumption rate in watts or kilowatts, the time they are on in hours, and the price per kilowatt-hour for your electric utility. Kilowatt-hours, similar all other specialized energy units such as nutrient calories, can be converted to joules. You lot can show to yourself that 1 kW ⋅ h = 3 . 6 × 10 ⁶ J .

The electric energy (E) used can be reduced either by reducing the fourth dimension of use or by reducing the ability consumption of that appliance or fixture. This volition non just reduce the price, but it volition also upshot in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical free energy used in a home or business. Well-nigh twenty% of a dwelling house's use of free energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more than efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 1(b).) Thus, a 60-West incandescent seedling can be replaced by a 15-West CFL, which has the same brightness and color. CFLs have a bent tube inside a earth or a screw-shaped tube, all connected to a standard spiral-in base that fits standard incandescent light sockets. (Original bug with colour, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat transfer from these CFLs is less, and they final upward to 10 times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of modest LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their toll is still high.

Making Connections: Energy, Power, and Time

The relationshipE = Pt is ane that you will detect useful in many dissimilar 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 fourth dimension information technology is applied. Even the radiation dose of an X-ray image is related to the ability and time of exposure.

Instance 2. Computing the Cost Effectiveness of Meaty Fluorescent Lights (CFL)

If the cost of electricity in your expanse is 12 cents per kWh, what is the full cost (capital plus operation) of using a 60-W incandescent bulb for m hours (the lifetime of that seedling) if the bulb cost 25 cents? (b) If we replace this seedling with a meaty fluorescent lite that provides the same light output, just at one-quarter the wattage, and which costs $ane.l but lasts ten times longer (10,000 hours), what will that full cost be?

Strategy

To find the operating cost, we start find the energy used in kilowatt-hours and then multiply by the toll per kilowatt-60 minutes.

Solution for (a)

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

E = Pt = (lx W)(grand h) = 60,000 W ⋅ h

In kilowatt-hours, this is

E= 60.0 kW ⋅ h.

At present the electricity cost is

price = (sixty.0 kW ⋅ h) ($0.12/kW ⋅ h) = $ 7.20.

The full toll volition be $seven.twenty for yard hours (virtually i-half year at five hours per day).

Solution for (b)

Since the CFL uses but 15 Westward and non 60 West, the electricity cost volition exist $7.20/4 = $ane.fourscore. The CFL will last 10 times longer than the incandescent, and so that the investment cost will be i/x of the seedling cost for that fourth dimension catamenia of use, or 0.1($1.l) = $0.xv. Therefore, the total cost will be $ane.95 for 1000 hours.

Word

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

Making Connections: Take-Home Experiment—Electrical Energy Employ Inventory

i) 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 college rating than a digital clock. Gauge the energy consumed by these appliances in an average day (by estimating their fourth dimension of utilise). Some appliances might only state the operating current. If the household voltage is 120 V, and so apply P = IV. 2) Check out the total wattage used in the rest rooms of your school's floor or edifice. (You might need to presume the long fluorescent lights in use are rated at 32 Westward.) 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 toll? How virtually for an entire year of weekends?

Section Summary

• Electric ability P is the rate (in watts) that energy is supplied past a source or dissipated by a device.
• Three expressions for electric power are

  [latex]P=\text{IV}\\[/latex]

  [latex]P=\frac{{V}^{2}}{R}\\[/latex]

  [latex]P={I}^{2}R\\[/latex].

• The energy used by a device with a powerP over a fourth dimensiont is E = Pt .

Conceptual Questions

ane. Why exercise incandescent lightbulbs grow dim belatedly in their lives, particularly just earlier their filaments pause?

The power prodigal in a resistor is given by P = Vⁱⁱ/R which means power decreases if resistance increases. Even so this power is also given past P = I ² R , which ways power increases if resistance increases. Explain why there is no contradiction here.

Problems & Exercises

1. What is the power of a i.00 × 10^twoMV lightning bolt having a current of  2.00 × ten^iv A?

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

3. A charge of 4.00 C of charge passes through a pocket estimator's solar cells in 4.00 h. What is the ability output, given the calculator's voltage output is 3.00 Five? (See Figure two.)

