Chapter 3: Electricity


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1 Chapter 3: Electricity Goals of Period 3 Section 3.1: To define electric charge, voltage, and energy Section 3.2: To define electricity as the flow of charge Section 3.3: To explain the generation electricity Section 3.4: To discuss the transmission of electricity Section 3.5: To calculate the cost of using electricity 3.1 Electric Charge, Voltage, and Energy Electric Charge The world around us is full of electric charge. Evidence for the existence of electric charge comes from the electric forces between charged objects. Because electric charges on two adjacent objects can cause the objects to move, we conclude that a force must exist between the electric charges. An electric force can move charged objects closer together (an attractive force) or move them apart (a repulsive force). Since electric forces can move charged objects together or apart, we conclude that two opposite types of electric charge must exist. We rarely notice electric charge because most objects contain equal amounts of these two opposite types of electric charge, which cancel one another. A charged object results when a quantity of one type of charge is separated from an equal quantity of the opposite type of charge. The sum of equal quantities of opposite charge is zero. For this reason, the two types of electric charge are called positive charge (+) and negative charge ( ). Objects with equal numbers of positive and negative charge have a total net charge of zero and are electrically neutral. Objects with more positive than negative charge have a net positive charge, and objects with more negative than positive charge have a net negative charge. Charge is measured in units of coulombs (coul). We will use the variable Q to represent charge. Electrical Potential Energy and Voltage Chapter 2 defined work as the product of the force applied to an object times the distance that force caused the object to move: W = F x D. If we ignore wasted energy, the joules of energy required to do this work is equal to the joules of work done. Work and energy are likewise xx in electric charge. When the attractive electric force between positive and negative charges pulls charge together, work must be done against this attractive force to separate the charges. When separated charges are allowed to come back together, we can get work back out. On the other hand, since two positive charges or two negative charges repel one another, we must do work against this repulsive electric force to bring the charges together. We can get work out as the two charges move apart. 24
2 Electrical potential energy is described in terms of charge, Q, and voltage, V, rather than in terms of force times distance. The energy per charge is called the voltage. Voltage is measured in units of volts (V), with 1 volt = 1 joule of energy per 1 coulomb of charge. Voltage is the ratio E pot /Q, that is, voltage is the potential energy per unit of charge as shown in Equation 3.1. E pot Q V (Equation 3.1) where E pot = the electrical potential energy (in joules) (Example 3.1) Q = the charge (in coulombs) V = the voltage (in volts) Electric charge is placed on a metal surface. How much electrical potential energy do 2 x 10 9 coulombs of electric charge have when maintained at a voltage of 2,000 volts? E pot = Q V = 2 x 10 9 coul x 2,000 V = 4 x 10 6 joules 3.2 Electric Current Moving Electric Charge Produces Electric Current When separated electric charge flows through conducting material, the moving charge constitutes an electric current in the conductor. Electric current usually consists of flowing electrons, which can move freely in a conducting material. The amount of electric current is the rate at which charge Q flows past a given point in the conductor. Since charge is measured in coulombs, current I is measured in units of coulombs per second. One coulomb per second is called one ampere (amp). The greater the amount of charge that flows, the larger the electric current. We can express this definition of current with the equation: or Current = Amount of Charge moved Elapsed Time Q I (Equation 3.2) t with I = current (in amperes) Q = charge (in coulombs) t = time elapsed (in seconds) 25
