Electric Current, Resistance, and Ohm’s Law
22 Electric Power and Energy
Learning Objectives
- Calculate the power dissipated by a resistor and the power supplied by a voltage source.
- Use the different forms of the electric power equation appropriately.
- Calculate the electrical energy consumed by a device and estimate its operating cost.
Power in Electric Circuits
Whenever electric current flows, energy is transferred. A battery supplies chemical energy to moving charges, a power plant delivers electrical energy to homes, and a medical device such as an electrosurgical unit converts electrical energy into heat to cut or cauterize tissue. The rate at which this energy is transferred is called electric power.
You have already encountered power in mechanics, where it is defined as the rate of doing work or transferring energy:
In an electric circuit, the energy transferred when a charge q moves through a potential difference V is
Substituting this expression into the definition of power gives
Since electric current is the rate at which charge flows,
the most general expression for electric power becomes
This equation is one of the most important relationships in electricity. It tells us that power depends on both the amount of current flowing and the potential difference that drives it. Increasing either the current or the voltage increases the rate at which energy is transferred.
The SI unit of power is the watt (W), named after the Scottish engineer James Watt. Since one watt is one joule transferred each second,
Power ratings appear on nearly every electrical device. A phone charger may supply about 20 W, a laptop charger around 65 W, a microwave oven about 1000 W, and a medical defibrillator can briefly deliver hundreds of joules of energy in only a few milliseconds, corresponding to an extremely large instantaneous power.
Health and Bioscience Connection
Many medical devices are classified by their electrical power requirements. Heating blankets, infant incubators, electrosurgical instruments, MRI systems, and X-ray generators all rely on controlled electrical power. Although these devices may operate at vastly different voltages and currents, the same equation,
[latex]P=IV[/latex], describes the rate at which electrical energy is delivered.
For a resistor, Ohm's law allows us to rewrite the power equation in two additional forms. Substituting
into [latex]P=IV[/latex] gives
Likewise, substituting
gives
Thus, electrical power can be written in three equivalent forms:
Although these equations are mathematically equivalent, they are not interchangeable in every situation. The expression [latex]P=IV[/latex] is always valid because it follows directly from the definitions of voltage, current, and power. The other two expressions assume that Ohm's law applies, so they should be used only for ohmic resistive elements.
Each form highlights a different physical idea:
- [latex]P=IV[/latex] emphasizes that power is the product of current and voltage.
- [latex]P=\frac{V^2}{R}[/latex] shows that for a fixed voltage source, reducing the resistance increases the power delivered.
- [latex]P=I^2R[/latex] shows that for a given current, increasing the resistance increases the thermal energy generated. This relationship explains why electrical heating elements are made from relatively high-resistance materials such as nichrome.
Consider two incandescent bulbs connected to the same household outlet. A 60-W bulb draws more current than a 25-W bulb and therefore transfers energy to its filament at a greater rate, making it brighter. If a bulb designed for 120 V is accidentally connected to 240 V, its power increases dramatically because power is proportional to the square of the voltage. The filament rapidly overheats and fails.
Example 22.1: Power Dissipation and Inrush Current in an Incandescent Headlight
An automobile headlight operates from a 12.0-V battery. When the tungsten filament reaches its normal operating temperature, it draws 2.50 A. However, when the bulb is first switched on, the filament is still cold and its resistance is only 0.350 Ω.
- Calculate the electrical power dissipated by the headlight when the filament is hot and when it is cold.
- Calculate the current drawn immediately after the bulb is switched on, before the filament has had time to heat up.
Strategy
Choose the equation that matches the known quantities for each part:
- For the hot filament, the voltage and current are known, so use
[latex]P=IV.[/latex]
- For the cold filament, the voltage and resistance are known, so use
[latex]P=\frac{V^2}{R}.[/latex]
- After finding the cold resistance, calculate the initial current using Ohm's law:
[latex]I=\frac{V}{R}.[/latex]
Solution
- Power dissipated
- Hot filament
[latex]P=IV=(12.0~\text{V})(2.50~\text{A})=30.0~\text{W}.[/latex]
- Cold filament
[latex]P=\frac{V^2}{R} =\frac{(12.0~\text{V})^2}{0.350~\Omega} =411~\text{W}.[/latex]
- Hot filament
- Initial current
Using Ohm's law,
[latex]I=\frac{V}{R} =\frac{12.0~\text{V}}{0.350~\Omega} =34.3~\text{A}.[/latex](The same result would be obtained from [latex]P=I^2R[/latex].)
