A 150-w Light Bulb Is Designed To Operate At 110 V Dc. How Much Current Does It Draw?

Learning Objectives

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

• Calculate the power dissipated by a resistor and power supplied by a power 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 utilisation or energy conversion, what is the expression for electric power? Power transmission lines might come to listen. We also retrieve of lightbulbs in terms of their ability ratings in watts. Permit us compare a 25-West seedling with a threescore-Westward bulb. (See Figure 1(a).) Since both operate on the same voltage, the lx-W bulb must describe more electric current to have a greater ability rating. Thus the 60-W seedling'due south resistance must be lower than that of a 25-Westward bulb. If we increment voltage, we also increase power. For example, when a 25-West bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric ability?

Figure 1. (a) Which of these lightbulbs, the 25-West bulb (upper left) or the 60-W bulb (upper correct), has the higher resistance? Which draws more current? Which uses the well-nigh energy? Tin can you tell from the colour that the 25-W filament is cooler? Is the brighter bulb a different color and if and so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This meaty fluorescent low-cal (CFL) puts out the aforementioned intensity of lite as the 60-W seedling, only at 1/four to i/10 the input power. (credit: dbgg1979, Flickr)

Electric energy depends on both the voltage involved and the charge moved. This is expressed nearly simply equally PE = qV, where q is the charge moved and V is the voltage (or more than precisely, the potential deviation the charge moves through). Power is the rate at which energy is moved, and so electric 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 production of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, ability has units of joules per 2nd, or watts. Thus, 1 A ⋅V= 1 W. For example, cars oft have one or more than auxiliary power outlets with which you lot can charge a cell phone or other electronic devices. These outlets may exist rated at 20 A, so that the excursion can deliver a maximum ability P = IV = (xx A)(12 V) = 240 Westward. In some applications, electrical ability may be expressed equally volt-amperes or even kilovolt-amperes (1 kA ⋅V = 1 kW). To see the relationship of power to resistance, nosotros combine Ohm's law withP = IV. Substituting I = V/R gives P= (V/R)V=5 ²/R. Similarly, substituting V = IR gives P = I(IR) = I²R. Three expressions for electrical 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].

Notation that the beginning equation is ever valid, whereas the other 2 can exist used only for resistors. In a simple excursion, with one voltage source and a unmarried resistor, the ability supplied by the voltage source and that dissipated past the resistor are identical. (In more complicated circuits, P can be the power prodigal by a single device and non the total ability in the circuit.) Unlike insights can exist gained from the three different expressions for electric power. For example, P=V ⁱⁱ/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=Five ⁱⁱ/R, the effect of applying a higher voltage is maybe greater than expected. Thus, when the voltage is doubled to a 25-Westward bulb, its power nearly quadruples to nigh 100 Westward, called-for information technology 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. Computing Ability Dissipation and Current: Hot and Cold Power

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

Strategy for (a)

For the hot headlight, nosotros know voltage and current, so nosotros can use P = IV to discover the power. For the cold headlight, we know the voltage and resistance, so nosotros can use P=5 ^two/R to detect the power.

Solution for (a)

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

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

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

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

Discussion for (a)

The xxx W dissipated by the hot headlight is typical. Just the 411 W when cold is surprisingly college. The initial ability chop-chop decreases as the bulb'southward temperature increases and its resistance increases.

Strategy and Solution for (b)

The electric current when the bulb is cold can be found several different ways. 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{ Westward}}{{0.350}\text{ }\Omega }}=34.iii\text{ A}\\[/latex].

Give-and-take for (b)

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

The Cost of Electricity

The more electrical appliances you utilize and the longer they are left on, the higher your electrical bill. This familiar fact is based on the human relationship between free energy and power. You pay for the energy used. Since P=Due east/t, we see that

E = Pt

is the energy used by a device using power P for a time interval t. For instance, 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 cost of operating electrical appliances if yous have some thought of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized free energy units such as food calories, can be converted to joules. You tin can evidence to yourself that i kW ⋅ h = 3 . 6 × 10 ^six J .

The electrical energy (East) used tin be reduced either by reducing the time of use or past reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but information technology volition also result in a reduced impact on the environs. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business organisation. About 20% of a dwelling's employ of free energy goes to lighting, while the number for commercial establishments is closer to forty%. 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). (Come across Effigy one(b).) Thus, a 60-Westward incandescent seedling can be replaced past a 15-W CFL, which has the aforementioned brightness and color. CFLs have a aptitude tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent calorie-free sockets. (Original problems with colour, flicker, shape, and high initial investment for CFLs take been addressed in contempo years.) The heat transfer from these CFLs is less, and they final 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-size LED bulbs) are even more efficient (twice that of CFLs) and last v times longer than CFLs. Nevertheless, their toll is still high.

