OpenStax-CNX module: m54446 1 Electric Power * OpenStax HS Physics This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 1 : By the end of this section, you will be able to: Dene electric power and describe the electric power equation Calculate electric power in circuits of resistors in series, parallel, and complex arrangements Teacher Support: following TEKS: The Learning Objectives in this section will help your students master the (5) Science concepts. The student knows the nature of forces in the physical world. The student is expected to: (5F)design, construct, and calculate in terms of current through, potential dierence across, resistance of, and power used by electric circuit elements connected in both series and parallel combinations Section Key Terms electric power Table 1 Power is associated by many people with electricity. Every day, we use electric power to run our modern appliances. Electric power transmission lines are visible examples of electricity providing power. We also use electric power to start our cars, to run our computers, or to light our homes. Power is the rate at which energy of any type is transferred; electric power is the rate at which electric energy is transferred in a circuit. In this section, we'll learn not only what this means, but also what factors determine electric power. To get started, let's think of light bulbs, which are often characterized in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb (see Figure 1). Although both operate at the same voltage, the 60-W bulb emits more light intensity than the 25-W bulb. This tells us that something other than voltage determines the power output of an electric circuit. Incandescent light bulbs, such as the two shown in Figure 1, are essentially resistors that heat up when current ows through them (they get so hot that they emit visible and invisible light). Thus the two light bulbs in the photo can be considered as two dierent resistors. In a simple circuit such as a light bulb with * Version 1.2: Sep 30, 2016 4:24 pm -0500 http://creativecommons.org/licenses/by/4.0/
OpenStax-CNX module: m54446 2 a voltage applied to it, the resistance determines the current (by Ohm's law), so we can see that current as well as voltage must determine the power. Figure 1: On the left is a 25-W light bulb, and on the right is a 60-W light bulb. Why are their power outputs dierent despite their operating on the same voltage? The formula for power may be found by dimensional analysis. Consider the units of power. In the SI system, power is given in watts (W), which is energy per unit time, or J/s: W = J s Recall now that a voltage is the potential energy per unit charge, which means that voltage has units of J/C: V = J C We can rewrite this equation as J = V C and substitute this into the equation for watts to get: W = J s = V C = V C (1) s s But a Coulomb per second (C/s) is an electric current, which we can see from the denition of electric current, I = Q t, where Q is the charge in coulombs and t is time in seconds. Thus, equation above tells us that electric power is voltage times current, or: P = IV (1) This equation gives the electric power consumed by a circuit with a voltage drop of V and a current of I. For example, consider the circuit in Figure 2. From Ohm's law, the current running through the circuit is:
OpenStax-CNX module: m54446 3 I = V R = 12 V 100 Ω = 0.12 A Thus, the power consumed by the circuit is: P = V I = (12 V) (0.12 A) = 1.4 W Where does this power go? In this circuit, the power goes primarily into heating the resistor in this circuit. Figure 2: A simple circuit that consumes electric power. In calculating the power in the circuit of Figure 2, we used the resistance and Ohm's law to nd the current. Ohm's law gives the current:i = V/R, which we can insert into the equation for electric power to obtain: P = IV = ( ) V V = V 2 R R This gives the power in terms of only the voltage and the resistance. We can also use Ohm's law to eliminate the voltage in the equation for electric power and obtain an expression for power in terms of just the current and the resistance. If we write Ohm's law as V = IR and use this to eliminate V in the equation P = IV, we obtain: P = IV = I (IR) = I 2 R (2) This gives the power in terms of only the current and the resistance. Thus, by combining Ohm's law with the equation P = IV for electric power, we obtain two more expressions for power: one in terms of voltage and resistance and one in terms of current and resistance. Note that only resistance (not capacitance or anything else), current, and voltage enter into the expressions for electric power. This means that the physical characteristic of a circuit that determines how much power it dissipates is its resistance. Any capacitors in the circuit do not dissipate electric poweron the contrary, capacitors either store electric energy or release electric energy back to the circuit. To clarify how voltage, resistance, current, and power are all related, consider Figure 3, which shows the formula wheel. The quantities in the center quarter circle are equal to the quantities in the corresponding outer quarter circle. For example, to express a potential V in terms of power and current, we see from the formula wheel that V = P/I. (2)
OpenStax-CNX module: m54446 4 Figure 3: The formula wheel shows how volts, resistance, current, and power are related. The quantities in the inner quarter circles equal the quantities in the corresponding outer quarter circles. Worked Example: 1 (Solution on p. 11.) A typical older incandescent lightbulb was 60 W. Assuming that 120 V is applied across the lightbulb, what is the current through the lightbulb? : We are given the voltage and the power output of a simple circuit containing a lightbulb, so we can use the equation P = IV to nd the current I that ows through the lightbulb.
