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ADVANCED CLASSICAL PHYSICS PREPARED BY: BRYAN QUINCENA RUEDAS DOCTOR OF PHILOSOPHY IN SCIENCE EDUCATION 1 In this module we will be concerned only with DC Circuits. We begin in EMF and Terminal Voltage with a discussion of basic principles of electric current and apply these principles to analyze dc circuits involving combinations of batteries, resistors, and capacitors. We also study the operation of some useful instruments. It is followed by Resistors in Series and in Parallel which deals with two or more resistors are connected end to end along a single path. The resistors could be simple resistors or they could be lightbulbs, or heating elements, or other resistive devices. This is interwoven with discussions of Kirchhoff’s Rule which based on the based on the conservation of electric charge and EMFs in Series and in Parallel Charging Battery that shows the algebraic sum of their respective voltages. The module also shows the Circuits Containing Capacitors in Series and in Parallel which shows that capacitors can be placed in series or in parallel in a circuit. The RC Circuits—Resistor and Capacitor in Series contains the application of capacitors and resistors are often found together in a circuit. Most importantly, the Electric Hazards are discussed in the module to have an awareness in terms of the excess electric current can overheat wires in buildings and cause fires. Ammeters and Voltmeters—Measurement Affects the Quantity Being Measured are being discussed by the module because Measurement is a fundamental part of physics, and is not as simple as you might think. Measuring instruments can not be taken for granted; their results are not perfect and often need to be interpreted. This module aims to help you to think about the laws of electrical circuitry in ways that will continue to be valid when you study more sophisticated circuits. Study comment: Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the Fast track questions. If not, proceed directly to Ready to study?. i 2 The development of electrical and electronic products in our society requires a sophisticated knowledge of circuits and devices: indeed some knowledge of circuitry is called for in the most elementary maintenance of plugs, fuses and connecting wires. Many measurements of physical quantities are made nowadays by sensors or transducers that result in current or voltage signals being developed in a detection circuit. Moreover, the electrical properties of materials give important clues to the underlying nature of the materials themselves. For all these reasons, the behavior of electrical circuits is an important area of physics. Modern life could not exist if it were not for electricity and electronics. The history of electricity starts more than two thousand years ago, with the Greek philosopher Thales being the earliest known researcher into electricity. But it was Alessandro Volta who created the most common DC power source. Direct current (also known as DC) is the flow of charged particles in one unchanging direction (most commonly found as electron flow through conductive materials). DC can be found in just about every home and electronic device, as it is more practical (compared to AC from power stations) for many consumer devices. Just a few of the places where you can find direct current are batteries, phones, computers, cars, TVs, calculators, and even lightning. Objectives At the end of this module, you are expected to: 1. Describe a DC Circuits. 2. Discuss the EMFs and Terminal Voltage. 3. Differentiate the resistors that connects in a series and parallel connection. 4. Discuss the Kirchhoff's rule. 5. Apply the different rules in DC circuits. 6. Derive formula and solve problems in DC circuits. 7. Describe the importance of electricity hazard. 8. Apply the learnings in measuring electricity using different devices. ii 3 Study comment Can you answer the following Fast track questions. If you answer the questions successfully you do not need to review the basic concepts in the module. If you are sure that you can meet each of these FT Questions, you may proceed to the Ready to Study Section. If you have difficulty with only one or two of the questions you are strongly advised to study the basic concepts in this module. Direction: Find the solution to the given problems. 1. How much current would a 12.0Ω resistor draw when connected to a 220V outlet? 2. An ammeter registers 1.5 A of current in a wire that is connected to a 12.0 V battery. What is the wire’s resistance? 3. The current in a circuit is 6.50 A. if the resistance circuit is 15.7Ω, what is the voltage across the oven? 4. Find the current in and voltage across each of the resistor in the following circuits: a. a 15Ω and a 18Ω resistor connected in series with a 12V source. b. a 15Ω and a 18Ω resistor connected in parallel with a 12V source. A string of 50V identical miniature decorative lights is wired in series. If it draws 0.75A of current when it is connected to a 220V emf source, what is the resistance of each miniature bulb when they are lighted? Answer: 1. 