What Is The Power Used By An Electric Motor That Draws A Current Of 15 A From A 240v Power Source

Learning Objectives

By the end of this section, you volition be able to:

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

Power in Electrical Circuits

Power is associated by many people with electricity. Knowing that power is the rate of free energy use or energy conversion, what is the expression for electric ability? Power transmission lines might come up to listen. We likewise think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-Due west bulb with a threescore-Westward bulb. (See Figure i(a).) Since both operate on the same voltage, the 60-Westward bulb must draw more current to have a greater power rating. Thus the threescore-W bulb's resistance must exist lower than that of a 25-West bulb. If we increase voltage, we also increase 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 then burns out. Precisely how are voltage, current, and resistance related to electrical power?

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

Electric energy depends on both the voltage involved and the charge moved. This is expressed virtually simply as PE = qV, where q is the accuse moved and V is the voltage (or more precisely, the potential deviation the charge moves through). Power is the rate at which energy is moved, and so electrical power is

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

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

P = Iv

Electric power (P) is simply 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, one A ⋅Five= 1 Westward. For case, cars often have one or more auxiliary power outlets with which you can charge a jail cell phone or other electronic devices. These outlets may exist rated at 20 A, and then that the circuit can deliver a maximum power P = IV = (twenty A)(12 V) = 240 W. In some applications, electric power may be expressed equally volt-amperes or fifty-fifty kilovolt-amperes (1 kA ⋅Five = one kW). To see the relationship of power to resistance, we combine Ohm's law withP = IV. Substituting I = 5/R gives P= (V/R)V=V ²/R. Similarly, substituting V = IR gives P = I(IR) = IⁱⁱR. Three expressions for electrical power are listed together hither for convenience:

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

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

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

Note that the outset equation is always valid, whereas the other ii can be used only for resistors. In a simple circuit, 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 complicated circuits, P can be the power dissipated by a unmarried device and not the total power in the circuit.) Different insights can be gained from the three different expressions for electric power. For example, P=5 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 ⁱⁱ/R, the upshot of applying a higher voltage is maybe greater than expected. Thus, when the voltage is doubled to a 25-W seedling, its ability nearly quadruples to about 100 Westward, called-for it out. If the bulb's resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.

Instance i. Calculating Ability Dissipation and Electric current: Hot and Cold Power

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

Strategy for (a)

For the hot headlight, nosotros know voltage and current, so we can employ P = IV to discover the ability. For the cold headlight, nosotros know the voltage and resistance, and then we can use P=Five ²/R to find the power.

Solution for (a)

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

P = Four = (2.fifty A)(12.0 V) = xxx.0 Due west.

The cold resistance was 0.350 Ω, and so the power it uses when showtime switched on is

[latex]P=\frac{{5}^{2}}{R}=\frac{{\left({12.0}\text{ V}\right)}^{2}}{0.350\text{ }\Omega }=411\text{ Due west}\\[/latex].

Give-and-take for (a)

The 30 W dissipated by the hot headlight is typical. Merely the 411 Due west when cold is surprisingly higher. The initial ability quickly decreases equally the bulb's temperature increases and its resistance increases.

Strategy and Solution for (b)

The electric current when the seedling is cold can be found several dissimilar means. We rearrange 1 of the power equations, P=I ² R, and enter known values, obtaining

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

Discussion for (b)

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

The Cost of Electricity

The more electric appliances you lot use and the longer they are left on, the college your electric nib. This familiar fact is based on the relationship between free energy and ability. You pay for the energy used. Since P=Eastward/t, we see that

East = Pt

is the energy used by a device using ability P for a time interval t. For case, the more than lightbulbs called-for, the greater P used; the longer they are on, the greater t is. The free energy unit on electric bills is the kilowatt-hour (kW ⋅ h), consequent with the human relationshipE = Pt. It is easy to approximate the cost of operating electrical appliances if you have some idea 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, like all other specialized free energy units such every bit food calories, can be converted to joules. Yous can testify to yourself that 1 kW ⋅ h = iii . half dozen × x ^six J .

The electrical energy (East) used can be reduced either by reducing the fourth dimension of utilize or past reducing the power consumption of that apparatus or fixture. This volition not only reduce the cost, only information technology volition also upshot in a reduced affect on the environment. Improvements to lighting are some of the fastest means to reduce the electrical energy used in a abode or concern. About twenty% of a home'southward use of free energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFL). (Meet Figure i(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-Westward CFL, which has the same brightness and color. CFLs take a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base of operations that fits standard incandescent calorie-free sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The oestrus transfer from these CFLs is less, and they last up 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 pocket-sized LED bulbs) are even more efficient (twice that of CFLs) and terminal 5 times longer than CFLs. However, their cost is nevertheless high.

Making Connections: Free energy, Power, and Time

The human relationshipDue east = Pt is one that yous will detect useful in many dissimilar contexts. The energy your body uses in exercise is related to the power level and duration of your activeness, for example. The amount of heating by a ability source is related to the power level and time it is applied. Even the radiation dose of an X-ray prototype is related to the power and time of exposure.