OpenStax-CNX module: m54446 5 Discussion This is a signicant current. Recall that household power is AC and not DC, so the 120 V supplied by household sockets is an alternating power, not a constant power. The 120 V is actually the time-averaged power provided by such sockets. Thus, the average current going through the light bulb over a period of time longer than a few seconds is 0.50 A. Worked Example: 2 (Solution on p. 11.) To warm your boots on cold days, you decide to sew a circuit with some resistors into the insole of your boots. You want 10 W of heat output from the resistors in each insole, and you want to run them from two 9-V batteries (connected in series). What total resistance should you put in each insole? : We know the desired power and the voltage (18 V, because we have two 9-V batteries connected in series), so we can use the equation P = V 2 /R to nd the requisite resistance. Discussion Let's see how much current would run through this circuit. We have 18 V applied across a resistance of 32 Ω, so Ohm's law gives: I = V R = 18 V 32 Ω = 0.56 A All batteries have labels that say how much charge they can deliver (in terms of a current multiplied by a time). A typical 9-V alkaline battery can deliver a charge of 565 ma h (so two 9 V batteries deliver 1130 ma h), so this heating system would function for a time of: t = 1130 10 3 A h 0.56 A = 2.0 h Worked Example: 3 (Solution on p. 11.) Each resistor in the circuit below is 30 Ω. What power is dissipated by the middle branch of the circuit?
OpenStax-CNX module: m54446 6 : The middle branch of the circuit contains resistors R 3 and R 5 in series. The voltage across this branch is 12 V. We will rst nd the equivalent resistance in this branch, and then use P = V 2 /R to nd the power dissipated in the branch. Discussion Let's see if energy is conserved in this circuit by comparing the power dissipated in the circuit to the power supplied by the battery. First, the equivalent resistance of the left branch is: R left = The power through the left branch is: 1 1 + R 4 = + 30 Ω = 45 Ω 1/R 1 + 1/R 2 1/30 Ω + 1/30 Ω P left = V 2 (12 V)2 = R left 45 Ω = 3.2 W The right branch contains only R 6, so the equivalent resistance is R right = R 6 = 30 Ω. The power through the right branch is: P right = V 2 (12 V)2 = R right 30 Ω = 4.8 W The total power dissipated by the circuit is the sum of the powers dissipated in each branch: P = P left + P middle + P right = 2.4 W + 3.2 W + 4.8 W = 10.4 W The power provided by the battery is: P = IV where I is the total current owing through the battery. We must therefore add up the currents going through each branch to obtain I. The branches contributes currents of: The total current is: V I left = R left = 12 V 45 Ω = 0.2667 A V I middle = R middle = 12 V 60 Ω = 0.20 A V I right = R right = 12 V 30 Ω = 0.40 A I = I left + I middle + I right = 0.2667 A + 0.20 A + 0.40 A = 0.87 A and the power provided by the battery is: P = IV = (0.87 A) (12 V) = 10.4 W This is the same power as is dissipated in the resistors of the circuit, which shows that energy is conserved in this circuit.