18.33 A 2. 8 Ω 3. 102. 05 V 4. (a.) 0.36 A (b.) 198.18 A 66.67 Ω 5. 5. iii 4 Overview of the Module i Introduction ii Fast Track Questions iii Basic Concept of Electricity Electric Charges 1 Electric Current 3 Resistance 4 Electric Circuit 5 Ohm’s Law 6 Electromotive Force: Terminal Voltage Electromotive Force 7 Internal Resistance 9 Terminal Voltage 12 Let’s Do It: Problem Solving 15 Kirchhoff’s Rule Kirchhoff’s Voltage Law (KVL) 17 KVL in Parallel Circuit 22 Kirchhoff’s Current Law (KCL) 26 Let’s Do It Together: Sample Problem 30 Let’s Check Your Understanding 32 Reference s 32 5 Electricity is a fundamental property of all matter. There are two types of electric charges: positive and negative. The proton and the electron are, respectively, the funda- mental positive and negative charges. If an object has the same number of protons and electrons, it is electrically neutral. Generally, charge is transferred by the gain or loss of electrons. Electrically charged objects attract or repel each other with the force that is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. The relationship is known as Coulomb’s law. Substances that allow charges to move freely through them are called conductors; substances that severely restrict this moment are insulators. The region of space in which an electric charge is subject to an electric force is known as an electric field. The electric-field concept is an alternative way of explaining how charged objects attract or repel each other. Constructions called field lines are an aid of visualizing the electric fields around various charge configurations. The charge that a conductor acquires is proportional to the potential difference across the conductor. The ratio of charge to potential difference is called capacitance. Devices that make use of this property are known as capacitors; they are used to store electric charge for a variety of applications. Electric Charges The fundamental positive charge is the proton, which is found in the nucleus of the atom along with uncharged neutrons. The fundamental negative charge is the electron, which is located outside the nucleus. 1 6 The properties of these three particles are compare in the table below: Particle Relative Charge Charge ( C ) Mass (kg) Proton +1 +1.6 x 10 –19 1.66 x 10 –27 Electron -1 -1.6 x 10 –19 9.11 x 10 –31 Neutron 0 0.00 1.67 x 10 –27 As we can see, proton and electron have equal, but opposite charges. Therefore, a neutral object has the same number of protons and electrons. The proton is nearly 2,000 times more massive than the electron and is tightly bound in the nucleus (along with the neutrons). If an object gains electrons, the excess of electrons gives the object a negative charge. If an object loses electrons. The deficiency of electrons give the object a positive charge. Key Ideas Conductors materials permit charged particles such as electrons or ions to move freely through them. Conductors cannot hold a charge if they are in contact with other materials since the charged particles move easily through them. Insulators materials that do not readily permit the free movement of charges. When an insulator is given an electric charge, the charge remains confined to the area where the charge was placed. Electric field exist in the region of space if an electric force is exerted on a charged particle. Potential difference is a scalar quantity, as is work. The unit of potential difference is the joule per coulomb (J/C) called the volt (V) in honor of Alessandro Volta, an Italian scientist. 2 7 Electric Current The word current means “flow” and electric current means “flow of charge.” when we use the term electric current, we are referring, not to the speed of the charged particles, but to the quantity of charge that passes a single point in time. The SI unit of current is ampere (A) which is equivalent to 1 coulomb per second. The symbol used to represent current is I, and we can write: Where: I = electric current ∆q = electric charge t = time Sample Problem What is the electric current in a conductor if 240 coulombs of charge pass through it in 1.0 minute? Given: ∆q = 240 C t = 60 s I =? Solution: 3 8 Resistance An effect of electrical resistance is the almost total conversion of the lost energy into heat. Resistance is the reason that wires become hot as they conduct a current. We measure the resistance of a material by placing a potential difference across it and then meas- uring the amount of current that passes through the material. Resistance is defined as the ratio of potential difference in current: Where: I = electric current V = voltage R = resistance The SI unit of resistance is the volt per ampere (V/A), which is called the ohm (Ω) in honor of German physicist Georg Ohm. Sample Problem When a conductor has a potential difference of 110 volts place across it, the current through it is 0.50 ampere. What is the resistance of the conductor? Given: V = 110 volts I = 0.50 A R=? Solution: The resistance of a material defends on (1) the nature of the material, (2) the geometry of the conductor, and (3) the temperature at which the resistance is measured. 