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| learn:courses:real-analog:start [2017/02/27 18:49] – [2. Introduction and Chapter Objectives] Martha | learn:courses:real-analog:start [2021/10/13 22:27] (current) – Arthur Brown | ||
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| - | * {{ : | + | ====== Real Analog: Circuits |
| - | * {{ :learn: | + | "Real Analog: Circuits |
| - | * {{ : | + | * A 12 chapter |
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| - | * {{ :learn: | + | * Homework assignments for every chapter |
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| - | | + | **Note:** The textbook can be downloaded as a PDF from the link at the bottom of the page, accessed online via the individual chapters below, or a [[https:// |
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| - | ====== 1. Introduction and Chapter Objectives ====== | ||
| - | In this chapter, we introduce all fundamental concepts associated with circuit analysis. Electrical circuits are constructed in order to direct the flow of electrons to perform a specific task. In other words, in circuit analysis and design, we are concerned with transferring electrical energy in order to accomplish a desired objective. For example, we may wish to use electrical energy to pump water into a reservoir; we can adjust the amount of electrical energy applied to the pump to vary the rate at which water is added to the reservoir. The electrical circuit, then, might be designed to provide the necessary electrical energy to the pump to create the desired water flow rate. | ||
| - | This chapter begins with introduction to the basic parameters which describe the energy in an electrical circuit: //charge//, // | ||
| - | Electrical circuits are composed of interconnected // | + | Design projects use Digilent' |
| - | Finally, we introduce | + | Real Analog: Circuits 1, the Analog Discovery 2, and Analog Parts Kit form the core of a world-class engineering educational program that can be used by themselves or in support of existing curricular materials. Students with their own design kits learn more, learn faster, retain information longer, and have a more enjoyable experience - now every student can take charge of their education for less than the cost of a traditional engineering textbook! |
| - | Please pay special attention to the //passive sign convention// | + | --> |
| + | In this chapter, we introduce all fundamental concepts associated with circuit analysis. Electrical circuits | ||
| + | * [[/ | ||
| - | In summary, this chapter introduces virtually all the basic concepts which will be used throughout this textbook. After this chapter, little information specific to electrical circuit analysis remains to be learned – the remainder of the textbook is devoted to developing analysis methods used to increase the efficiency of our circuit analysis and | + | <-- |
| - | introducing additional circuit components such as capacitors, inductors, and operational amplifiers. The student should be aware, however, that all of our circuit analysis is based on energy transfer among circuit components; this energy transfer is governed by Kirchhoff’s Current Law and Kirchhoff’s Voltage Law and the circuit components are modeled by their voltage-current relationships. | + | |
| - | ---- | + | -->Chapter 2: Circuit Reduction# |
| + | In this chapter, we introduce analysis methods based on circuit reduction. Circuit reduction consists of combining resistances in a circuit to a smaller number of resistors, which are (in some sense) equivalent to the original resistive network. Reducing the number of resistors, of course, reduces the number of unknowns in a circuit. | ||
| + | * [[/ | ||
| - | ==== After Completing this Chapter, You Should be Able to: ==== | + | <-- |
| - | * Define voltage and current in terms of electrical charge | + | |
| - | * State common prefixes and the symbols used in scientific notation | + | |
| - | * State the passive sign convention from memory | + | |
| - | * Determine the power absorbed or generated by an circuit element, based on the current and voltage provided to it | + | |
| - | * Write symbols for independent voltage and current sources | + | |
| - | * State from memory the function of independent voltage and current sources | + | |
| - | * Write symbols for dependent voltage and current sources | + | |
| - | * State governing equations for the four types of dependent sources | + | |
| - | * State Ohm’s Law from memory | + | |
| - | * Use Ohm’s Law to perform voltage and current calculations for resistive circuit elements | + | |
| - | * Identify nodes in an electrical circuit | + | |
| - | * Identify loops in an electrical circuit | + | |
| - | * State Kirchhoff’s current law from memory, both in words and as a mathematical expression | + | |