Figure two. The strip of solar cells just above the keys of this calculator catechumen light to electricity to supply its free energy needs. (credit: Evan-Amos, Wikimedia Commons)

4. How many watts does a flashlight that has6.00 × 10 ² pass through information technology in 0.500 h employ if its voltage is three.00 V?

5. Detect the power dissipated in each of these extension cords: (a) an extension cord having a 0.0600 Ω resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of 0.300 Ω.

6. Verify that the units of a volt-ampere are watts, as unsaid past the equation P = 4.

7. Show that the units 1V^two/Ω = 1W as implied by the equation P = V² /R.

viii. Prove that the units one A ⁱⁱ ⋅ Ω = 1 W , equally implied by the equation P = I ² R .

9. Verify the free energy unit equivalence that 1 kW ⋅ h = three.60 × x⁶J.

ten. Electrons in an X-ray tube are accelerated through1.00 × 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 fifteen.0 mA.

xi. An electric water heater consumes 5.00 kW for ii.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents/kW ⋅ h? See Figure 3.

Effigy 3. On-demand electric hot water heater. Heat is supplied to h2o but when needed. (credit: aviddavid, Flickr)

12. With a 1200-W toaster, how much electric energy is needed to brand a slice of toast (cooking fourth dimension = 1 infinitesimal)? At 9.0 cents/kW · h, how much does this toll?

13. What would be the maximum cost of a CFL such that the total price (investment plus operating) would be the aforementioned for both CFL and incandescent 60-Westward bulbs? Presume the toll of the incandescent bulb is 25 cents and that electricity costs 10 cents/kWh. Calculate the toll for one thousand hours, equally in the cost effectiveness of CFL example.

14. Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a xxx.0-West headlight in such a automobile? (b) What electric current flows through information technology?

15. Alkaline batteries accept the advantage of putting out constant voltage until very nearly the cease of their life. How long volition an element of group i battery rated at 1.00 A ⋅ h and 1.58 V keep a ane.00-W flashlight seedling burning?

sixteen. A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at fifteen.0 kV. (a) What is its power output? (b) What is the resistance of the path?

17. The average tv set is said to be on 6 hours per solar day. Judge the yearly toll of electricity to operate 100 million TVs, assuming their power consumption averages 150 West and the cost of electricity averages 12.0 cents/kW ⋅ h.

18. An old lightbulb draws just fifty.0 Due west, rather than its original lx.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning forth its length? Neglect whatever effects caused past temperature differences.

19. 00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a kilometer of such wire when information technology carries ane.00 × xⁱⁱA.

twenty.Integrated Concepts

Common cold vaporizers pass a electric current through water, evaporating information technology with only a small increase in temperature. Ane such habitation device is rated at 3.fifty A and utilizes 120 5 Air-conditioning with 95.0% efficiency. (a) What is the vaporization charge per unit in grams per minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Effigy 4.)

Effigy 4. This cold vaporizer passes current straight through water, vaporizing information technology directly with relatively fiddling temperature increase.

21. Integrated Concepts(a) What free energy is dissipated by a lightning bolt having a twenty,000-A current, a voltage of 1.00 × 10^twoMV and a length of 1.00 ms? (b) What mass of tree sap could exist raised from 18ºC to its boiling point and then evaporated past this free energy, assuming sap has the same thermal characteristics every bit water?

22. Integrated ConceptsWhat current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 1000 of glass, 250 m of babe formula, and 3.00×10² of aluminum from 20º C to 90º in 5.00 min?

23. Integrated ConceptsHow much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from 37º to 100and and then boil away 0.500 1000 of h2o, if information technology puts out ii.00 mA at 15.0 kV? Ignore heat transfer to the surround.

24. Integrated ConceptsHydroelectric generators (run across Figure 5) at Hoover Dam produce a maximum current of 8.00 × x³ A at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the organization at depression speed (thus its kinetic free energy does not alter) but loses 160 yard in distance. How many cubic meters per 2d are needed, assuming 85.0% efficiency?

Figure 5. Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan)

25. Integrated Concepts(a) Bold 95.0% efficiency for the conversion of electrical ability past the motor, what current must the 12.0-V batteries of a 750-kg electric machine be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00 × 10²-m-high-hill in 2.00 min at a abiding 25.0-thou/s speed while exerting 5.00 × ten²N of force to overcome air resistance and friction? (c) To travel at a abiding 25.0-m/s speed, exerting a 5.00 × 10^twoNorthward force to overcome air resistance and friction? See Figure half-dozen.