4 9 Electric Circuits The word circuit means “closed path.” By an electric circuit we mean an arrangement where electric charges can flow in a closed path. The simplest electric circuit consist of a source of potential difference. (a battery or a power source), a single resistance and con- necting wires. A device that provides resistance to a circuit is called a resistor. Most circuits contains more than one load. The load in a circuit can be connected in two different ways—in series or in parallel. Series Circuits A series circuit has only one current path and if that path is interrupted, the entire circuit ceases to operate. Mathematically, the relationship for any series connection may be express as: IT = I1 = I2 = I3 = . . . V T = V 1 + V2 + V 3 + . . . RT = R1 + R2 + R3 + . . . The first equation deals with the current through each of the component is the same. The second one pertains to the total voltages across all the components is the sum of the voltages across each of them. The last equation shows the total resistance is the sum of individual resistance. 5 10 Parallel Circuits A parallel circuit is a circuit in which different loads are located on separate branches. Because there are separate branches, the charges travel through more than one path. When resistors or other components are in parallel: • The voltage across each of the component is the same. V T = V1 + V 2 + V 3 + . . . • The total current in the main circuit is the sum of the currents in the branches. IT = I1 = I2 = I3 = . . . • The total resistance in the sum of the reciprocal of the individual resistance. Ohm’s Law The relationship among voltage, current, and resistance is given by George Simon Ohm. He determined the relationship between current (I), voltage (V), and resistance (R) could be expressed with the equation known as Ohm’s law. Where: I = electric current V = voltage R = resistance Sample Problem How much current flows across a 3 V battery connected to resistor of 30Ω? Given: Solution: V = 3V R = 30Ω I=? 6 11 LESSON OBJECTIVES • Compare and contrast the voltage and the electromagnetic force of an electric power source. • Describe what happens to the terminal voltage, current, and power delivered to a load as internal resistance of the voltage source increases (due to aging of batteries, for example). • Explain why it is beneficial to use more than one voltage source connected in parallel. When you forget to turn off your car lights, they slowly dim as the battery runs down. Why don't they simply blink off when the battery's energy is gone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted. Furthermore, if you connect an excessive number of 12-V lights in parallel to a car battery, they will be dim even when the battery is fresh and even if the wires to the lights have very low resistance. This implies that the battery's output voltage is reduced by the overload. The reason for the decrease in output voltage for depleted or overloaded batteries is that all volt- age sources have two fundamental parts—a source of electrical energy and an internal resistance. Let us examine both. You can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences. 7 12 A few voltage sources are shown in Figure 1. All such devices create a potential difference and can supply current if connected to a resistance. On the small scale, the potential difference creates an electric eld that exerts force on charges, causing current. We thus use the name electromotive force, abbreviated emf. Emf is not a force at all; it is a special type of potential difference. To be precise, the electromotive force (emf) is the potential difference of a source when no current is owing. Units of emf are volts. Figure 1: A variety of voltage sources (clockwise from top left): the Brazos Wind Farm in Fluvanna, Texas (credit: Lea et, Wikimedia Commons); the Krasnoyarsk Dam in Russia (credit: Alex Polezhaev); a solar farm (credit: U.S. Department of Energy); and a group of nickel metal hydride batteries (credit: Tiaa Monto). The voltage output of each depends on its construction and load, and equals emf only if there is no load. Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. However, emf differs from the voltage output of the device when current ows. The voltage across the terminals of a battery, for example, is less than the emf when the battery supplies current, and it declines further as the battery is depleted or loaded down. However, if the device's output voltage can be measured without drawing current, then output voltage will equal emf (even for a very de- pleted battery). 