| - | * State Kirchhoff’s voltage law from memory, both in words and as a mathematical expression | + | |
| - | * Apply Kirchhoff’s voltage and current laws to electrical circuits | + | |
| - | ---- | + | -->Chapter 3: Nodal and Mesh Analysis# |
| + | In cases where circuit reduction is not feasible, approaches are still available to reduce the total number of unknowns in the system. Nodal analysis and mesh analysis are two of these. | ||
| + | * [[/ | ||
| - | ===== 1.1 Basic Circuit Parameters and Sign Conventions ===== | + | <-- |
| - | This section introduces the basic engineering parameters for electric circuits: // | + | |
| - | This section also introduces the //passive sign convention// | ||
| - | ==== Electrical Charge ==== | + | --> |
| - | Electron flow is fundamental to operation of electric circuits; | + | In this chapter, we introduce |
| + | * [[/learn/courses/real-analog-chapter-4/start|Real Analog - Chapter Four]] | ||
| - | ==== Voltage ==== | + | <-- |
| - | //Voltage// is energy per unit charge. Energy is specified relative to some reference level; thus, voltages are more accurately specified as voltage // | + | |
| - | voltage are volts, abbreviated V. The voltage difference between two points indicates the energy necessary to move a unit charge from one of the points to the other. Voltage differences can be either positive or negative. | + | |
| - | Mathematically, | ||
| - | $$ v=\frac{dw}{dq}\ | ||
| - | where //v// is the voltage difference | + | --> |
| + | Operational amplifiers | ||
| + | * [[/learn/courses/real-analog-chapter-5/start|Real Analog - Chapter Five]] | ||
| - | ==== Current ==== | + | <-- |
| - | Current is the rate at which charge is passing a given point. Current is specified at a particular point in the circuit, and is not relative to a reference. Since current is caused by charge in motion, it can be thought of as indicating //kinetic// energy. | + | |
| - | Mathematically, | ||
| - | $$ i=\frac{dq}{dw}\ | + | --> |
| + | This chapter begins with an overview of the basic concepts associated with energy storage. This discussion focuses not on electrical systems, but instead introduces the topic qualitatively in the context of systems with which the reader is already familiar. The goal is to provide a basis for the mathematics, | ||
| + | * [[/ | ||
| - | Where //i// is the current in amperes, //q// is the charge in coulombs, and //t// is the time in seconds. Thus, current is the time rate of change of charge and units of charge are coulombs per second, or //amperes// (abbreviated as // | + | <-- |
| - | ==== Power ==== | ||
| - | An electrical system is often used to drive a non-electrical system (in an electric stove burner, for example, electric energy is converted to heat). Interactions between electrical and non-electrical systems are generally described in terms of //power//. Electrical power associated with a particular circuit element is the product of the current passing through the element and the voltage difference across the element. This is often written as: | ||
| - | $$ p(t)=v(t) \cdot i(t) (Eq. 1.3) $$ | + | --> |
| + | First order systems are, by definition, systems whose input-output relationship is a first order differential equation. A first order differential equation contains a first order derivative but no derivative higher than first order - the order of a differential equation is the order of the highest order derivative present in the equation. | ||
| + | * [[/ | ||
| - | Where //p(t)// is the // | + | <-- |
| - | ==== International System of Units and Prefixes ==== | ||
| - | We will use the international system of units (SI). The scales of parameters that are of interest to engineers can vary over many orders of magnitudes. For example, voltages experienced during lightning strikes can be on the order of 10< | ||
| - | ^ Multiple | + | -->Chapter 8: Second Order Circuits# |
| - | | 10<sup>9</ | + | Second order systems are, by definition, systems whose input-output relationship is a second order differential equation. A second order differential equation contains a second order derivative but no derivative higher than second order. |
| - | | 10< | + | * [[/learn/courses/real-analog-chapter-8/start|Real Analog |
| - | | 10< | + | |
| - | | 10< | + | |
| - | | 10< | + | |
| - | | 10< | + | |
| - | | 10< | + | |
| - | //Table 1.1. SI prefixes.// | + | <-- |
| - | ==== Passive Sign Convention ==== | ||
| - | A general two-terminal electrical circuit element is shown in Fig. 1.1. In general, there will be some current, //i//, flowing through the element and some voltage difference, //v//, across its terminals. Note that we are currently representing both voltage and current as constants, but none of the assertions made in this chapter change if they are functions of time. | ||
| - | |||
| - | {{ : | ||
