5.8: Electric Power and Energy
By the end of this section, you will be able to:
- Calculate the power dissipated by a resistor and power supplied by a power supply.
- Calculate the cost of electricity under various circumstances.
- Express electrical power in terms of the voltage and the current
- Describe the power dissipated by a resistor in an electric circuit
- Calculate the energy efficiency and cost effectiveness of appliances and equipment
In an electric circuit, electrical energy is continuously converted into other forms of energy. For example, when a current flows in a conductor, electrical energy is converted into thermal energy within the conductor. The electrical field, supplied by the voltage source, accelerates the free electrons, increasing their kinetic energy for a short time. This increased kinetic energy is converted into thermal energy through collisions with the ions of the lattice structure of the conductor. Power is defined as the rate at which work is done by a force and is measured in watts. Power can also be defined as the rate at which energy is transferred. In this section, we discuss the time rate of energy transfer, or power, in an electric circuit.
Power in Electric Circuits
Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power ?
Let us compare a 25-W bulb with a 60-W bulb (Figure \(\PageIndex{1a}\)). The 60-W bulb glows brighter than the 25-W bulb. Although it is not shown, a 60-W light bulb is also warmer than the 25-W bulb. The heat and light is produced by from the conversion of electrical energy. The kinetic energy lost by the electrons in collisions is converted into the internal energy of the conductor and radiation. How are voltage, current, and resistance related to electric power?
Electric energy depends on both the voltage involved and the charge moved. This is expressed most 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). Power is the rate at which energy is moved, and so electric power is
\[P = \frac{PE}{t} = \frac{qV}{t}.\label{20.5.1}\]
Recognizing that current is \(I = q/t\) (note that \(\Delta t = t\) here), the expression for power becomes
\[P = IV.\label{20.5.2}\]
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, \(1 A \cdot V = 1 W\). For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power \(P = IV = \left(20A\right) \left(12V\right) = 240W\). In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes ( \(1kA \cdot V = 1kW\) ).
To see the relationship of power to resistance, we combine Ohm’s law with \(P = IV\). Substituting Ohm Law (\(I = V/R\)) into Equation \ref{20.5.2} gives
\[P = \left( V/R \right) V = V^{2} / R.\]
Similarly, substituting \(V = IR\) gives
\[P = \left( IR \right) = I^{2}R.\]
Three expressions for electric power are listed together here for convenience:
\[P = IV \label{20.5.3}\]
\[P = \frac{V^{2}}{R}\label{20.5.4}\]
\[P = I^{2}R.\label{20.5.5}\]
Note that the first equation is always valid, whereas the other two 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 single device and not the total power in the circuit.)
The electric power gained or lost by any device has the form
\[P = IV.\]
The power dissipated by a resistor has the form
\[P = I^2 R = \dfrac{V^2}{R}.\]
Different insights can be gained from the three different expressions for electric power. For example, \(P = V^{2} / R\) 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 effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning 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.
(a) Consider the examples given in 20.3 and 20.4. Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold.
Strategy
For the hot headlight, we know voltage and current, so we can use \(P = IV\) to find the power. For the cold headlight, we know the voltage and resistance, so we can use \(P=V^{2}/R\) to find the power.
Solution
Entering the known values of current and voltage for the hot headlight, we obtain \[P = IV = \left(2.50 A\right)\left(12.0 V\right) = 30.0 W.\] The cold resistance was \(0.350 \Omega\), and so the power it uses when first switched on is
\[P = \frac{V^{2}}{R} = \frac{\left(12.0 V\right)^{2}}{0.350 \Omega} = 411 W.\]
Discussion
The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb’s temperature increases and its resistance increases.
(b) What current does it draw when cold?
Solution
The 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
\[I = \sqrt{\frac{P}{R}} = \sqrt{\frac{411 W}{0.350 \Omega}} = 34.3 A.\]
Discussion
The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb’s temperature increases. Most fuses and circuit 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 current remains high for several seconds, necessitating special “slow blow” fuses.
A DC winch motor is rated at 20.00 A with a voltage of 115 V. When the motor is running at its maximum power, it can lift an object with a weight of 4900.00 N a distance of 10.00 m, in 30.00 s, at a constant speed.
- What is the power consumed by the motor?
- What is the power used in lifting the object? Ignore air resistance. (c) Assuming that the difference in the power consumed by the motor and the power used lifting the object are dissipated as heat by the resistance of the motor, estimate the resistance of the motor?
Strategy
- The power consumed by the motor can be found using \(P = IV\).
- The power used in lifting the object at a constant speed can be found using \(P = Fv\), where the speed is the distance divided by the time. The upward force supplied by the motor is equal to the weight of the object because the acceleration is zero. (c) The resistance of the motor can be found using \(P = I^2 R\).