8 13 As noted before, a 12-V truck battery is physically larger, contains more charge and energy, and can deliver a larger current than a 12-V motorcycle battery. Both are lead-acid batteries with identical emf, but, because of its size, the truck battery has a smaller internal resistance r. Internal resistance is the inherent resistance to the ow of current within the source itself. Figure 2 is a schematic representation of the two fundamental parts of any voltage source. The emf (represented by a script E in the figure) and internal resistance r are in series. The smaller the internal resistance for a given emf, the more current and the more power the source can supply. Figure 2: Any voltage source (in this case, a carbon-zinc dry cell) has an emf related to its source of potential difference, and an internal resistance r related to its construction. (Note that the script E stands for emf.). Also shown are the output terminals across which the terminal voltage V is measured. Since V = emf − Ir, terminal voltage equals emf only if there is no current flowing. The internal resistance r can behave in complex ways. As noted, r increases as a battery is depleted. But internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. 9 14 Various types of batteries are available, with emfs determined by the combination of chemicals involved. We can view this as a molecular reaction (what much of chemistry is about) that separates charge. The lead-acid battery used in cars and other vehicles is one of the most common types. A single cell (one of six) of this battery is seen in Figure 3. The cathode (positive) terminal of the cell is connected to a lead oxide plate, while the anode (negative) terminal is connected to a lead plate. Both plates are immersed in sulfuric acid, the electrolyte for the system. Figure 3: Artist's conception of a lead-acid cell. Chemical reactions in a lead-acid cell separate charge, sending negative charge to the anode, which is connected to the lead plates. The lead oxide plates are connected to the positive or cathode terminal of the cell. Sulfuric acid conducts the charge as well as participating in the chemical reaction. The details of the chemical reaction are left to the reader to pursue in a chemistry text, but their results at the molecular level help explain the potential created by the battery. Figure 4 shows the result of a single chemical reaction. Two electrons are placed on the anode, making it negative, provided that the cathode supplied two electrons. This leaves the cathode positively charged, because it has lost two electrons. In short, a separa- tion of charge has been driven by a chemical reaction. Note that the reaction will not take place unless there is a complete circuit to allow two electrons to be supplied to the cathode. Under many circumstances, these electrons come from the anode, ow through a resistance, and return to the cathode. Note also that since the chemical reactions involve substances with resistance, it is not possible to create the emf without an internal resistance. 10 15 Figure 4: Artist's conception of two electrons being forced onto the anode of a cell and two electrons being removed from the cathode of the cell. The chemical reaction in a lead-acid battery places two electrons on the anode and removes two from the cathode. It requires a closed circuit to proceed, since the two electrons must be supplied to the cathode. Why are the chemicals able to produce a unique potential difference? Quantum mechanical descriptions of molecules, which take into account the types of atoms and numbers of electrons in them, are able to predict the energy states they can have and the ener- gies of reactions between them. In the case of a lead-acid battery, an energy of 2 eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy divided by charge: V = PE/q . An electron volt is the energy given to a single electron by a voltage of 1 V. So the voltage here is 2 V, since 2 eV is given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a different energy and, hence, a different voltage. 11 16 The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage V . Terminal voltage is given by: V = emf − Ir where r is the internal resistance and I is the current flowing at the time of the measurement. I is positive if current flows away from the positive terminal, as shown in Figure 2. You can see that the larger the current, the smaller the terminal voltage. And it is likewise true that the larger the internal resistance, the smaller the terminal voltage. Suppose a load resistance Rload is connected to a voltage source, as in Figure 5. Since the resistances are in series, the total resistance in the circuit is Rload + r. Thus the current is given by Ohm's law to be: Figure 5: Schematic of a voltage source and its load Rload. Since the internal resistance r is in series with the load, it can significantly affect the terminal voltage and current delivered to the load. (Note that the script E stands for emf.) We see from this expression