Discussion
- The bulb normally operates at only 30.0 W, but immediately after it is switched on it briefly dissipates more than 400 W.
- The initial current of 34.3 A is much larger than the steady operating current of 2.50 A because a cold tungsten filament has a much lower resistance than a hot one.
- As the filament heats, its resistance increases rapidly, causing both the current and the power to decrease to their normal operating values.
- This brief surge, known as an inrush current, is common in incandescent lamps and many electric motors. Circuit breakers and time-delay ("slow-blow") fuses are designed to tolerate these short-lived current spikes without unnecessarily interrupting the circuit.
The Cost of Electricity
You pay your electric utility for the energy you use, not simply for the power of your appliances. Recall that power is the rate at which energy is transferred or converted:
Rearranging this expression gives the energy consumed during a time interval:
This equation shows that electrical energy depends on both the power rating of a device and the length of time it operates. A high-power appliance running for a short time may use less energy than a lower-power appliance left on for many hours.
Electric utility companies typically bill customers in kilowatt-hours (kWh), where one kilowatt-hour is the energy used by a device with a power of 1 kW operating for 1 hour. Although it is called a "kilowatt-hour," it is a unit of energy, not power.
Knowing a device's power rating, operating time, and the local electricity rate makes it easy to estimate its operating cost. For example, a 100-W appliance running for 10 hours uses
If electricity costs $0.18 per kWh, operating that appliance costs about $0.18.
Reducing electricity use can be accomplished in two ways:
- Operate devices for less time.
- Use devices that require less power to perform the same task.
Lighting provides an excellent example. Traditional incandescent bulbs convert only a small fraction of electrical energy into visible light; most of the energy becomes heat. Compact fluorescent lamps (CFLs) greatly improved efficiency, but today light-emitting diode (LED) bulbs are the standard technology. LEDs typically produce the same amount of light while using only about one-sixth to one-eighth the power of an equivalent incandescent bulb and generally last much longer.
Reducing electrical energy consumption not only lowers utility bills but also decreases the demand placed on electrical power plants, reducing fuel consumption and greenhouse gas emissions in regions where electricity is generated from fossil fuels.
Health and Bioscience Connection
The relationship [latex]E=Pt[/latex] appears throughout medicine and biology. Medical devices such as heating pads, electrosurgical units, MRI systems, X-ray equipment, and therapeutic ultrasound all deliver energy at a certain power over a specific time. Controlling both the power level and the duration of exposure allows clinicians to deliver the desired treatment or diagnostic image while minimizing unnecessary heating or radiation dose.
Example 22.2: Comparing the Cost of Incandescent and LED Light Bulbs
An electric utility charges $0.12 per kWh for electricity.
- A 60-W incandescent light bulb costs $0.25 and lasts for 1000 hours. What is its total cost (purchase price plus electricity) over its lifetime?
- An LED bulb that produces the same amount of light uses only 10.0 W, costs $3.00, and lasts 25,000 hours. What is its total cost over the same 1000-hour period?
Strategy
First calculate the electrical energy consumed using
Convert the energy from watt-hours to kilowatt-hours, determine the electricity cost, and then add the appropriate purchase cost of the bulb.
Solution for (a)
The incandescent bulb consumes
The cost of the electricity is therefore
Adding the purchase price of the bulb gives
Solution for (b)
The LED bulb consumes
The electricity cost is therefore
Because the LED lasts 25,000 hours, only
of its lifetime is used during these 1000 hours. Therefore, only 4.0% of its purchase cost should be attributed to this period:
The total cost is therefore
Discussion
Although the LED bulb has a much higher purchase price, it uses far less electrical energy and lasts much longer. Over the same 1000-hour period, the LED costs only about $1.32, compared with $7.45 for the incandescent bulb. This example illustrates why LEDs have largely replaced incandescent and compact fluorescent bulbs in homes, hospitals, and commercial buildings: they reduce both electricity consumption and maintenance costs.