Making Connections: Energy, Power, and Time

The relationshipEastward = Pt is one that you volition detect useful in many different contexts. The energy your trunk uses in do is related to the ability level and elapsing of your activeness, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiations dose of an 10-ray image is related to the ability and time of exposure.

Instance ii. Calculating the Cost Effectiveness of Meaty Fluorescent Lights (CFL)

If the cost of electricity in your area is 12 cents per kWh, what is the total toll (majuscule plus performance) of using a 60-Westward incandescent bulb for yard hours (the lifetime of that bulb) if the seedling toll 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the aforementioned light output, but at one-quarter the wattage, and which costs $1.50 simply lasts ten times longer (ten,000 hours), what will that total cost be?

Strategy

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

Solution for (a)

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

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

In kilowatt-hours, this is

E= 60.0 kW ⋅ h.

Now the electricity price is

cost = (sixty.0 kW ⋅ h) ($0.12/kW ⋅ h) = $ 7.xx.

The total cost volition exist $7.20 for 1000 hours (about one-half yr at five hours per day).

Solution for (b)

Since the CFL uses only fifteen Westward and not sixty W, the electricity price will be $7.twenty/4 = $1.80. The CFL will terminal 10 times longer than the incandescent, then that the investment cost volition exist 1/10 of the bulb cost for that time period of use, or 0.1($1.fifty) = $0.15. Therefore, the total toll will exist $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 hither.

Making Connections: Take-Habitation Experiment—Electrical Energy utilisation Inventory

i) Make a list of the ability ratings on a range of appliances in your home or room. Explain why something similar a toaster has a higher rating than a digital clock. Approximate the free energy consumed past these appliances in an average day (by estimating their time of utilize). Some appliances might only state the operating current. If the household voltage is 120 V, then apply P = IV. ii) Check out the total wattage used in the rest rooms of your schoolhouse's floor or edifice. (Y'all might need to assume the long fluorescent lights in utilise are rated at 32 Due west.) Suppose that the building was airtight all weekend and that these lights were left on from 6 p.k. Friday until 8 a.m. Mon. What would this oversight price? How about for an entire year of weekends?

Section Summary

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

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

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

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

• The free energy used by a device with a abilityP over a timet is E = Pt .

Conceptual Questions

1. Why do incandescent lightbulbs abound dim late in their lives, particularly simply earlier their filaments pause?

The ability dissipated in a resistor is given by P = 5^two/R which ways power decreases if resistance increases. Notwithstanding this power is also given by P = I ² R , which means power increases if resistance increases. Explain why there is no contradiction hither.

Problems & Exercises

1. What is the power of a 1.00 × 10ⁱⁱMV lightning bolt having a current of  two.00 × 10⁴ A?

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

3. A accuse of 4.00 C of charge passes through a pocket calculator'due south solar cells in 4.00 h. What is the power output, given the calculator's voltage output is iii.00 Five? (Run into Figure ii.)

Figure ii. The strip of solar cells just to a higher place the keys of this calculator catechumen calorie-free to electricity to supply its free energy needs. (credit: Evan-Amos, Wikimedia Eatables)

4. How many watts does a flashlight that has6.00 × 10 ⁱⁱ pass through information technology in 0.500 h use if its voltage is 3.00 V?

5. Find the power dissipated in each of these extension cords: (a) an extension string 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.

six. Verify that the units of a volt-ampere are watts, equally unsaid by the equation P = Four.

vii. Show that the units 1V²/Ω = 1W as implied past the equation P = V² /R.

viii. Show that the units 1 A ⁱⁱ ⋅ Ω = 1 W , as unsaid by the equation P = I ² R .

9. Verify the energy unit equivalence that 1 kW ⋅ h = 3.threescore × 10^{half dozen}J.

10. Electrons in an X-ray tube are accelerated throughone.00 × 10 ^two kV and directed toward a target to produce X-rays. Summate the power of the electron beam in this tube if it has a current of 15.0 mA.

11. An electric water heater consumes v.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? Encounter Effigy 3.

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

12. With a 1200-Westward toaster, how much electrical energy is needed to brand a piece of toast (cooking time = ane infinitesimal)? At 9.0 cents/kW · h, how much does this cost?

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

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

15. Alkaline batteries have the advantage of putting out constant voltage until very well-nigh the cease of their life. How long will an alkaline battery rated at i.00 A ⋅ h and ane.58 V keep a 1.00-Westward flashlight bulb burning?