that the smaller the internal resistance r, the greater the current the voltage source supplies to its load Rload. As batteries are depleted, r increases. If r becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates. 12 17 Example 1: Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage and Load A certain battery has a 12.0-V emf and an internal resistance of 0.100 Ω. (a) Calculate its terminal voltage when connected to a 10.0-Ω load. (b) What is the terminal voltage when connected to a 0.500-Ω load? (c) What power does the 0.500-Ω load dissipate? (d) If the internal resistance grows to 0.500 Ω, find the current, terminal voltage, and power dissipated by a 0.500-Ω load. Strategy: The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation V = emf − Ir. Once current is found, the power dissipated by a resistor can also be found. Solution for (a): Entering the given values for the emf, load resistance, and internal resistance into the expression above yields I = 1.188 A Enter the known values into the equation V = emf − Ir to get the terminal voltage: V = emf − Ir = 12.0 V − (1.188 A) (0.100 Ω) = 11.9 V. Discussion for (a): The terminal voltage here is only slightly lower than the emf, implying that 10.0 Ω is a light load for this particular battery. 13 18 Solution for (b): Similarly, with Rload = 0.500 Ω, the current is: I = 20.0 A Discussion for (b): This terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 Ω is a heavy load for this battery. Solution for (c): The power dissipated by the 0.500 - Ω load can be found using the formula P = I2R. Entering the known values gives: Pload = I2Rload = (20.0 A)2 (0.500 Ω) = 2.00 × 102 W. Discussion for (c): Note that this power can also be obtained using the expressions V 2/R or IV, where V is the terminal voltage (10.0 V in this case). Solution for (d): Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding; I = 12.0 A Now the terminal voltage is; V = emf − Ir = 12.0 V − (12.0 A) (0.500 Ω) = 6.00 V, and the power dissipated by the load is; Pload = I2Rload = (12.0 A)2 (0.500 Ω) = 72.0 W. Discussion for (d): We see that the increased internal resistance has significantly de- creased terminal voltage, current, and power delivered to a load. 14 19 Problem Solving 1. Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell? 2. Carbon-zinc dry cells (sometimes referred to as non-alkaline cells) have an emf of 1.54 V, and they are produced as single cells or in various combinations to form other voltages. (a) How many 1.54-V cells are needed to make the common 9-V battery used in many small electronic devices? (b) What is the actual emf of the approximately 9-V battery? (c) Discuss how internal resistance in the series connection of cells will a ect the terminal voltage of this approximately 9-V battery. 3. What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell's internal resistance is 2.00 Ω? 15 20 To deal with complicated circuits, we use Kirchhoff’s rules, devised by G. R. Kirchhoff (1824–1887) in the mid-nineteenth century. There are two rules, and they are simply con- venient applications of the laws of conservation of charge and energy. Kirchhoff’s first rule or junction rule is based on the conservation of electric charge (we already used it to derive the equation for parallel resistors). It states that at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction. Kirchhoff’s second rule or loop rule is based on the conservation of energy. It states that the sum of the changes in potential around any closed loop of a circuit must be zero. To see why this rule should hold, consider a rough analogy with the potential energy of a roller coaster on its track. When the roller coaster starts from the station, it has a particular potential energy. As it is pulled up the first hill, its gravitational potential energy increases and reaches a maximum at the top. As it descends the other side, its potential energy decreases and reaches a local minimum at the bottom of the hill. As the roller coaster continues on its up and down path, its potential energy goes through more changes. But when it arrives back at the starting point, it has exactly as much potential energy as it had when it started at this point. Another way of saying this is that there was as much uphill as there was downhill. 