How Much Power Is Used By A 9.0-v Battery That Draws 2 Amps Of Current

Learning Objectives

By the end of this department, you will be able to:

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

Power in Electric Circuits

Power is associated past many people with electricity. Knowing that ability is the rate of free energy utilise or free energy conversion, what is the expression for electric ability? Ability manual lines might come up to listen. We as well think of lightbulbs in terms of their power ratings in watts. Allow us compare a 25-W bulb with a lx-W seedling. (See Effigy one(a).) Since both operate on the same voltage, the threescore-W seedling must depict more electric current to have a greater power rating. Thus the 60-W seedling's resistance must exist lower than that of a 25-W bulb. If we increment 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, information technology briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric ability?

Effigy 1. (a) Which of these lightbulbs, the 25-West seedling (upper left) or the sixty-W bulb (upper right), has the higher resistance? Which draws more than current? Which uses the about energy? Can y'all tell from the color that the 25-Due west filament is libation? Is the brighter seedling a dissimilar color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the aforementioned intensity of low-cal equally the 60-Due west bulb, but at 1/4 to i/10 the input power. (credit: dbgg1979, Flickr)

Electric free energy depends on both the voltage involved and the accuse moved. This is expressed near simply as PE = qV, where q is the charge 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 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 ability becomes

P = IV

Electric power (P) is merely 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 2d, or watts. Thus, 1 A ⋅V= 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a prison cell telephone or other electronic devices. These outlets may be rated at xx A, so that the excursion can deliver a maximum power P = Iv = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (i kA ⋅V = 1 kW). To see the relationship of power to resistance, nosotros combine Ohm'southward police force withP = IV. Substituting I = V/R gives P= (V/R)5=V ²/R. Similarly, substituting V = IR gives P = I(IR) = I²R. Iii expressions for electric power are listed together here for convenience:

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

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

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

Notation that the first equation is always valid, whereas the other two can be used merely for resistors. In a uncomplicated circuit, with one voltage source and a single resistor, the ability supplied by the voltage source and that dissipated past the resistor are identical. (In more complicated circuits, P can be the ability dissipated past a single device and non the total power in the circuit.) Different insights can be gained from the three dissimilar expressions for electric ability. For instance, 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 ²/R, the event of applying a college voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its ability well-nigh quadruples to about 100 W, burning information technology out. If the bulb's resistance remained abiding, its power would exist exactly 100 Due west, just at the higher temperature its resistance is higher, too.

Example i. Calculating Power Dissipation and Current: Hot and Cold Power

(a) Consider the examples given in Ohm's Police: Resistance and Simple Circuits and Resistance and Resistivity. So detect the power dissipated by the car headlight in these examples, both when information technology is hot and when it is cold. (b) What current does information technology describe when cold?

Strategy for (a)

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

Solution for (a)

Inbound the known values of current and voltage for the hot headlight, nosotros obtain

P = IV = (2.50 A)(12.0 V) = 30.0 Due west.

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

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

Discussion for (a)

The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial ability quickly decreases as the bulb'southward temperature increases and its resistance increases.

Strategy and Solution for (b)

The current when the seedling is cold can be found several different ways. Nosotros 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{ West}}{{0.350}\text{ }\Omega }}=34.3\text{ A}\\[/latex].

Word for (b)

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

The Cost of Electricity

The more electric appliances yous apply and the longer they are left on, the higher your electric pecker. This familiar fact is based on the human relationship between energy and power. Yous pay for the energy used. Since P=E/t, we see that

E = Pt

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

The electric energy (E) used can be reduced either by reducing the time of use or by reducing the ability consumption of that appliance or fixture. This will not only reduce the price, only it will also result in a reduced bear upon on the environment. Improvements to lighting are some of the fastest ways to reduce the electric free energy used in a home or business. Most twenty% of a home'due south use of energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about iv times more efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure ane(b).) Thus, a lx-W incandescent bulb can exist replaced by a xv-W CFL, which has the same effulgence and colour. CFLs have a bent tube inside a world or a spiral-shaped tube, all connected to a standard screw-in base of operations that fits standard incandescent calorie-free 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 concluding up to x times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are fifty-fifty more than efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their price is notwithstanding high.

Making Connections: Energy, Power, and Time

The human relationshipE = Pt is 1 that you lot will notice useful in many different contexts. The energy your torso uses in practice is related to the power level and elapsing of your activity, for example. The corporeality of heating past a power source is related to the ability level and time it is applied. Even the radiation dose of an 10-ray paradigm is related to the ability and time of exposure.