3 (Example 3.2) How much charge must flow to provide a current of 10 amps for 20 seconds? Solve equation 3.2 for Q by multiplying both sides by t and canceling: t Q t I t 10 amps x 20 sec = 200 coul Concept Check 3.1 a) How much current is present if 10 coulombs of charge flow through a conductor every 5 seconds? b) How long must a 5 amp current flow to provide 200 coul of charge? 3.3 Generating Electricity Moving Electric Charges and Magnetic Fields When an electric charge, such as an electron, moves, the charge is surrounded by a magnetic field. The magnetic field produced by an electric current can exert a force and do work on nearby permanent magnets or on other moving electric charges. Likewise, a nearby magnetic field can do work on a currentcarrying wire. But an additional effect also occurs a changing magnetic field can produce a current in a nearby conductor such as a piece of metal or a wire that is part of a closed or complete circuit. The changing magnetic field can be produced by moving a magnet or by moving the conductor. The process of generating a current in a conductor is known as inducing a current. In class, we induce a current by moving a magnet into and out of a coil of wire. Moving the magnet into the coil induces a current in one direction in the wire. When the magnet stops moving, the current stops flowing. Pulling the magnet out of the coil produces a current flowing in the opposite direction. To produce this current, only the relative motion of the coil and wire is important the same current is induced whether the coil moves or the magnet moves. Electric Generating Plants The principle that a magnet moving with respect to a wire induces a current in the wire is used in power plants to generate electricity. Generators use magnets and coils of wire to convert kinetic energy into electrical energy. Electric generating plants convert the kinetic energy of rotating magnets into electrical energy by spinning large magnets near coils of conducting wire. To rotate the magnets, generating plants use 26
4 kinetic energy from the sources described below to turn the blades of turbines. Turbines are wheels with blades attached, similar in principle to waterwheels. A shaft attached to the rotating turbine causes the magnets to spin. The most common mechanism for turning turbines is steam pressure. In steam generating plants, water is heated in a closed container. As the water changes to steam, the volume and pressure increase. The steam exerts pressure on the turbine blades, turning them. In many generating plants, water is heated by burning coal, oil, natural gas, or other fuels. In nuclear power plants, water is heated by the thermal energy that results from reactions in radioactive fuels. Fig. 3.1 Turbine In hydroelectric plants, falling water rotates the turbines. The gravitational potential energy of the water is converted into kinetic energy of motion as the water falls and turns the turbines. In addition to the many hydroelectric power plants built along rivers, a tidal powered generating plant along the ocean shore can use tides to turn turbines. In tidal plants, water flows in during high tide and is trapped behind gates. As the tide recedes, the trapped water is left at a higher level than the surrounding ocean. The trapped water returning to the ocean spills over the turbines, turning them. Wind turbines use the kinetic energy of moving air molecules to spin large blades that are attached to a turbine. Wind turbines are most effective when placed in a near constant source of wind, such as along the shore of large bodies of water or on flat plains. 3.4 Transmitting Electricity Joule Heating As electric current moves through wires, electrons collide with atoms of the wire material and some electrical energy is converted into thermal energy. Conversion of electrical energy into thermal energy in a resistor is known as joule heating. Unless the purpose of transmitting current is to generate heat in a wire, such as in a toaster s filament, this thermal energy is wasted. The watts of joule heating power are the product of the current squared times the resistance of the wire. Resistance is the ability of a material to resist the flow of electric current. Resistance is measured in units of ohms ( ). (Equation 3.3) P joule = I 2 R with P joule = power (in watts) I = current (in amperes) R = resistance (in ohms) 27