| - | The __assumed__ direction of the current, //i//, passing through | + | --> |
| + | In this chapter, we will provide a very brief introduction to the topic of state variable modeling. The brief presentation provided here is intended simply | ||
| + | * [[/learn/courses/real-analog-chapter-9/start|Real Analog - Chapter Nine]] | ||
| + | <-- | ||
| - | ---- | ||
| - | === Example 1.1: === | + | --> |
| - | Three amperes (3 A) of current is passing through a circuit element connecting nodes a and b. The current is flowing from node a to node b. The physical situation can be represented schematically by any of the figures shown below - all four figures represent the same current flow and direction. | + | In this chapter we will study dynamic systems which are subjected |
| + | * [[/ | ||
| - | {{ : | + | <-- |
| - | ---- | ||
| - | The __assumed__ polarity | + | --> |
| + | In this chapter we discuss representation | ||
| + | * [[/learn/courses/real-analog-chapter-11/start|Real Analog - Chapter Eleven]] | ||
| + | <-- | ||
| - | ---- | ||
| - | === Example 1.2: === | + | -->Chapter 12: Steady-state Sinusoidal Power# |
| - | A 5 volt (5 V) voltage potential difference is applied across a circuit element connecting nodes a and b. The voltage at node a is positive relative to the voltage at node b. The physical situation can be represented schematically by either of the figures shown below - both figures represent the same voltage potential difference. | + | In this chapter we will address |
| - | + | * [[/ | |
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | The assumed voltage polarity and current direction are not individually significant - the assumed direction of voltage polarity __relative__ to current direction is important. To satisfy our sign convention, we will assume that positive current enters the node at which the positive voltage polarity is defined. This sign convention is called the //passive sign convention// | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | === Example 1.3: === | + | |
| - | The passive sign convention is satisfied for either of the two voltage-current definitions shown below - the current is assumed to enter the positive voltage node. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The passive sign convention is __not__ satisfied for either of the two voltage current definitions shown below - the current is assumed to enter the negative voltage node. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | **Note**: | + | |
| - | Many students attempt to choose current directions and voltage polarities so that their calculations result in positive values for voltages and currents. In general, this is a wast of time - it is best to arbitrarily assume __either__ a voltage polarity of current direction for each circuit element. | + | |
| - | + | ||
| - | Choice of a positive direction for current dictates the choice of positive voltage polarity, per Fig. 1.1. Choice of a positive voltage polarity dictates the choice of positive current direction, per Fig. 1.1. | + | |
| - | + | ||
| - | Analysis of the circuit is performed using the above assumed signs for voltage and current. The sign of the results indicates whether the assumed choice of voltage polarity and current direction was correct. A positive magnitude of a calculated voltage indicates that the assumed sign convention is correct; a negative magnitude indicates that the actual voltage polarity is opposite to the assumed polarity. Likewise, a positive magnitude of a calculated current indicates that the assumed current direction is correct; a negative magnitude indicates that the current direction is opposite to that assumed. | + | |
| - | + | ||
| - | + | ||
| - | ==== Voltage Subscript and Sign Conventions ==== | + | |
| - | The assumed sign convention for voltage potentials is sometimes expressed by using subscripts. The first subscript denotes the node at which the __positive__ voltage polarity is assumed and the second subscript is the __negative__ voltage polarity. For example, //v<sub>ab</ | + | |
| - | + | ||
| - | ==== Reference Voltages and Ground ==== | + | |
| - | For convenience, | + | |
| - | + | ||
| - | {{ :learn: | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | === Example 1.4: === | + | |
| - | The two figures below show identical ways of specifying the voltage across a circuit element. In the circuit to the left, the voltage //v// is the voltage potential between nodes a and b, with the voltage at node a being assumed positive relative to voltage at node b. This can be equivalently specified as // | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Power and Sign Conventions ==== | + | |
| - | The sign of the voltage across an element relative to the sign of the current through the element governs the sign of the power. Equation (1.3) above defines power as the product of the voltage times current: | + | |