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

17. The average television is said to be on half-dozen hours per day. Judge the yearly cost of electricity to operate 100 one thousand thousand TVs, assuming their power consumption averages 150 West and the toll of electricity averages 12.0 cents/kW ⋅ h.

18. An old lightbulb draws simply l.0 Due west, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is its bore reduced, assuming uniform thinning along its length? Neglect any 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 × 10ⁱⁱA.

20.Integrated Concepts

Cold vaporizers pass a electric current through h2o, evaporating information technology with only a pocket-size increase in temperature. One 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 minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Figure 4.)

Figure iv. This cold vaporizer passes electric current directly through h2o, vaporizing it directly with relatively little temperature increment.

21. Integrated Concepts(a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of 1.00 × 10ⁱⁱMV and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18ºC to its humid point and then evaporated past this free energy, assuming sap has the aforementioned thermal characteristics as water?

22. Integrated ConceptsWhat current must be produced by a 12.0-5 battery-operated bottle warmer in order to oestrus 75.0 thousand of glass, 250 g of baby formula, and 3.00×10^two 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 ane.00 g of tissue from 37º to 100and then boil abroad 0.500 g of water, if information technology puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the environment.

24. Integrated ConceptsHydroelectric generators (run into Figure 5) at Hoover Dam produce a maximum current of viii.00 × x^three A at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the organisation at depression speed (thus its kinetic free energy does non change) merely loses 160 m in distance. How many cubic meters per second are needed, assuming 85.0% efficiency?

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

25. Integrated Concepts(a) Bold 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-Five batteries of a 750-kg electric machine exist able to supply: (a) To accelerate from rest to 25.0 grand/s in i.00 min? (b) To climb a 2.00 × tenⁱⁱ-1000-high-hill in 2.00 min at a abiding 25.0-m/s speed while exerting 5.00 × x²N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-chiliad/s speed, exerting a five.00 × 10ⁱⁱNorthward 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-rails commuter railroad train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does information technology take to reach 20.0 m/southward starting from rest if its loaded mass is five.xxx × 10^ivkg, assuming 95.0% efficiency and constant power? (c) Find its average dispatch. (d) Discuss how the acceleration you lot found for the calorie-free-rail train compares to what might be typical for an automobile.

27. Integrated Concepts(a) An aluminum ability transmission 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 past lowering the resistance.

28. Integrated Concepts(a) An immersion heater utilizing 120 5 can heighten the temperature of a 1.00 × 10²-thou aluminum cup containing 350 g 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. Hash out the applied limits to speeding the heating by lowering the resistance.

29. Integrated Concepts(a) What is the toll 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 surroundings? The cost of electricity is nine cents/kW ⋅ h. (b) What current was used by the 220-V Air-conditioning electric heater, if this took 4.00 h?

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

31. Unreasonable Results(a) What current is needed to transmit 1.00 × 10ⁱⁱMW of power at 10.0 kV? (b) Observe the resistance of one.00 km of wire that would cause a 0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (due east) Which assumptions are unreasonable, or which premises are inconsistent?

32. Construct Your Own ProblemConsider an electric immersion heater used to oestrus a cup of water to make tea. Construct a problem in which y'all calculate the needed resistance of the heater so that it increases the temperature of the h2o and cup in a reasonable amount of time. Also calculate the cost of the electrical energy used in your process. Among the things to be considered are the voltage used, the masses and heat capacities involved, heat losses, and the time over which the heating takes place. Your instructor may wish for you lot to consider a thermal safety switch (peradventure bimetallic) that will halt the procedure before damaging temperatures are reached in the immersion unit of measurement.

Glossary

electric power:

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

Selected Solutions to Problems & Exercises

1. 2 . 00 × 10 ¹² W

5. (a) one.50 W (b) vii.fifty W

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

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

11. $438/y

13. $6.25

15. i.58 h

17. $3.94 billion/year

19. 25.v Westward

21.(a) 2.00 × 10⁹J (b) 769 kg

23. 45.0 s

25. (a) 343 A (b) 2.17 × ten³A (c) 1.ten × 10³A

27. (a) 1.23 × 10³kg (b) ii.64 × 10³kg

29. (a) 2.08 × 10^fiveA
(b) four.33 × 10⁴MW
(c) The manual lines dissipate more power than they are supposed to transmit.
(d) A voltage of 480 Five is unreasonably depression for a transmission voltage. Long-altitude transmission lines are kept at much higher voltages (often hundreds of kilovolts) to reduce power losses.

Source: https://courses.lumenlearning.com/austincc-physics2/chapter/20-4-electric-power-and-energy/

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