5 Joule heating in power transmission lines represents both a waste of energy resources and a loss of revenue for the power generating company. Even a voltage drop of only 18 volts between the generating plant and the user represents a large amount of power waste from joule heating. Example 3.3 calculates the amount of wasted power. (Example 3.3) How much power would be wasted as joule heating of the transmission wires if 1.67 x 10 6 amps of current was transmitted with the very small resistance of 1.1 x 105 ohms? P joule = I 2 R = (1.67 x 10 6 amps) 2 x (1.1 x 105 ohms) = (2.79 x amps 2 ) x (1.1 x 105 ohms) = 3.1 x 10 7 watts As shown in Example 3.3, transmitting large amounts of current even at very low resistance would wastes large amounts of power as joule heating of the wires. Fortunately, there is another alternative power transmission at high voltages. Transformers One way to reduce the amount of current in transmission wires is to increase the voltage of that current. Power is directly proportional to the product of current and voltage as shown in Equation 3.4. A higher voltage results in a smaller current for a given amount of power transmitted. (Equation 3.4) P = I V with P = power (in watts) I = current (in amperes) V = voltage (in volts) Transformers make it possible to transmit electricity at a higher voltage by changing the voltage of the electric current. The purpose of a transformer is to trade high voltage for low current or low voltage for high current. From Equation 3.4, we see that as the voltage is increased, the current must decrease to provide the same power. Likewise, if the voltage is decreased, the current increases. Stepup transformers increase voltage and decrease current for long distance transmission. While high voltage reduces joule heating, it poses a safety hazard for consumers. Stepdown transformers are used to decrease voltages to safe levels at the point of use of the electricity. 28
6 Stepdown transformers reduce the voltage of transmitted power in stages to progressively lower voltages. Power is transmitted over long distances at up to 700,000 volts. Over shorter distances or to small towns with low power requirements, power is transmitted at 25,000 volts. Power lines along city streets typically operate at about 2,000 volts. Finally, a transformer about the size of a garbage can, often located on a utility pole, reduces the voltage for household use. Households are normally supplied 240 volts, which is delivered in such a way as to provide 120 volts to regular outlets and 240 volts to outlets for devices that require more power, such as electric stoves, water heaters, and clothes dryers. (Example 3.5) If a power company transmits electricity at 500,000 volts (5 x 10 5 V) rather than 120 volts, how much current would be needed to provide 200 megawatts of power (2 x 10 8 watts) to a city? Solve P = I V for I by dividing both sides by V and canceling: P V I 2 x x 10 5 watts volts 400 amps Just 400 amps of current could supply 200 megawatts of power when transmitted at a high voltage, compared to the 2.79 x amps of current required to transmit at 120 volts in Example 3.4. Transmitting at high voltage and low current greatly reduces the power wasted as joule heating of the wires. Transformers make it possible to transmit electricity at high voltages, then step the electricity down to lower voltages that are safer for consumer use. Concept Check 3.2 a) A 400 amp current is transmitted through wires with a resistance of 0.8 ohms for every mile of wire. How much power is wasted as joule heating of a transmission wire 200 miles in length? b) If 200 megawatts (2 x 10 8 watts) is the total power input, what is the efficiency of the transmission process? 29
7 3.5 Cost of Electricity Power Measurements and Requirements Power companies provide electrical energy, or electricity, and charge for it in units of kilowatthours (kwh). To measure the electrical energy supplied to consumers, power companies use kilowatthour meters attached to residences and businesses. The meter is a small electric motor with a rotating disc. Because the rotation speed of the disc is proportional to the power being used at the time, electricity can be measured by the number of rotations of the disc. The measurement of electricity may be shown by a digital display or by an analog display with rotating pointers on dials. The difference between the current month's dial reading and previous month's reading measures the total energy provided, which is used to determine the electric bill. To calculate the cost of electricity from an electric bill, we must know how many kilowatts of electric power were used and for how many hours. Power is the energy transferred or work done divided by the time required, as shown in Equation 3.5. with P = E = W t t P = power (in watts) E = energy transferred (in joules) W = work done (in joules) t = time elapsed (in seconds) (Equation 3.5) From Equation 3.5, you see that the faster energy is transferred or work is done, the greater the power required. (Example 3.6) How many kilowatthours (kwh) of energy are required to operate a 1,000 watt electric heater for 2 hours? Find the energy used by an appliance by solving equation 3.5 for the energy requirement, E. P E t or E P t E = 1,000 watts x 2 hours = 2,000 watthours. Use a ratio to convert from watts to kilowatts: 2,000 watthours x 1 kw = 2 kilowatthours = 2 kwh 1,000 watts 30