| - | + | ||
| - | $$ P=vi $$ | + | |
| - | + | ||
| - | The power is // | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | === Example 1.5: === | + | |
| - | In Fig. (a) below, the element agrees with the passive sign convention since a positive current is entering the positive voltage node. Thus, the element of Fig. (a) is absorbing energy. In Fig. (b), the element is absorbing power - positive current is leaving the negative voltage node, which implies that positive current enters the positive voltage node. The element of Fig. (c) generates power; negative current enters the positive voltage node, which disagrees with the passive sign convention. Fig. (d) also illustrates an element which is generating power, since positive current is entering a negative voltage node. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * In this text, we will be primarily concerned with the movement | + | |
| - | * Voltage is an energy difference between two physically separated points. The polarity of a voltage is used to indicate which point is to be assumed to be at the higher energy level. The positive terminal (+) is assumed to be at a higher voltage than the negative terminal (-). A negative voltage value simply indicates that the actual voltage polarity is opposite to the assumed polarity. | + | |
| - | * The sign of the current indicates the assumed direction of charge motion past a point. A change in the sign of the current value indicates that the current direction is opposite to the assumed direction. | + | |
| - | * The assumed polarity of the voltage across a passive circuit element must be consistent with the __assumed__ current direction through the element. The assumed positive direction for current must be such that positive current enters the positive voltage terminal of the element. Since this sign convention is applied only to passive elements, it is known as the //passive sign convention// | + | |
| - | * The assumed current direction __or__ the assumed voltage polarity can be chosen arbitrarily, | + | |
| - | * The power absorbed or generated by an electrical circuit component is the product of the voltage difference across the element and the current through the element: $ p=iv $. The relative sign of the voltage and current are set according to the passive sign convention. Positive power implies that the voltage and current are consistent with the passive sign convention | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | === Exercises === | + | |
| - | 1. Assign reference voltage and current directions to the circuit elements represented by the shaded boxes in the circuits below. | + | |
| - | {{ : | + | |
| - | 2. Either the reference voltage polarity or the reference current direction is provided for the circuit elements below. Provide the appropriate sign convention for the missing parameters. | + | |
| - | {{ : | + | |
| - | 3. Determine the magnitude and direction of the current in the circuit element below if the element absorbs 10W. | + | |
| - | {{ : | + | |
| - | 4. Determine the power absorbed or supplied by the circuit element below. State whether the power is absorbed or supplied. | + | |
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 1.2 Power Sources ===== | + | |
| - | Circuit elements are commonly categorized as either //passive// or //active//. A circuit element is passive if the __total__ amount of energy it delivers to the rest of the circuit (over all time) is non-positive (passive elements can __temporarily__ deliver energy to a circuit, but only if the energy was previously stored in the passive element by the circuit). An active circuit element has the ability to create and provide power to a circuit from mechanisms __external__ to the circuit. Examples of active circuit elements are batteries (which create electrical energy from chemical processes) and generators (which create electrical energy from mechanical processes, such as spinning a turbine). | + | |
| - | + | ||
| - | In this section we consider some very important active circuit elements: voltage and current sources. We will discuss two basic types of sources: // | + | |
| - | + | ||
| - | ==== Independent Voltage Sources ==== | + | |
| - | An independent voltage source maintains a specified voltage across its terminals. The symbol used to indicate a voltage source delivering a voltage // | + | |
| - | + | ||
| - | Note that the sign of the voltage being applied by the source is provided on the source symbol - there is no need to assume a voltage polarity for voltage sources. The current direction, however, is unknown and must be determined (if necessary) from an analysis of the overall circuit. | + | |
| - | + | ||