Making Connections: Take-Home Investigation—Electrical Energy Use Inventory
Make a list of several electrical appliances in your home or workplace and record their power ratings. Which devices have the largest power ratings, and why? Estimate how many hours each device is used during a typical day, then calculate the electrical energy each consumes using [latex]E=Pt[/latex].
If an appliance lists only its operating current, estimate its power using [latex]P=IV[/latex], assuming a household voltage of 120 V (or 230 V in countries that use that standard).
As an additional exercise, estimate the cost of leaving the lights on in a classroom or office over an entire weekend. If each fixture contains several LED lamps, determine the total electrical energy consumed and estimate the resulting electricity cost using your local utility rate.
Section Summary
- Electric power is the rate at which electrical energy is transferred or converted. It is measured in watts (W), where [latex]1~\text{W}=1~\text{J/s}[/latex]. In a circuit, power may be supplied by a voltage source or dissipated by a circuit element such as a resistor.
- For any electrical device, power is given by
[latex]P=IV[/latex]
where [latex]I[/latex] is the current through the device and [latex]V[/latex] is the voltage across it.
- For an ohmic resistor, combining Ohm's law with the power equation gives
[latex]P=\frac{V^{2}}{R}[/latex]
and
[latex]P=I^{2}R[/latex]These equations are valid for resistive elements that obey Ohm's law.
- The electrical energy consumed by a device operating at constant power for a time [latex]t[/latex] is
[latex]E=Pt[/latex]
- Electric utility companies bill energy in kilowatt-hours (kWh), where
[latex]1~\text{kWh}=3.6\times10^{6}~\text{J}.[/latex]
Conceptual Questions
- 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 [latex]P={V}^{2}/R[/latex], which means power decreases if resistance increases. Yet this power is also given by [latex]P={I}^{2}R[/latex], which means power increases if resistance increases. Explain why there is no contradiction here.
Problems & Exercises
- What is the power of a [latex]1.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{MV}[/latex] lightning bolt having a current of [latex]{2.00 × 10}^{\text{4}}\phantom{\rule{0.25em}{0ex}}\text{A}[/latex]?
- 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?
- 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 3.00 V? (See Figure 22.2.)
Figure 22.2. The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons) - How many watts does a flashlight that has [latex]6.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{C}[/latex] pass through it in 0.500 h use if its voltage is 3.00 V?
- Find the power dissipated in each of these extension cords: (a) an extension cord having a [latex]0\text{.}\text{0600}\phantom{\rule{0.25em}{0ex}}\text{-}\phantom{\rule{0.25em}{0ex}}\Omega[/latex] resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of [latex]0\text{.}\text{300}\phantom{\rule{0.25em}{0ex}}\Omega .[/latex]
- Verify that the units of a volt-ampere are watts, as implied by the equation [latex]P=\text{IV}[/latex].
- Show that the units [latex]1\phantom{\rule{0.25em}{0ex}}{\text{V}}^{2}/\Omega =1\text{W}[/latex], as implied by the equation [latex]P={V}^{2}/R[/latex].
- Show that the units [latex]1\phantom{\rule{0.25em}{0ex}}{\text{A}}^{2}\cdot \Omega =1\phantom{\rule{0.25em}{0ex}}\text{W}[/latex], as implied by the equation [latex]P={I}^{2}R[/latex].
- Verify the energy unit equivalence that [latex]1\phantom{\rule{0.25em}{0ex}}\text{kW}\cdot \text{h = 3}\text{.}\text{60}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}[/latex].
- Electrons in an X-ray tube are accelerated through [latex]1.00\times10^{2}\ \text{kV}[/latex] 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 [latex]15.0\ \text{mA}[/latex].
- An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs [latex]\text{12.0 cents}\text{/kW}\cdot \text{h}[/latex]? See Figure 22.3.
Figure 22.3. On-demand electric hot water heater. Heat is supplied to water only when needed. (credit: aviddavid, Flickr) - With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At [latex]\text{9.0 cents/kW · h}[/latex], how much does this cost?
- What would be 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 [latex]\text{10 cents/kWh}[/latex]. Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example.
- 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 car? (b) What current flows through it?
- Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at [latex]1\text{.}\text{00 A}\cdot \text{h}[/latex] and 1.58 V keep a 1.00-W flashlight bulb burning?