Effigy 6. This REVAi, an electrical car, gets recharged on a street in London. (credit: Frank Hebbert)

26. Integrated ConceptsA lite-runway commuter railroad train draws 630 A of 650-Five DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to reach 20.0 m/s starting from remainder if its loaded mass is five.xxx × 10⁴kg, assuming 95.0% efficiency and constant power? (c) Notice its average acceleration. (d) Hash out how the acceleration you constitute for the light-rail train compares to what might exist typical for an automobile.

27. Integrated Concepts(a) An aluminum power manual line has a resistance of 0.0580 Ω/km. What is its mass per kilometer? (b) What is the mass per kilometer of a copper line having the same resistance? A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.

28. Integrated Concepts(a) An immersion heater utilizing 120 Five can raise the temperature of a 1.00 × 10²-g aluminum cup containing 350 one thousand of h2o from 20º C to 95º C in 2.00 min. Observe its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Hash out the practical limits to speeding the heating past lowering the resistance.

29. Integrated Concepts(a) What is the price of heating a hot tub containing 1500 kg of water from 10º C to 40º C, assuming 75.0% efficiency to account for estrus transfer to the environment? The toll of electricity is 9 cents/kW ⋅ h. (b) What electric current was used by the 220-V AC electric heater, if this took 4.00 h?

thirty. Unreasonable Results(a) What electric current is needed to transmit one.00 × 10²MW of power at 480 Five? (b) What power is dissipated by the transmission lines if they take a one.00 – Ω resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which bounds are inconsistent?

31. Unreasonable Results(a) What electric current is needed to transmit 1.00 × ten²MW of power at 10.0 kV? (b) Find the resistance of ane.00 km of wire that would cause a 0.0100% ability loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (east) Which assumptions are unreasonable, or which premises are inconsistent?

32. Construct Your Ain ProblemConsider an electrical immersion heater used to heat a loving cup of h2o to make tea. Construct a problem in which you calculate the needed resistance of the heater so that information technology increases the temperature of the water and cup in a reasonable amount of time. Also calculate the price of the electrical energy used in your process. Amongst the things to exist considered are the voltage used, the masses and heat capacities involved, estrus losses, and the fourth dimension over which the heating takes place. Your instructor may wish for y'all to consider a thermal rubber switch (maybe bimetallic) that will halt the process before damaging temperatures are reached in the immersion unit.

Glossary

electric ability:

the rate at which electrical energy is supplied by a source or dissipated past a device; it is the product of current times voltage

Selected Solutions to Bug & Exercises

1. 2 . 00 × 10 ¹² W

5. (a) 1.50 W (b) 7.l W

7. [latex]\frac{{V}^{2}}{\Omega }=\frac{{V}^{2}}{\text{5/A}}=\text{AV}=\left(\frac{C}{south}\right)\left(\frac{J}{C}\right)=\frac{J}{s}=1\text{Westward}\\[/latex]

9. [latex]ane\text{kW}\cdot \text{h=}\left(\frac{ane\times {\text{10}}^{3}\text{J}}{\text{one s}}\correct)\left(one h\right)\left(\frac{\text{3600 s}}{\text{1 h}}\correct)=3\text{.}\text{60}\times {\text{x}}^{half dozen}\text{J}\\[/latex]

xi. $438/y

xiii. $6.25

15. ane.58 h

17. $iii.94 billion/yr

19. 25.five W

21.(a) 2.00 × ten⁹J (b) 769 kg

23. 45.0 s

25. (a) 343 A (b) 2.17 × 10^threeA (c) 1.10 × 10ⁱⁱⁱA

27. (a) i.23 × x³kg (b) 2.64 × tenⁱⁱⁱkg

29. (a) 2.08 × 10⁵A
(b) 4.33 × 10⁴MW
(c) The transmission lines misemploy more power than they are supposed to transmit.
(d) A voltage of 480 V is unreasonably low for a transmission voltage. Long-distance transmission lines are kept at much college voltages (often hundreds of kilovolts) to reduce power losses.

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