16 21 What is Kirchhoff’s Voltage Law (KVL)? The principle known as Kirchhoff’s Voltage Law (discovered in 1847 by Gustav R. Kirchhoff, a German physicist) can be stated as such: “The algebraic sum of all voltages in a loop must equal zero” By algebraic, I mean accounting for signs (polarities) as well as magnitudes. By loop, I mean any path traced from one point in a circuit around to other points in that circuit, and finally back to the initial point. Demonstrating Kirchhoff’s Voltage Law in a Series Circuit Let’s take another look at our example series circuit, this time numbering the points in the circuit for voltage reference: If we were to connect a voltmeter between points 2 and 1, red test lead to point 2 and the black test lead to point 1, the meter would register +45 volts. Typically the “+” sign is not shown but rather implied, for positive readings in digital meter displays. However, for this lesson, the polarity of the voltage reading is very important and so I will show positive numbers explicitly: E2-1 = +45 V When a voltage is specified with a double subscript (the characters “2-1” in the notation “E2-1”), it means the voltage at the first point (2) as measured in reference to the second point (1). A voltage specified as “Ecd” would mean the voltage as indicated by a digital meter with the red test lead on point point “d”: the voltage at “c” in reference to “d”. “c” and the black test lead on 17 22 If we were to take that same voltmeter and measure the voltage drop across each resistor, stepping around the circuit in a clockwise direction with the red test lead of our meter on the point ahead and the black test lead on the point behind, we would obtain the following readings: E3-2 = -10 V E4-3 = -20 V E1-4 = -45 V 18 23 We should already be familiar with the general principle for series circuits stating that individual voltage drops add up to the total applied voltage, but measuring voltage drops in this manner and paying attention to the polarity (mathematical sign) of the readings reveals another facet of this principle: that the voltages measured as such all add up to zero: In the above example, the loop was formed by the following points in this order: 1-2-3 -4-1. It doesn’t matter which point we start at or which direction we proceed in tracing the loop; the voltage sum will still equal zero. To demonstrate, we can tally up the voltages in loop 3-2-1-4-3 of the same circuit: This may make more sense if we re-draw our example series circuit so that all components are represented in a straight line: 19 24 It’s still the same series circuit, just with the components arranged in a different form. Notice the polarities of the resistor voltage drops with respect to the battery: the battery’s voltage is negative on the left and positive on the right, whereas all the resistor voltage drops are oriented the other way: positive on the left and negative on the right. This is because the resistors are resisting the flow of electric charge being pushed by the battery. In other words, the “push” exerted by the resistors against the flow of electric charge must be in a direction opposite the source of electromotive force. Here we see what a digital voltmeter would indicate across each component in this circuit, black lead on the left and red lead on the right, as laid out in horizontal fashion: 20 25 If we were to take that same voltmeter and read voltage across combinations of components, starting with the only R1 on the left and progressing across the whole string of components, we will see how the voltages add algebraically (to zero): The fact that series voltages add up should be no mystery, but we notice that the polarity of these voltages makes a lot of difference in how the figures add. While reading voltage across R1—R2, and R1—R2—R3(I’m using a “double-dash” symbol “—” to represent the series connection between resistors R1, R2, and R3), we see how the voltages measure successively larger (albeit negative) magnitudes, because the polarities of the individual voltage drops are in the same orientation (positive left, negative right). 21 26 The sum of the voltage drops across R1, R2, and R3 equals 45 volts, which is the same as the battery’s output, except that the battery’s polarity is opposite that of the resistor voltage drops (negative left, positive right), so we end up with 0 volts measured across the whole string of components. That we should end up with exactly 0 volts across the whole string should be no mystery, either. Looking at the circuit, we can see that the far left of the string (left side of R1: point number 2) is directly connected to the far right of the string (right side of battery: point number 2), as necessary to complete the circuit. Since these two points are directly connected, they are electrically common to each other. And, as such, the voltage between those two electrically common points must be zero. Demonstrating Kirchhoff’s Voltage Law in a Parallel Circuit Kirchhoff’s Voltage Law (sometimes denoted as KVL for short) will work for any circuit configuration at all, not just simple series. Note how it works for this parallel circuit: 22 27 Being a parallel circuit, the voltage across every resistor is the same as the supply voltage: 6 volts. Tallying up voltages around loop 2-3-4-5-6-7-2, we get: Note how I label the final (sum) voltage as E2-2. Since we began our loop-stepping se- quence at point 2 and ended at point 2, the algebraic sum of those voltages will be the same as the voltage measured between the same point (E 2-2), which of course must be zero. The Validity of Kirchhoff’s Voltage Law, Regardless of Circuit Topology