Solution
- The power consumed by the motor is equal to \(P = IV\) and the current is given as 20.00 A and the voltage is 115.00 V: \[P = IV = (20.00 \, A) 115.00 \, V = 2300.00 \, W.\]
- The power used lifting the object is equal to \(P = Fv\) where the force is equal to the weight of the object (1960 N) and the magnitude of the velocity is \[v = \dfrac{10.00 \, m}{30.00 \, s} = 0.33 \dfrac{m}{s}\] \[P = Fv = (4900 \, N)0.33 \, m/s = 1633.33 \, W.\]
- The difference in the power equals \(2300.00 \, W - 1633.33 \, W = 666.67 \, W\) and the resistance can be found using \(P = I^2R\): \[R = \dfrac{P}{I^2} = \dfrac{666.67 \, W}{(20.00 \, A)^2} = 1.67 \, \Omega.\]
Significance The resistance of the motor is quite small. The resistance of the motor is due to many windings of copper wire. The power dissipated by the motor can be significant since the thermal power dissipated by the motor is proportional to the square of the current \((P = I^2R)\).
A fuse (Figure \(\PageIndex{3}\)) is a device that protects a circuit from currents that are too high. A fuse is basically a short piece of wire between two contacts. As we have seen, when a current is running through a conductor, the kinetic energy of the charge carriers is converted into thermal energy in the conductor. The piece of wire in the fuse is under tension and has a low melting point. The wire is designed to heat up and break at the rated current. The fuse is destroyed and must be replaced, but it protects the rest of the circuit. Fuses act quickly, but there is a small time delay while the wire heats up and breaks.
Circuit breakers are also rated for a maximum current, and open to protect the circuit, but can be reset. Circuit breakers react much faster. The operation of circuit breakers is not within the scope of this chapter and will be discussed in later chapters. Another method of protecting equipment and people is the ground fault circuit interrupter (GFCI), which is common in bathrooms and kitchens. The GFCI outlets respond very quickly to changes in current. These outlets open when there is a change in magnetic field produced by current-carrying conductors, which is also beyond the scope of this chapter and is covered in a later chapter.
The Cost of Electricity
The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since \(P = E/t\), we see that \[E = Pt\label{20.5.6}\] is the energy used by a device using power \(P\) 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 energy unit on electric bills is the killowatt-hour ( \(kW \cdot h\) ), consistent with the relationship \(E = Pt\). It is easy to estimate the cost of operating electric appliances if you have some idea 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 energy units such as food calories, can be converted to joules. You can prove to yourself that \(1 kW \cdot h = 3.6 \times 10^{6} J\).
The electric energy ( \(E\) ) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About 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 true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 1b.) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has the same brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat 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 small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their cost is still high.
The relationship \(E = Pt\) is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiation dose of an X-ray image is related to the power and time of exposure.
(a) If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents?
Strategy
To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour.
Solution
The energy used in kilowatt-hours is found by entering the power and time into the expression for energy: \[E = Pt = \left(60 W\right)\left(1000 h\right) = 60,000 W \cdot h.\] In kilowatt-hours, this is \]E = 60.0 kW \cdot h.\] Now the electricity cost is \[cost = \left(60.0 kW \cdot h\right)\left($0.12/kW \cdot h\right) = $7.20.\] The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day).
(b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be?
Solution
Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $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 here.
The heat transfer from these CFLs is less, and they last up to 10 times longer than incandescent bulbs. 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 even more efficient (twice that of CFLs) and last five times longer than CFLs.
The typical replacement for a 100-W incandescent bulb is a 20-W LED bulb. The 20-W LED bulb can provide the same amount of light output as the 100-W incandescent light bulb. What is the cost savings for using the LED bulb in place of the incandescent bulb for one year, assuming $0.10 per kilowatt-hour is the average energy rate charged by the power company? Assume that the bulb is turned on for three hours a day.
Strategy
- Calculate the energy used during the year for each bulb, using \(E = Pt\).
- Multiply the energy by the cost.
Solution
- Calculate the power for each bulb. \[E_{Incandescent} = Pt = 100 \, W \left(\dfrac{1 \, kW}{1000 \, W}\right) \left(\dfrac{3 \, h}{day}\right) (365 \, days) = 109.5 \, kW \cdot h\] \[E_{LED} = Pt = 20 \, W \left(\dfrac{1 \, kW}{1000 \, W}\right) \left(\dfrac{3 \, h}{day}\right) (365 \, days) = 21.9 \, kW \cdot h\]
- Calculate the cost for each. \[cost_{Incandescent} = 109.5 \, kW \cdot h \left(\dfrac{\$0.10}{kW \cdot h}\right) = \$10.95\] \[cost_{LED} = 21.90 \, kW \cdot h \left(\dfrac{\$0.10}{kW \cdot h}\right) = \$2.19\]
Significance
A LED bulb uses 80% less energy than the incandescent bulb, saving $8.76 over the incandescent bulb for one year. The LED bulb can cost $20.00 and the 100-W incandescent bulb can cost $0.75, which should be calculated into the computation. A typical lifespan of an incandescent bulb is 1200 hours and is 50,000 hours for the LED bulb. The incandescent bulb would last 1.08 years at 3 hours a day and the LED bulb would last 45.66 years. The initial cost of the LED bulb is high, but the cost to the home owner will be $0.69 for the incandescent bulbs versus $0.44 for the LED bulbs per year. (Note that the LED bulbs are coming down in price.) The cost savings per year is approximately $8.50, and that is just for one bulb.