- A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path?
- The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages [latex]\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{cents/kW}\cdot \text{h}[/latex].
- An old lightbulb draws only 50.0 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.
- 00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a kilometer of such wire when it carries [latex]1.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{A}[/latex].
- Integrated Concepts Cold vaporizers pass a 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 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 22.4.)
Figure 22.4. This cold vaporizer passes current directly through water, vaporizing it directly with relatively little temperature increase. - Integrated Concepts (a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of [latex]1.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{MV}[/latex], and a length of 1.00 ms? (b) What mass of tree sap could be raised from [latex]\text{18}\text{.}0º\text{C}[/latex] to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water?
- Integrated Concepts What current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and [latex]3.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{g}[/latex] of aluminum from [latex]\text{20}\text{.}0º\text{C}[/latex] to [latex]\text{90}\text{.}0º\text{C}[/latex] in 5.00 min?
- Integrated Concepts How much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from [latex]\text{37}\text{.}0º\text{C}[/latex] to [latex]\text{100º}\text{C}[/latex] and then boil away 0.500 g of water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings.
- Integrated Concepts Hydroelectric generators (see Figure 22.5) at Hoover Dam produce a maximum current of [latex]8.00×{\text{10}}^{\text{3}}\phantom{\rule{0.25em}{0ex}}\text{A}[/latex] 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 change) but loses 160 m in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency?
Figure 22.5. Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan) - 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 car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a [latex]2.00×{\text{10}}^{\text{2}}\text{-m}[/latex]-high hill in 2.00 min at a constant 25.0-m/s speed while exerting [latex]5.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a [latex]5.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] force to overcome air resistance and friction? See Figure 22.6.
Figure 22.6. This REVAi, an electric car, gets recharged on a street in London. (credit: Frank Hebbert) - Integrated Concepts A 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 reach 20.0 m/s starting from rest if its loaded mass is [latex]5\text{.}\text{30}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex], assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d) Discuss how the acceleration you found for the light-rail train compares to what might be typical for an automobile.
- Integrated Concepts (a) An aluminum power transmission line has a resistance of [latex]0\text{.}\text{0580}\phantom{\rule{0.25em}{0ex}}\Omega /\text{km}[/latex]. 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.
- Integrated Concepts (a) An immersion heater utilizing 120 V can raise the temperature of a [latex]1.00×{\text{10}}^{\text{2}}\text{-g}[/latex] aluminum cup containing 350 g of water from [latex]\text{20}\text{.}0º\text{C}[/latex] to [latex]\text{95}\text{.}0º\text{C}[/latex] in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.
- Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of water from [latex]\text{10}\text{.}0º\text{C}[/latex] to [latex]\text{40}\text{.}0º\text{C}[/latex], assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is [latex]\text{9}\phantom{\rule{0.25em}{0ex}}\text{cents/kW}\cdot \text{h}[/latex]. (b) What current was used by the 220-V AC electric heater, if this took 4.00 h?
- Unreasonable Results (a) What current is needed to transmit [latex]1.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{MW}[/latex] of power at 480 V? (b) What power is dissipated by the transmission lines if they have a [latex]1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{-}\phantom{\rule{0.25em}{0ex}}\Omega[/latex] resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent?
- Unreasonable Results (a) What current is needed to transmit [latex]1.00×{\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{MW}[/latex] of power at 10.0 kV? (b) Find the resistance of 1.00 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? (e) Which assumptions are unreasonable, or which premises are inconsistent?
- Construct Your Own Problem Consider an electric immersion heater used to heat a cup of water to make tea. Construct a problem in which you calculate the needed resistance of the heater so that it increases the temperature of the water 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 to consider a thermal safety switch (perhaps bimetallic) that will halt the process before damaging temperatures are reached in the immersion unit.
Glossary
- electric power
- The rate at which electrical energy is transferred, supplied by a source, or converted by an electrical device. It is measured in watts (W) and is given by the product of current and voltage, [latex]P=IV[/latex].
The rate at which electrical energy is transferred, supplied by a source, or converted by an electrical device. It is measured in watts (W) and is given by the product of current and voltage, [latex]P=IV[/latex].
