If The Power Rating Of A 400-ï‰ Resistor Is 0.80 W, What Is The Maximum Current It Can Safely Draw?

Learning Objectives

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

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

Ability in Electric Circuits

Ability is associated by many people with electricity. Knowing that power is the charge per unit of energy utilisation or energy conversion, what is the expression for electrical power? Power transmission lines might come to listen. We also call up of lightbulbs in terms of their power ratings in watts. Let us compare a 25-West seedling with a lx-West bulb. (See Figure 1(a).) Since both operate on the aforementioned voltage, the lx-W bulb must draw more than current to have a greater ability rating. Thus the 60-Due west seedling's resistance must be lower than that of a 25-W seedling. If we increase voltage, we too increase power. For case, when a 25-Westward seedling that is designed to operate on 120 Five is continued to 240 5, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electrical power?

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

Electric energy depends on both the voltage involved and the accuse moved. This is expressed most but as PE = qV, where q is the charge moved and V is the voltage (or more precisely, the potential difference 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 hither), the expression for power becomes

P = Four

Electric power (P) is just 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 2d, or watts. Thus, i A ⋅V= 1 W. For example, cars oft have ane or more auxiliary ability outlets with which you tin can charge a prison cell phone or other electronic devices. These outlets may be rated at xx A, so that the circuit can deliver a maximum power P = 4 = (xx A)(12 5) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (i kA ⋅5 = 1 kW). To meet the relationship of power to resistance, nosotros combine Ohm's law withP = Four. Substituting I = V/R gives P= (V/R)V=Five ²/R. Similarly, substituting 5 = IR gives P = I(IR) = I^twoR. Three 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}^{2}R\\[/latex].

Annotation that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with 1 voltage source and a unmarried resistor, the power supplied past the voltage source and that prodigal by the resistor are identical. (In more complicated circuits, P tin can be the power dissipated by a unmarried device and non the total power in the excursion.) Different insights can exist gained from the three unlike expressions for electric power. For example, P=5 ²/R implies that the lower the resistance continued to a given voltage source, the greater the ability delivered. Furthermore, since voltage is squared in P=V ²/R, the effect of applying a higher voltage is possibly greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to near 100 Due west, called-for it out. If the seedling'southward resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.

Instance one. Calculating Power Dissipation and Electric current: Hot and Cold Power

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

Strategy for (a)

For the hot headlight, we know voltage and current, so we can utilize P = Four to discover the power. For the common cold headlight, we know the voltage and resistance, and so we can apply P=V ^two/R to detect the power.

Solution for (a)

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

P = IV = (two.fifty A)(12.0 V) = 30.0 West.

The cold resistance was 0.350 Ω, and and then the power it uses when showtime 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 30 Westward dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the seedling's temperature increases and its resistance increases.

Strategy and Solution for (b)

The current when the seedling is cold tin exist constitute several dissimilar ways. Nosotros rearrange i 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.three\text{ A}\\[/latex].

Discussion for (b)

The cold electric current is remarkably higher than the steady-country value of 2.50 A, merely the electric current will chop-chop reject 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 loftier currents briefly as a device comes on. In some cases, such every bit with electric motors, the current remains high for several seconds, necessitating special "irksome accident" fuses.

The Cost of Electricity

The more electric appliances you lot use and the longer they are left on, the higher your electric beak. This familiar fact is based on the relationship between energy and ability. You pay for the free energy used. Since P=E/t, nosotros run across that

East = 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 on electric bills is the kilowatt-hour (kW ⋅ h), consistent with the human relationshipEast = Pt. It is easy to estimate the cost of operating electrical appliances if yous have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hr for your electric utility. Kilowatt-hours, like all other specialized free energy units such as food calories, tin be converted to joules. You can prove to yourself that 1 kW ⋅ h = 3 . six × 10 ⁶ J .