The fact that this circuit is parallel instead of series has nothing to do with the validity of Kirchhoff’s Voltage Law. For that matter, the circuit could be a “black box”—its component configuration completely hidden from our view, with only a set of exposed terminals for us to measure the voltage between—and KVL would still hold true: Try any order of steps from any terminal in the above diagram, stepping around back to the original terminal, and you’ll find that the algebraic sum of the voltages always equals zero. 23 28 Furthermore, the “loop” we trace for KVL doesn’t even have to be a real current path in the closed-circuit sense of the word. All we have to do to comply with KVL is to begin and end at the same point in the circuit, tallying voltage drops and polarities as we go between the next and the last point. Consider this absurd example, tracing “loop” 2-3-6-3-2 in the same parallel resistor circuit: Lets take a short review before we go to our next lesson. Kirchhoff’s Voltage Law (KVL) is defined as: The preferred version for KVL is the sum of voltages around a closed loop is equal to zero because it is sometimes difficult to know whether a voltage source causes a voltage drop or rise in a circuit. The term “around a closed loop” means that you travel around the circuit and start and stop at the same location. The sum of the voltages across the elements that you travel through (any path you choose) will be equal to zero if the starting and ending points are the same. 24 29 When performing a KVL the following steps are followed: 1. Write the polarity (+ and – signs) across every element in the circuit. a. For voltage sources the polarity is set by the battery symbol (+ sign on the long end line) b. For current sources the head side of the arrow has the + sign. c. For resistors the + rent flow. and – signs are written in the direction of the assumed cur 2. Start at a point in the circuit and travel around any path you choose until you get back to the same starting point. As you are travelling around the path you write down the voltage of the elements (I*R for resistors and Vs for sources) you go across using the sign that you travel through. If you travel into the + sign the voltage term is positive and if you travel into the – sign the voltage term is negative. 3. If you have only one unknown, solve the KVL equation for that unknown. If you have multiple unknowns, you will need to obtain additional KVL and/or KCL circuit equations (or other information) to solve for them. If your calculation yields a negative current that means you assumed the current in the wrong direction. The current magnitude is correct, but the direction needs to be changed. Re-solving the circuit is NOT necessary if the wrong direction is assumed. 25 30 What Is Kirchhoff’s Current Law? Kirchhoff’s Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.” This law is used to describe how a charge enters and leaves a wire junction point or node on a wire. Armed with this information, let’s now take a look at an example of the law in practice, why it’s important, and how it was derived. Parallel Circuit Review Let’s take a closer look at that last parallel example circuit: 26 31 Solving for all values of voltage and current in this circuit: At this point, we know the value of each branch current and of the total current in the circuit. We know that the total current in a parallel circuit must equal the sum of the branch currents, but there’s more going on in this circuit than just that. Taking a look at the currents at each wire junction point (node) in the circuit, we should be able to see something else: Currents Entering and Exiting a Node At each node on the positive “rail” (wire 1-2-3-4) we have current splitting off the main flow to each successive branch resistor. At each node on the negative “rail” (wire 8-7-6-5) we have current merging together to form the main flow from each successive branch resistor. This fact should be fairly obvious if you think of the water pipe circuit analogy with every branch node acting as a “tee” fitting, the water flow splitting or merging with the main piping as it travels from the output of the water pump toward the return reservoir or sump. 27 32 If we were to take a closer look at one particular “tee” node, such as node 6, we see that the current entering the node is equal in magnitude to the current exiting the node: From the top and from the right, we have two currents entering the wire connection labeled as node 6. To the left, we have a single current exiting the node equal in magnitude to the sum of the two currents entering. To refer to the plumbing analogy: so long as there are no leaks in the piping, what flow enters the fitting must also exit the fitting. This holds true for any node (“fitting”), no matter how many flows are entering or exiting. Mathematically, we can express this general relationship as such: Kirchhoff’s Current Law Mr. Kirchhoff decided to express it in a slightly different form (though mathematically equivalent), calling it Kirchhoff’s Current Law (KCL): Summarized in a phrase, Kirchhoff’s Current Law reads as such: “The algebraic sum of all currents entering and exiting a node must equal zero” That is, if we assign a mathematical sign (polarity) to each current, denoting whether they enter (+) or exit (-) a node, we can add them together to arrive at a total of zero, guar- anteed. 