OpenStax-CNX module: m54446 7 2 Practice Problems Exercise 4 1 Exercise 5 2 3 Check your Understanding Exercise 6 3 Exercise 7 4 4 Section Summary Electric power is dissipated in the resistances of a circuit. Capacitors do not dissipate electric power. Electric power is proportional to the voltage and the current in a circuit. Ohm's law provides two extra expressions for electric power: one that does not involve current and one that does not involve voltage. 5 Key Equations For a given current I owing through a potential dierence V, the electric power dissipated For a given current I owing through a resistance R, the electric power dissipated For a given voltage dierence V across a resistor R, the electric power dissipated P = IV P = I 2 R P = V 2 R Table 2 6 Concept Items Exercise 8 5 1 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex082 2 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex083 3 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex084 4 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex085 5 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex088
OpenStax-CNX module: m54446 8 7 Critical Thinking Exercise 9 6 8 Problems Exercise 10 7 Exercise 11 8 9 Test Prep Multiple Choice Exercise 12 9 Exercise 13 10 10 Test Prep Short Answer Exercise 14 11 Exercise 15 12 Exercise 16 13 Exercise 17 14 11 Performance Task Exercise 18 Materials 1. an incandescent light bulb (i.e., and old-fashioned light bulb with a little wire in it) 2. a lightbulb socket to hold the light bulb 3. a variable voltage source 6 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex089 7 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex086 8 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex087 9 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex090 10 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex091 11 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex092 12 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex093 13 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex094 14 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex095
OpenStax-CNX module: m54446 9 4. an ammeter Procedure Screw the lightbulb into its socket. Connect the positive terminal of the voltage source to the input of the ammeter. Connect the output of the ammeter to one connection of the socket. Connect the other connection of the socket to the negative terminal of the voltage source. Ensure that the voltage source is set to supply DC voltage and that the ammeter is set to measure DC amperes. The desired circuit is shown below. On a piece of paper, make a two-column table with 10 rows. Label the left column volts and the right column current. Adjust the voltage source so that it supplies from between 1 and 10 volts DC. For each voltage, write the voltage in the volts column and the corresponding amperage measured by the ammeter in the current column. Make a plot of volts versus
OpenStax-CNX module: m54446 10 current, that is, a plot with volts on the vertical axis and current on the horizontal axis. Use this data and the plot to answer the following questions: 1. What is the resistance of the lightbulb? 2. What is the range of possible error in your result for the resistance? 3. In a single word, how would you describe the curve formed by the data points? Teacher Support: This performance task supports NGSS Practice 4 : Analyzing and interpreting data, and TEKS 2K : Communicate valid conclusions supported by the data through various methods such as lab reports, labeled drawings, graphic organizers, journals, summaries, oral reports, and technology-based reports Once collected, data must be presented in a form that can reveal any patterns and relationships and that allows results to be communicated to others. Because raw data as such have little meaning, a major practice of scientists is to organize and interpret data through tabulating, graphing, or statistical analysis. Such analysis can bring out the meaning of data and their relevance so that they may be used as evidence. Engineers, too, make decisions based on evidence that a given design will work; they rarely rely on trial and error. Engineers often analyze a design by creating a model or prototype and collecting extensive data on how it performs, including under extreme conditions. Analysis of this kind of data not only informs design decisions and enables the prediction or assessment of performance but also helps dene or clarify problems, determine economic feasibility, evaluate alternatives, and investigate failures. The curve should be linear and approximately go through the origin of the plot. The resistance of a 100-W, 120-V incandescent lightbulb should be around 144 Ω. 12 Test Prep Extended Response Exercise 19 15 Exercise 20 16 15 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex096 16 http://cnx.org/content/m54446/latest/#ost/api/ex/k12phys-ch19-ex097
OpenStax-CNX module: m54446 11 Solutions to Exercises in this Module to Exercise (p. 4): Solution Solving P = IV for the current and inserting the given values for voltage and power gives: P = IV I = P V = 60 W 120 V = 0.50 A Thus, a half ampere ows through the lightbulb when 120 V is applied across it. to Exercise (p. 5): Solution Solving P = V 2 /R for the resistance and inserting the given voltage and power, we obtain: P = V 2 R R = V 2 P = (18 V)2 10 W = 32 Ω Thus, the total resistance in each insole should be 32 Ω. to Exercise (p. 5): Solution The equivalent resistance is R middle = R 3 + R 5 = 30 Ω + 30 Ω = 60 Ω. The power dissipated by the middle branch of the circuit is: Glossary P middle = V 2 (12 V)2 = R middle 60 Ω = 2.4 W Denition 3: electric power rate at which electric energy is transferred in a circuit