The electric energy (E) used can exist 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 it volition also outcome in a reduced touch on on the environment. Improvements to lighting are some of the fastest ways to reduce the electric energy used in a home or business. Near 20% of a home's use of 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 truthful for both the long tubes and the compact fluorescent lights (CFL). (Meet Figure 1(b).) Thus, a 60-W incandescent seedling tin can be replaced by a 15-W CFL, which has the same brightness and color. CFLs take a bent tube within a world or a spiral-shaped tube, all continued to a standard spiral-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs accept been addressed in contempo years.) The heat transfer from these CFLs is less, and they last upwards to ten times longer. The significance of an investment in such bulbs is addressed in the next case. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last five times longer than CFLs. However, their toll is still high.

Making Connections: Free energy, Power, and Time

The relationshipEastward = Pt is ane that you will find useful in many dissimilar contexts. The energy your body uses in exercise is related to the power level and duration of your action, for example. The corporeality of heating by a power source is related to the ability level and fourth dimension it is applied. Even the radiation dose of an X-ray epitome is related to the ability and time of exposure.

Example 2. Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL)

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

Strategy

To find the operating cost, we showtime notice the energy used in kilowatt-hours and and then 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 energy:

East = Pt = (60 Due west)(1000 h) = threescore,000 West ⋅ h

In kilowatt-hours, this is

E= lx.0 kW ⋅ h.

Now the electricity cost is

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

The full toll will exist $7.20 for grand hours (about ane-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 $seven.20/four = $1.80. The CFL will last 10 times longer than the incandescent, then that the investment cost volition exist 1/10 of the seedling toll for that time menstruation of employ, or 0.1($1.50) = $0.fifteen. Therefore, the total cost will exist $1.95 for 1000 hours.

Give-and-take

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

Making Connections: Take-Abode Experiment—Electrical Energy use Inventory

1) 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 college rating than a digital clock. Guess the energy consumed by these appliances in an average solar day (past estimating their time of utilize). Some appliances might only state the operating current. If the household voltage is 120 5, and then utilise P = IV. 2) Check out the full wattage used in the balance rooms of your school's flooring or building. (Y'all might need to presume the long fluorescent lights in apply are rated at 32 Due west.) Suppose that the building was closed all weekend and that these lights were left on from half-dozen p.m. Friday until viii a.m. Monday. What would this oversight toll? How about for an entire twelvemonth of weekends?

Section Summary

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

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

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

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

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

Conceptual Questions

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

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

Problems & Exercises

1. What is the ability of a 1.00 × 10²MV lightning bolt having a electric current of  2.00 × 10⁴ 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 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 ability output, given the reckoner'due south voltage output is 3.00 V? (Come across Figure two.)

Figure 2. The strip of solar cells merely above the keys of this figurer convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons)

four. How many watts does a flashlight that has6.00 × x ² pass through information technology in 0.500 h use if its voltage is 3.00 5?

v. Find the power prodigal 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.

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

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

8. Show that the units 1 A ⁱⁱ ⋅ Ω = 1 West , as implied by the equation P = I ⁱⁱ R .

9. Verify the energy unit of measurement equivalence that 1 kW ⋅ h = 3.60 × 10⁶J.

10. Electrons in an X-ray tube are accelerated through1.00 × 10 ² 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.

xi. An electrical water heater consumes 5.00 kW for 2.00 h per 24-hour interval. What is the cost of running it for 1 year if electricity costs 12.0 cents/kW ⋅ h? See Figure 3.

Figure iii. On-demand electrical hot h2o heater. Estrus 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 fourth dimension = i infinitesimal)? At ix.0 cents/kW · h, how much does this cost?

13. 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 cost of the incandescent bulb is 25 cents and that electricity costs 10 cents/kWh. Calculate the price for 1000 hours, every bit in the cost effectiveness of CFL example.

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

xv. Element of group i batteries take the advantage of putting out abiding voltage until very most the stop of their life. How long will an alkali metal battery rated at 1.00 A ⋅ h and 1.58 Five keep a one.00-Due west flashlight bulb burning?

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

17. The average goggle box is said to be on half-dozen hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their ability consumption averages 150 W and the price of electricity averages 12.0 cents/kW ⋅ h.

18. An old lightbulb draws but 50.0 W, rather than its original 60.0 West, due to evaporative thinning of its filament. By what factor is its diameter reduced, bold compatible thinning along its length? Fail any furnishings caused by 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 × 10²A.