Changing light bulbs from incandescent bulbs to CFL or LED bulbs is a simple way to reduce energy consumption in homes and commercial sites. CFL bulbs operate with a much different mechanism than do incandescent lights. The mechanism is complex and beyond the scope of this chapter, but here is a very general description of the mechanism. CFL bulbs contain argon and mercury vapor housed within a spiral-shaped tube. The CFL bulbs use a “ballast” that increases the voltage used by the CFL bulb. The ballast produce an electrical current, which passes through the gas mixture and excites the gas molecules. The excited gas molecules produce ultraviolet (UV) light, which in turn stimulates the fluorescent coating on the inside of the tube. This coating fluoresces in the visible spectrum, emitting visible light. Traditional fluorescent tubes and CFL bulbs had a short time delay of up to a few seconds while the mixture was being “warmed up” and the molecules reached an excited state. It should be noted that these bulbs do contain mercury, which is poisonous, but if the bulb is broken, the mercury is never released. Even if the bulb is broken, the mercury tends to remain in the fluorescent coating. The amount is also quite small and the advantage of the energy saving may outweigh the disadvantage of using mercury.
The CFL light bulbs are being replaced with LED light bulbs, where LED stands for “light-emitting diode.” The diode was briefly discussed as a nonohmic device, made of semiconducting material, which essentially permits current flow in one direction. LEDs are a special type of diode made of semiconducting materials infused with impurities in combinations and concentrations that enable the extra energy from the movement of the electrons during electrical excitation to be converted into visible light. Semiconducting devices will be explained in greater detail in Condensed Matter Physics .
Commercial LEDs are quickly becoming the standard for commercial and residential lighting, replacing incandescent and CFL bulbs. They are designed for the visible spectrum and are constructed from gallium doped with arsenic and phosphorous atoms. The color emitted from an LED depends on the materials used in the semiconductor and the current. In the early years of LED development, small LEDs found on circuit boards were red, green, and yellow, but LED light bulbs can now be programmed to produce millions of colors of light as well as many different hues of white light.
1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use \(P = IV\).
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 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends?
Comparison of Incandescent, CFL, and LED Light Bulbs
The energy savings can be significant when replacing an incandescent light bulb or a CFL light bulb with an LED light. Light bulbs are rated by the amount of power that the bulb consumes, and the amount of light output is measured in lumens. The lumen (lm) is the SI -derived unit of luminous flux and is a measure of the total quantity of visible light emitted by a source. A 60-W incandescent light bulb can be replaced with a 13- to 15-W CFL bulb or a 6- to 8-W LED bulb, all three of which have a light output of approximately 800 lm. A table of light output for some commonly used light bulbs appears in Table \(\PageIndex{1}\).
The life spans of the three types of bulbs are significantly different. An LED bulb has a life span of 50,000 hours, whereas the CFL has a lifespan of 8000 hours and the incandescent lasts a mere 1200 hours. The LED bulb is the most durable, easily withstanding rough treatment such as jarring and bumping. The incandescent light bulb has little tolerance to the same treatment since the filament and glass can easily break. The CFL bulb is also less durable than the LED bulb because of its glass construction. The amount of heat emitted is 3.4 btu/h for the 8-W LED bulb, 85 btu/h for the 60-W incandescent bulb, and 30 btu/h for the CFL bulb. As mentioned earlier, a major drawback of the CFL bulb is that it contains mercury, a neurotoxin, and must be disposed of as hazardous waste. From these data, it is easy to understand why the LED light bulb is quickly becoming the standard in lighting.
| Light Output (lumens) | LED Light Bulbs (watts) | Incandescent Light Bulbs (watts) | CFL Light Bulbs (watts) |
|---|---|---|---|
| 450 | 4−5 | 40 | 9−13 |
| 800 | 6−8 | 60 | 13−15 |
| 1100 | 9−13 | 75 | 18−25 |
| 1600 | 16−20 | 100 | 23−30 |
| 2600 | 25−28 | 150 | 30−55 |
Summary of Relationships
In this chapter, we have discussed relationships between voltages, current, resistance, and power. Figure \(\PageIndex{4}\) shows a summary of the relationships between these measurable quantities for ohmic devices. (Recall that ohmic devices follow Ohm’s law \(V = IR\).) For example, if you need to calculate the power, use the pink section, which shows that \(P = VI, \, P = \dfrac{V^2}{R}\), and \(P = I^2R\).
Which equation you use depends on what values you are given, or you measure. For example if you are given the current and the resistance, use \(P = I^2R\). Although all the possible combinations may seem overwhelming, don’t forget that they all are combinations of just two equations, Ohm’s law \((V = IR)\) and power \((P = IV)\).
Summary
- Electric power \(P\) is the rate (in watts) that energy is supplied by a source or dissipated by a device.
- Three expressions for electrical power are \[P = IV,\] \[P = \frac{V^{2}}{R},\] and \[P = I^{2}R.\]
- The energy used by a device with a power \(P\) over a time \(t\) is \(E = Pt\).
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