Case 2. Calculating the Price Effectiveness of Compact Fluorescent Lights (CFL)

If the cost of electricity in your area is 12 cents per kWh, what is the full cost (capital plus functioning) of using a 60-W incandescent bulb for m hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this seedling with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.fifty merely lasts ten times longer (10,000 hours), what will that full cost exist?

Strategy

To find the operating cost, we get-go observe the energy used in kilowatt-hours and so multiply by the toll per kilowatt-60 minutes.

Solution for (a)

The energy used in kilowatt-hours is constitute by entering the power and fourth dimension into the expression for energy:

Due east = Pt = (60 W)(1000 h) = 60,000 W ⋅ h

In kilowatt-hours, this is

E= lx.0 kW ⋅ h.

Now the electricity cost is

toll = (60.0 kW ⋅ h) ($0.12/kW ⋅ h) = $ seven.20.

The full price will be $vii.twenty for 1000 hours (about half year at five hours per day).

Solution for (b)

Since the CFL uses only 15 W and non threescore West, the electricity toll will be $7.20/4 = $1.eighty. The CFL volition concluding x times longer than the incandescent, so that the investment cost will exist 1/x of the bulb cost for that time period of use, or 0.ane($one.50) = $0.fifteen. Therefore, the total toll will exist $1.95 for 1000 hours.

Discussion

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

Making Connections: Have-Domicile Experiment—Electrical Energy Use Inventory

1) Brand a listing of the power ratings on a range of appliances in your home or room. Explicate why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average twenty-four hours (by estimating their time of utilize). Some appliances might only state the operating electric current. If the household voltage is 120 V, so utilise P = 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 Due west.) Suppose that the building was airtight all weekend and that these lights were left on from 6 p.thou. Friday until 8 a.grand. Monday. What would this oversight cost? How about for an entire twelvemonth of weekends?

Department Summary

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

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

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

i. 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 = V²/R which ways power decreases if resistance increases. All the same this power is too given by P = I ² R , which means ability increases if resistance increases. Explicate why there is no contradiction here.

Issues & Exercises

1. What is the power of a 1.00 × ten²MV lightning bolt having a current of  2.00 × 10^iv A?

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

three. 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 iii.00 V? (Run across Figure 2.)

Figure two. The strip of solar cells just to a higher place the keys of this calculator catechumen low-cal to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons)

4. How many watts does a flashlight that hassix.00 × 10 ^two pass through it in 0.500 h utilise if its voltage is three.00 V?

5. Discover 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 implied by the equation P = IV.

7. Bear witness that the units 1V²/Ω = 1W as implied by the equation P = Five^two /R.

8. Testify that the units one A ² ⋅ Ω = 1 W , every bit implied by the equation P = I ² R .

9. Verify the energy unit of measurement equivalence that one kW ⋅ h = 3.60 × 10^half-dozenJ.

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

11. An electrical water heater consumes v.00 kW for 2.00 h per day. What is the price of running information technology for one twelvemonth if electricity costs 12.0 cents/kW ⋅ h? See Figure iii.

Figure 3. On-demand electrical hot h2o heater. Rut is supplied to water 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 = 1 minute)? 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 cost (investment plus operating) would be the aforementioned for both CFL and incandescent 60-Due west bulbs? Assume the cost of the incandescent bulb is 25 cents and that electricity costs x cents/kWh. Calculate the cost for 1000 hours, as in the toll effectiveness of CFL example.

fourteen. 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 auto? (b) What current flows through it?

fifteen. Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline bombardment rated at 1.00 A ⋅ h and 1.58 V go on a 1.00-West flashlight bulb burning?

xvi. A cauterizer, used to terminate bleeding in surgery, puts out ii.00 mA at xv.0 kV. (a) What is its power output? (b) What is the resistance of the path?

17. The boilerplate television is said to exist on 6 hours per day. Judge 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.

18. An old lightbulb draws only 50.0 West, rather than its original 60.0 Westward, due to evaporative thinning of its filament. By what cistron is its bore reduced, assuming compatible thinning along its length? Neglect any effects caused by temperature differences.

xix. 00-guess copper wire has a diameter of 9.266 mm. Calculate the ability loss in a kilometer of such wire when it carries 1.00 × 10²A.

twenty.Integrated Concepts

Cold vaporizers pass a current through water, evaporating information technology with only a small increase in temperature. One such habitation device is rated at 3.50 A and utilizes 120 Five AC with 95.0% efficiency. (a) What is the vaporization rate in grams per minute? (b) How much h2o must you put into the vaporizer for eight.00 h of overnight operation? (See Figure 4.)