20.Integrated Concepts

Cold vaporizers pass a electric current through water, evaporating it with only a small increase in temperature. One such home device is rated at 3.50 A and utilizes 120 Five 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 eight.00 h of overnight functioning? (See Figure 4.)

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

21. Integrated Concepts(a) What free energy is dissipated past a lightning bolt having a 20,000-A current, a voltage of 1.00 × 10ⁱⁱMV and a length of i.00 ms? (b) What mass of tree sap could be raised from 18ºC to its boiling indicate and then evaporated past this energy, assuming sap has the same thermal characteristics as water?

22. Integrated ConceptsWhat electric current must be produced by a 12.0-V battery-operated bottle warmer in society to heat 75.0 g of glass, 250 1000 of infant formula, and three.00×ten² 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 i.00 thou of tissue from 37º to 100and and then boil away 0.500 g of h2o, if information technology puts out 2.00 mA at 15.0 kV? Ignore oestrus transfer to the environs.

24. Integrated ConceptsHydroelectric generators (see Effigy 5) at Hoover Dam produce a maximum current of 8.00 × 10³ A at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not alter) but loses 160 m in altitude. How many cubic meters per 2nd are needed, assuming 85.0% efficiency?

Figure five. 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-V batteries of a 750-kg electrical car be able to supply: (a) To accelerate from residue to 25.0 m/due south in ane.00 min? (b) To climb a two.00 × 10²-m-loftier-hill in 2.00 min at a abiding 25.0-m/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 five.00 × 10²N strength to overcome air resistance and friction? See Figure vi.

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

26. Integrated ConceptsA low-cal-rail commuter railroad train draws 630 A of 650-V DC electricity when accelerating. (a) What is its ability consumption rate in kilowatts? (b) How long does it take to reach 20.0 grand/s starting from balance if its loaded mass is 5.xxx × 10^ivkg, assuming 95.0% efficiency and constant power? (c) Find its average dispatch. (d) Talk over how the dispatch y'all found for the lite-runway railroad train compares to what might be typical for an car.

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 tin raise the temperature of a 1.00 × 10²-thousand aluminum cup containing 350 g of water from 20º C to 95º C in two.00 min. Find its resistance, assuming information technology is abiding during the process. (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 cost of heating a hot tub containing 1500 kg of water from 10º C to 40º C, assuming 75.0% efficiency to account for heat transfer to the environs? The price of electricity is ix cents/kW ⋅ h. (b) What electric current was used by the 220-Five AC electric heater, if this took iv.00 h?

xxx. Unreasonable Results(a) What current is needed to transmit 1.00 × 10²MW of ability at 480 V? (b) What power is dissipated by the transmission lines if they have a ane.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 electric current is needed to transmit ane.00 × 10²MW of power at 10.0 kV? (b) Find the resistance of i.00 km of wire that would crusade a 0.0100% power loss. (c) What is the diameter of a i.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 loving cup of h2o to make tea. Construct a trouble in which y'all calculate the needed resistance of the heater then that it increases the temperature of the water and cup in a reasonable amount of time. Likewise calculate the cost of the electric 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 teacher may wish for you to consider a thermal condom switch (maybe bimetallic) that will halt the process before dissentious temperatures are reached in the immersion unit.

Glossary

electric power:

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

Selected Solutions to Bug & Exercises

1. ii . 00 × x ¹² Westward

5. (a) ane.50 W (b) seven.l W

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

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

11. $438/y

13. $6.25

15. 1.58 h

17. $3.94 billion/twelvemonth

19. 25.5 Westward

21.(a) 2.00 × x^ixJ (b) 769 kg

23. 45.0 s

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

27. (a) 1.23 × 10³kg (b) 2.64 × x^threekg

29. (a) 2.08 × 10⁵A
(b) 4.33 × 10⁴MW
(c) The transmission lines dissipate more ability than they are supposed to transmit.
(d) A voltage of 480 5 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/physics/chapter/20-4-electric-power-and-energy/

Posted by: leveyoverects36.blogspot.com

Share this post