Case 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 performance) of using a 60-West incandescent seedling for m hours (the lifetime of that seedling) if the bulb cost 25 cents? (b) If nosotros replace this bulb with a compact fluorescent light that provides the same light output, only at one-quarter the wattage, and which costs $1.50 simply lasts x times longer (10,000 hours), what will that total cost be?

Strategy

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

Solution for (a)

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

E = Pt = (60 Due west)(1000 h) = sixty,000 W ⋅ h

In kilowatt-hours, this is

E= threescore.0 kW ⋅ h.

Now the electricity price is

cost = (threescore.0 kW ⋅ h) ($0.12/kW ⋅ h) = $ 7.20.

The total cost will be $7.twenty for 1000 hours (about one-one-half year at 5 hours per day).

Solution for (b)

Since the CFL uses only 15 West and not 60 W, the electricity cost volition be $7.20/four = $1.80. The CFL volition final ten times longer than the incandescent, so that the investment cost volition be one/10 of the seedling toll for that time menstruum of use, or 0.1($1.l) = $0.15. Therefore, the total price volition be $ane.95 for grand hours.

Word

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

Making Connections: Take-Domicile Experiment—Electrical Energy Apply Inventory

1) Brand a listing of the power ratings on a range of appliances in your home or room. Explicate why something similar a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average mean solar day (past estimating their time of use). Some appliances might simply state the operating electric current. If the household voltage is 120 V, then use P = Four. 2) Cheque out the total wattage used in the rest rooms of your school's flooring or building. (You might need to presume the long fluorescent lights in use are rated at 32 Westward.) Suppose that the building was airtight all weekend and that these lights were left on from 6 p.1000. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends?

Section Summary

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

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

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

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

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

Conceptual Questions

one. Why do incandescent lightbulbs abound dim late in their lives, particularly just before their filaments intermission?

The power dissipated in a resistor is given by P = V²/R which means power decreases if resistance increases. Yet this ability is also given by P = I ² R , which means power increases if resistance increases. Explicate why there is no contradiction here.

Issues & Exercises

1. What is the power of a one.00 × ten²MV lightning bolt having a current of  2.00 × x⁴ A?

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

iii. A charge of 4.00 C of charge passes through a pocket calculator'south solar cells in 4.00 h. What is the ability output, given the calculator's voltage output is 3.00 V? (See Figure 2.)

Effigy 2. The strip of solar cells just above the keys of this figurer convert calorie-free 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 it in 0.500 h use if its voltage is 3.00 Five?

5. Find the ability prodigal in each of these extension cords: (a) an extension cord having a 0.0600 Ω resistance and through which v.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of 0.300 Ω.

half dozen. Verify that the units of a volt-ampere are watts, as implied by the equation P = IV.

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

8. Bear witness that the units i A ^two ⋅ Ω = i W , as implied past the equation P = I ⁱⁱ R .

9. Verify the energy unit equivalence that ane kW ⋅ h = three.lx × 10^sixJ.

10. Electrons in an X-ray tube are accelerated through1.00 × 10 ² kV and directed toward a target to produce X-rays. Calculate the ability of the electron beam in this tube if information technology has a electric current of 15.0 mA.

11. An electrical water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running information technology for ane year if electricity costs 12.0 cents/kW ⋅ h? See Figure three.

Figure 3. On-demand electrical hot water heater. Estrus is supplied to h2o only when needed. (credit: aviddavid, Flickr)

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

xiii. What would exist the maximum cost of a CFL such that the total cost (investment plus operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the toll of the incandescent bulb is 25 cents and that electricity costs 10 cents/kWh. Summate the cost for one thousand hours, as in the cost effectiveness of CFL example.

fourteen. Some makes of older cars accept half dozen.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a motorcar? (b) What electric current flows through it?

fifteen. Alkaline metal batteries accept the reward of putting out constant voltage until very nearly the finish of their life. How long will an alkaline bombardment rated at one.00 A ⋅ h and 1.58 5 keep a 1.00-Westward flashlight bulb burning?

16. A cauterizer, used to stop haemorrhage in surgery, puts out two.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path?

17. The average television set is said to be on 6 hours per twenty-four hours. Estimate the yearly cost of electricity to operate 100 meg TVs, assuming their power consumption averages 150 W and the toll of electricity averages 12.0 cents/kW ⋅ h.

18. An one-time lightbulb draws only 50.0 W, rather than its original sixty.0 W, due to evaporative thinning of its filament. By what gene is its diameter reduced, bold uniform thinning along its length? Neglect whatever effects caused past temperature differences.

19. 00-gauge copper wire has a diameter of 9.266 mm. Calculate the ability loss in a kilometer of such wire when it carries one.00 × ten^twoA.

20.Integrated Concepts

Cold vaporizers pass a current through water, evaporating information technology with only a small increase in temperature. 1 such home device is rated at 3.50 A and utilizes 120 V Ac with 95.0% efficiency. (a) What is the vaporization rate in grams per infinitesimal? (b) How much water must yous put into the vaporizer for 8.00 h of overnight performance? (See Figure 4.)