8 The total kilowatthours of energy used during a month is the sum of the power requirement of each appliance multiplied by the hours that appliance operated. (Example 3.7) How many kilowatthours of energy are used when a 100 watt light bulb burns for 7 hours, a 1,000 watt hair dryer operates for 15 minutes, and a 1,500 watt electric heater operates for ½ hour? Light bulb: 1kilowatt 100 watts x x 7 hours 1,000 watts 0.70 kilowatt hours Dryer: 1kilowatt 1 hour 1,000 watts x x 15 min x 1,000 watts 60 min 0.25 kilowatt hours Heater: 1 kilowatt 1 hour 1,500 watts x x 30 min x 1,000 watts 60 min 0.75 kilowatt hours The total energy used is 0.70 kwh kwh kwh = 1.7 kilowatthours. (Example 3.8) If the power company charges $0.10 per kilowatthour of energy, how much would you pay for the energy used by the light bulb, hair dryer, and heater in Example 3.7? Energy Cost = number of kwh used x cost per kwh $ kilowatt hours x 1 kilowatt hour $0.17 Concept Check 3.3 a) How many kilowatthours are required to use a 600 watt oven for 45 minutes? b) If electricity costs $0.10 per kilowatthour, how much would you pay to operate the oven for 45 minutes? 31
9 Table 3.1: Metric and English Units and Their Symbols Quantity Symbol Metric Units Metric Unit Abbreviation English Unit English Unit Abbrev. Force F newton N (1 N = 1 kg m/s 2 ) pound lb Energy E joule J (1 J = 1 kg m 2 /s 2 ) footpound ftlb Work W joule J footpound ftlb Power P watt W (1 W = 1 J/s) horsepower hp (1 hp = 550 ftlb/s) 32
10 Chapter 3 Summary 3.1 Electric charge Q is measured in coulombs. The voltage V is the energy per charge. The electrical potential energy of charge Epot = Q V 3.2: Moving electric charge is electric current I. Current is measured in units of amperes or amps. Since current is the rate of flow of charge, I = Q/t 3.3: Electric current is generated, or induced, when magnets and coils of wire move relative to one another. In power generating plants, kinetic energy from steam, flowing water, or wind turns turbines. The moving turbines spin magnets near coils of wire. 3.4: Electric generating plants are often located far from heavily populated areas, thus the electricity they generate must be transmitted long distances to its point of use. To transmit electricity to consumers, the electricity is transmitted at high voltage to minimize the power wasted as joule heating of the transmission wires. Joule heating (P joule = I 2 R) occurs when some electrical energy transmitted through wires is wasted as thermal energy heating the wires. Transformers trade current and voltage, while keeping the power nearly constant. (P = I V). Stepup transformers are used to reduce the current and increase the voltage for efficient long distance transmission of electricity. Less current transmitted reduces power wasted as joule heating. Near the point of electricity consumption, stepdown transformers reduce the voltage and increase the current for safer use of electricity. 3.5: Power is the rate at which work is done or energy is transferred. P = W/t or P = E/t Power is measured in joules/second, or watts, in the metric system and in footpounds/second, or horsepower, in the English system. Electrical energy provided to homes is measured and billed to consumers in units of kilowatthours. To find the cost of electricity for using an appliance: 1) Convert watts to kilowatts, by dividing watts by 1,000. 2) Find the total time of use in hours. 3) Multiply kilowatts times hours to find the total kwh. 4) Multiply kwh by the cost per kwh. 33
11 Solutions to Chapter 3 Concept Checks 3.1 a) I = Q = 10 coul. = 2 amps t 5 sec b) I = Q or t = Q = 200 coul. = 40 sec t I 5 amps 3.2 a) P joule = I 2 R = (400 amps) 2 x 0.8 ohms = 128,000 watts/mile 128,000 watts/mile x 200 miles = 25,600,000 watts = 2.56 x 10 7 watts b) 2 x 10 8 watts = total power input total power input = useful power out + wasted power out useful power out = total power input wasted power out useful power out =2 x x 10 7 = 2 x x 10 8 = 1.74 x 10 8 (Note that 2.56 x 10 7 was converted to x Adding or subtracting numbers in scientific notation requires each number to have the same power of ten.) Efficiency = useful power out = 1.74 x 10 8 watts = 0.87 = 87% total power in 2 x 10 8 watts 3.3 a) E = P t = 600 watts x 0.75 hours = 450 watthours. Use a ratio to convert from watts to kilowatts: 450 watthours x 1 kw = 0.45 kilowatthours = 0.45 kwh 1,000 watts b) 0.45 kwh x $0.10 = $0.045 kwh 34
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