| - | //Ideal// voltage sources provide a specified voltage regardless of the current flowing through the device. Ideal sources can, obviously, provide infinite power; all real sources will provide only limited power to the circuit. We will discuss approaches for modeling non-ideal sources in later chapters. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Independent Current Sources ==== | + | |
| - | An independent current source maintains a specified current. This current | + | |
| - | + | ||
| - | Note that the sign of the current being applied by the source is provided on the source symbol - there is no need to assume a current direction. The voltage polarity, however, is unknown and must be determined (if necessary) from an analysis of the overall circuit. | + | |
| - | + | ||
| - | //Ideal// current sources provide a specified current regardless of the voltage difference across the device. Ideal current sources can, like ideal voltage sources, provide infinite power; all real sources will provide only limited power to the circuit. We will discuss approaches for modeling non-ideal current sources in later chapters. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Dependent Sources ==== | + | |
| - | Dependent sources can be either voltage or current sources; Fig. 1.5(a) shows the symbol for a dependent voltage source and Fig. 1.5(b) shows the symbol for a dependent current source. Since each type of source can be controlled by either a voltage or current, there are four types of dependent current sources: | + | |
| - | * Voltage-controlled voltage source (VCVS) | + | |
| - | * Current-controlled voltage source (CCVS) | + | |
| - | * Voltage-controlled current source (VCCS) | + | |
| - | * Current-controlled current source (CCCS) | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Figure 1.6 illustrates the voltage-controlled dependent sources, and Fig. 1.7 illustrates the current-controlled dependent sources. In all cases, some electrical circuit exists which has some voltage and current combination at its terminals. Either the voltage or current at these terminals is used to set the voltage or current of the dependent source. The parameters µ and β in Figs. 1.6 and 1.7 are dimensionless constants. //µ// is the //voltage gain// of a VCVS and //β// is the //current gain// of a CCCS. The parameter //r// is the voltage-to-current ratio of a CCVS and has units of volts/ | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * Circuit elements can be either active or passive. Active elements provide electrical energy from a circuit from sources outside the circuit; active elements can be considered to create energy (from the standpoint of the circuit, anyway). Passive elements will be discussed in section 1.3, when we introduce resistors. Active circuit elements introduced in this section are ideal independent and dependent voltage and current sources. | + | |
| - | * Ideal independent sources presented in this section are voltage and current sources. Independent voltage sources deliver the specified voltage, regardless of the current demanded of them. Independent current sources provide the specified current, regardless of the voltage levels required to provide this current. Devices such as batteries are often modeled as independent sources. | + | |
| - | * Dependent sources provide a voltage or current which is controlled by a voltage or current elsewhere in the circuit. Devices such as operational amplifiers and transistors are often modeled as dependent sources. We will revisit the subject of dependent sources in chapter 5 of this text, when we discuss operational amplifier circuits. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. The ideal voltage source shown in the circuit below delivers 12V to the circuit element shown. What is the current //I// through the circuit element? | + | |
| - | {{ : | + | |
| - | 2. The ideal current source shown in the circuit below delivers 2A to the circuit element shown. What is the voltage difference //V// across the circuit element? | + | |
| - | {{ : | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 1.3 Resistors and Ohm's Law ===== | + | |
| - | Resistance is a property of all materials - this property characterizes the loss of energy associated with passing an electrical current through some conductive element. Resistors are circuit elements whose characteristics are dominated by this energy loss. Since energy is always lost when current is passed through an electrical circuit element, all electrical elements exhibit resistive properties which are characteristic of resistors. Resistors are probably the simplest and most commonly used circuit elements. | + | |
| - | + | ||
| - | All materials impede the flow of current through them to come extent. Essentially, | + | |
| - | + | ||
| - | The relationship between voltage and current for a resistor is given by // | + | |
| - | + | ||