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

21. Integrated Concepts(a) What energy is prodigal past a lightning commodities having a 20,000-A current, a voltage of 1.00 × ten²MV and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18ºC to its boiling signal and then evaporated past this energy, assuming sap has the aforementioned thermal characteristics equally water?

22. Integrated ConceptsWhat current must be produced by a 12.0-V battery-operated bottle warmer in society to heat 75.0 grand of drinking glass, 250 chiliad of baby formula, and 3.00×tenⁱⁱ of aluminum from 20º C to 90º in v.00 min?

23. Integrated ConceptsHow much time is needed for a surgical cauterizer to raise the temperature of 1.00 yard of tissue from 37º to 100and then boil abroad 0.500 thou of water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings.

24. Integrated ConceptsHydroelectric generators (run across Figure 5) at Hoover Dam produce a maximum electric current of viii.00 × 10³ 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 change) only loses 160 yard in distance. How many cubic meters per second 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 electric power by the motor, what electric current must the 12.0-Five batteries of a 750-kg electric car be able to supply: (a) To accelerate from balance to 25.0 m/due south in 1.00 min? (b) To climb a 2.00 × 10²-chiliad-high-loma in 2.00 min at a constant 25.0-m/southward speed while exerting 5.00 × 10²N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/due south speed, exerting a 5.00 × tenⁱⁱNorthward force to overcome air resistance and friction? See Figure vi.

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

26. Integrated ConceptsA light-rail commuter 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 achieve 20.0 m/southward starting from residuum if its loaded mass is 5.30 × 10⁴kg, assuming 95.0% efficiency and constant power? (c) Find its boilerplate acceleration. (d) Discuss how the acceleration you institute for the lite-rail train compares to what might exist typical for an automobile.

27. Integrated Concepts(a) An aluminum power 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 by lowering the resistance.

28. Integrated Concepts(a) An immersion heater utilizing 120 V can raise the temperature of a 1.00 × 10²-g aluminum loving cup containing 350 chiliad of water from 20º C to 95º C in 2.00 min. Find its resistance, bold it is constant during the procedure. (b) A lower resistance would shorten the heating time. Discuss the applied 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 oestrus transfer to the surroundings? The cost of electricity is 9 cents/kW ⋅ h. (b) What current was used by the 220-Five AC electrical heater, if this took 4.00 h?

thirty. Unreasonable Results(a) What current is needed to transmit 1.00 × x²MW of power at 480 5? (b) What power is prodigal by the transmission lines if they have a one.00 – Ω resistance? (c) What is unreasonable virtually this result? (d) Which assumptions are unreasonable, or which bounds are inconsistent?

31. Unreasonable Results(a) What current is needed to transmit 1.00 × x²MW 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 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent?

32. Construct Your Own ProblemConsider an electric immersion heater used to heat a cup of h2o to brand tea. Construct a trouble in which you calculate the needed resistance of the heater so that it increases the temperature of the h2o and cup in a reasonable corporeality of fourth dimension. Also calculate the price of the electrical energy used in your procedure. Among the things to be considered are the voltage used, the masses and estrus capacities involved, heat 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 dissentious temperatures are reached in the immersion unit.

Glossary

electric power:

the charge per unit 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. 2 . 00 × x ¹² W

5. (a) one.50 W (b) 7.l W

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

9. [latex]one\text{kW}\cdot \text{h=}\left(\frac{ane\times {\text{10}}^{3}\text{J}}{\text{1 s}}\right)\left(ane h\right)\left(\frac{\text{3600 south}}{\text{1 h}}\right)=3\text{.}\text{60}\times {\text{10}}^{6}\text{J}\\[/latex]

xi. $438/y

thirteen. $half dozen.25

fifteen. i.58 h

17. $3.94 billion/yr

xix. 25.5 Due west

21.(a) two.00 × 10^nineJ (b) 769 kg

23. 45.0 s

25. (a) 343 A (b) 2.17 × ten³A (c) ane.ten × x^threeA

27. (a) 1.23 × 10³kg (b) 2.64 × 10³kg

29. (a) 2.08 × 10⁵A
(b) 4.33 × 10^fourMW
(c) The manual lines dissipate more than 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 (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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