Figure 4. This cold vaporizer passes current directly through water, vaporizing information technology directly with relatively little temperature increase.

21. Integrated Concepts(a) What free energy is dissipated by a lightning commodities having a xx,000-A electric current, a voltage of 1.00 × 10²MV and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18ºC to its humid point and so evaporated past this energy, assuming sap has the same thermal characteristics as water?

22. Integrated ConceptsWhat current must be produced by a 12.0-Five bombardment-operated bottle warmer in club to heat 75.0 g of glass, 250 g 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 one.00 g of tissue from 37º to 100and then boil away 0.500 g of h2o, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings.

24. Integrated ConceptsHydroelectric generators (encounter Figure 5) at Hoover Dam produce a maximum electric current of viii.00 × ten³ A at 250 kV. (a) What is the ability output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does non alter) but loses 160 grand in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency?

Effigy five. Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan)

25. Integrated Concepts(a) Assuming 95.0% efficiency for the conversion of electrical power by 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/south in 1.00 min? (b) To climb a two.00 × 10^two-m-loftier-hill in 2.00 min at a constant 25.0-m/southward speed while exerting 5.00 × 10²N of strength to overcome air resistance and friction? (c) To travel at a constant 25.0-grand/s speed, exerting a 5.00 × 10²Northward force to overcome air resistance and friction? See Figure 6.

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

26. Integrated ConceptsA light-rail driver train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to attain xx.0 m/due south starting from balance if its loaded mass is five.30 × x⁴kg, assuming 95.0% efficiency and abiding power? (c) Find its average acceleration. (d) Discuss how the acceleration y'all constitute for the low-cal-track train compares to what might be 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. Hash out the practical limits to speeding the heating by lowering the resistance.

28. Integrated Concepts(a) An immersion heater utilizing 120 5 can enhance the temperature of a ane.00 × 10ⁱⁱ-1000 aluminum cup containing 350 thousand of h2o from 20º C to 95º C in two.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Talk over the practical limits to speeding the heating past lowering the resistance.

29. Integrated Concepts(a) What is the cost of heating a hot tub containing 1500 kg of water from 10º C to 40º C, bold 75.0% efficiency to account for heat transfer to the surroundings? The price of electricity is ix cents/kW ⋅ h. (b) What current was used by the 220-V Ac electric heater, if this took four.00 h?

30. Unreasonable Results(a) What current is needed to transmit 1.00 × 10²MW of power at 480 Five? (b) What power is prodigal past the transmission lines if they have a i.00 – Ω resistance? (c) What is unreasonable about this issue? (d) Which assumptions are unreasonable, or which premises are inconsistent?

31. Unreasonable Results(a) What electric current is needed to transmit 1.00 × x^twoMW of power at 10.0 kV? (b) Find the resistance of 1.00 km of wire that would crusade a 0.0100% power loss. (c) What is the diameter of a one.00-km-long copper wire having this resistance? (d) What is unreasonable most these results? (eastward) Which assumptions are unreasonable, or which bounds are inconsistent?

32. Construct Your Own ProblemConsider an electric immersion heater used to estrus a loving cup of water to make tea. Construct a problem in which you summate the needed resistance of the heater so that it increases the temperature of the water and cup in a reasonable amount of time. Besides calculate the cost of the electric free energy used in your process. Amid the things to exist considered are the voltage used, the masses and heat capacities involved, heat losses, and the fourth dimension over which the heating takes place. Your teacher may wish for you to consider a thermal safety switch (perhaps bimetallic) that volition halt the process before damaging temperatures are reached in the immersion unit.

Glossary

electric power:

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

Selected Solutions to Problems & Exercises

1. two . 00 × 10 ¹² Westward

5. (a) 1.l W (b) seven.l W

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

9. [latex]1\text{kW}\cdot \text{h=}\left(\frac{1\times {\text{ten}}^{three}\text{J}}{\text{1 s}}\correct)\left(1 h\right)\left(\frac{\text{3600 southward}}{\text{1 h}}\right)=3\text{.}\text{threescore}\times {\text{10}}^{6}\text{J}\\[/latex]

eleven. $438/y

13. $half dozen.25

15. 1.58 h

17. $3.94 billion/year

xix. 25.5 W

21.(a) two.00 × x^nineJ (b) 769 kg

23. 45.0 south

25. (a) 343 A (b) 2.17 × 10³A (c) 1.x × xⁱⁱⁱA

27. (a) ane.23 × 10³kg (b) 2.64 × ten³kg

29. (a) ii.08 × 10⁵A
(b) four.33 × 10⁴MW
(c) The transmission lines misemploy more power than they are supposed to transmit.
(d) A voltage of 480 Five is unreasonably depression for a transmission voltage. Long-distance transmission lines are kept at much higher voltages (frequently hundreds of kilovolts) to reduce power losses.

Source: https://courses.lumenlearning.com/physics/chapter/20-4-electric-power-and-energy/

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