28 33 Taking our example node (number 6), we can determine the magnitude of the current exiting from the left by setting up a KCL equation with that current as the unknown value: The negative (-) sign on the value of 5 milliamps tells us that the current is exiting the node, as opposed to the 2 milliamp and 3 milliamp currents, which must both be positive (and therefore entering the node). Whether negative or positive denotes current entering or exiting is entirely arbitrary, so long as they are opposite signs for opposite directions and we stay consistent in our notation, KCL will work. Together, Kirchhoff’s Voltage and Current Laws are a formidable pair of tools useful in analyzing electric circuits. Their usefulness will become all the more apparent in a later chapter (“Network Analysis”), but suffice it to say that these Laws deserve to be memorized by the electronics student every bit as much as Ohm’s Law. 29 34 Sample Problem lifted from Giancoli, 2016 Using Kirchhoff’s rules: Calculate the currents I1, I2, and I3 in the three branches of the circuit. APPROACH and SOLUTION 1. Label the currents and their directions. Figure uses the labels I1, I2, and I3 for the current in the three separate branches. Since (positive) current tends to move away from the positive terminal of a battery, we choose I2 and I3 to have the directions shown in the figure The direction of I1 is not obvious in advance, so we arbitrarily chose the direction indicated. If the current actually flows in the opposite direction, our answer will have a negative sign. 2. Identify the unknowns. We have three unknowns I1, I2, and I3 and therefore we need three equations, which we get by applying Kirchhoff’s junction and loop rules. 3. Junction rule: We apply Kirchhoff’s junction rule to the currents at point a, where I3 enters and I2 and I1 leave: I3 = I2 + I1 (Equation 1) This same equation holds at point d, so we get no new information by writing an equation for point d. 4. Loop rule: We apply Kirchhoff’s loop rule to two different closed loops. First we apply it to the upper loop ahdcba. We start (and end) at point a. From a to h we have a potential decrease Vha = (-I1) (30Ω). From h to d there is no change, but from d to c the potential increases by 45 V: that is, Vcd = + 45V. From c to a the potential decreases through the two resistances by an amount Vac = -(I3) (40Ω + 1Ω) = -(41Ω) I3. Thus we have Vha + Vcd + Vac = 0, or –30I1 + 45 - 41I3 = 0, (Equation 2) where we have omitted the units (volts and amps) so we can more easily see the algebra. 30 35 For our second loop, we take the outer loop ahdefga. (We could have chosen the lower loop abcdefga instead.) Again we start at point a, and going to point h we have Vha = (-I1) (30Ω). Next, Vdh = 0. But when we take our positive test charge from d to e, it actually is going uphill, against the current—or at least against the assumed direction of the current, which is what counts in this calculation. Thus, Ved = +I1 (20Ω) has a positive sign. Similarly, Vfe = +I1 (1Ω) From f to g there is a decrease in potential of 80 V because we go from the high potential terminal of the battery to the low. Thus Vgf = - 80V. Finally, Vag = 0 and the sum of the potential changes around this loop is; –30I1 + (20 + 1)I2 - 80 = 0. (Equation 3) Our major work is done. The rest is algebra. We have three equations and three unknowns. From Equation 3 we have; (Equation 4) From Equation 2 we have; (Equation 5) We substitute equation 4 and equation 5 int o equation 1; We solve for I1 , collecting term: 3.1 I1 = - 2.7 I1 = - 0.87 A The negative sign indicates that the direction of I1 is actually opposite to that initially assumed. The answer automatically comes out in amperes because our voltages and resistances were in volts and ohms. From Equation 4 we have; I2 = 3.8 + 1.4I1 = 3.8 + 1.4(–0.87) = 2.6 A, and from equation 5; I3 = 1.1 - 0.73I1 = 1.1 - 0.73(–0.87) = 1.7 A. This completes the solution. 31 36 1. Write the KVL equation for the circuit below and solve for the current in the loop and the voltage across R4. Note: The Multisim current and voltage combo probe is only added to this circuit to show the calculations agree with Multisim. You cannot use Multisim probe values in your calculations unless it is specifically stated to do so in the problem statement. 2. Calculate the current in the circuit and show that the sum of all the voltage changes around the circuit is zero. Giancoli, Douglas, Physics with application, 2016 www.allaboutcircuits.com Shareok.com/DC circuits 32 37