| - | $$ v(t)=Ri(t) | + | |
| - | + | ||
| - | Where voltage and current are explicitly denoted as functions of time. Note that in Fig. 1.8, the current is flowing from a higher voltage potential to a lower potential, as indicated by the polarity (+ and -) of the voltage and the arrow indicating direction of the current flow. The relative polarity between voltage and current for a resistor __must__ be as shown in Fig. 1.8; the current enters the node at which the voltage potential is highest. Values of resistance, //R// are __always__ positive, and resistors __always__ absorb power. | + | |
| - | + | ||
| - | **Note:** The voltage-current relationship for resistors always agrees with the passive sign convention. Resistors always absorb power. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Figure 1.9 shows a graph of //v// vs. //i// according to equation (1.5); the resulting plot is a straight line with slope //R//. Equation (1.5) thus describes the voltage-current relationship for a //linear// resistor. Linear resistors do not exist in reality - all resistors are // | + | |
| - | + | ||
| - | **Note:** | + | |
| - | For the most part, we will consider only linear resistors in this text. These resistors obey the linear voltage-current relationship shown in equation (1.5). All real resistors are nonlinear to some extent, but can often be assumed to operate as linear resistors over some reange of voltages and currents. | + | |
| - | + | ||
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Conductance ==== | + | |
| - | // | + | |
| - | + | ||
| - | $$ G=\frac{1}{R} | + | |
| - | + | ||
| - | The unit for conductance is siemens, abbreviated //S//. Ohm's law, written in terms of conductance, | + | |
| - | + | ||
| - | $$ i(t)=Gv(t) | + | |
| - | + | ||
| - | Some circuit analyses can be performed more easily and interpreted more readily if the elements' | + | |
| - | + | ||
| - | + | ||
| - | **Note:** | + | |
| - | In section 1.2, we characterized a current-controlled voltage source in terms of a parameter with units of ohms, since it had units of volts/amp. We characterized a voltage-controlled current source in terms of a parameter with units of siemens, since it had units of amps/volts. | + | |
| - | + | ||
| - | + | ||
| - | ==== Power Dissipation ==== | + | |
| - | Instantaneous power was defined by equation (1.3) in section 1.1 as: | + | |
| - | + | ||
| - | $$ P(t)=v(t) \cdot i(t)$$ | + | |
| - | + | ||
| - | For the special case of a resistor, we can re-write this (by substituting equation (1.5) into the above) as: | + | |
| - | + | ||
| - | $$ P(t)=Ri^2(t)= \frac{v^2(t)}{R} | + | |
| - | + | ||
| - | Likewise, we can write the power dissipation in terms of the conductance of a resistor as: | + | |
| - | + | ||
| - | $$ P(t)= \frac{i^2(t)}{G}=Gv^2(t) | + | |
| - | + | ||
| - | + | ||
| - | **Note:** | + | |
| - | We can write the power dissipation from a resistor in terms of the resistance or conductance of the resistor and __either__ the current through the resistor __or__ the voltage drop across the resistor. | + | |
| - | + | ||
| - | + | ||
| - | ==== Practical Resistors ==== | + | |
| - | All materials have some resistance, so all electrical components have non-zero resistance. However, circuit design often relies on implementing a specific, desired resistance at certain locations in a circuit; resistors are often placed in the circuit at these points to provide the necessary resistance. Resistors can be purchased in certain standard values. Resistors are manufactured in a variety of ways, though most commonly available commercial resistors are carbon composition or wire-wound. Resistors can have either a fixed or variable resistance. | + | |
| - | + | ||
| - | //Fixed// resistors provide a single specified resistance value and have two terminals, as shown in Fig. 1.5 above. // | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Resistors, which are physically large enough, will generally have their resistance value printed directly on them. Smaller resistors generally will use a color code to identify their resistance value. The color coding scheme is provided in Fig. 1.12. The resistance values indicated on the resistor will provide a //nominal// resistance value for the component; the actual resistance value for the component will vary from this by some amount. The expected tolerance between the allowable actual resistance values and the nominal resistance is also provided on the resistor, either printed directly on the resistor or provided as an additional color band. The color-coding scheme for resistor tolerances is also provided in Fig. 1.12. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.6 ==== | + | |
| - | A resistor has the following color bands below. Determine the resistance value and tolerance: | + | |
| - | + | ||
| - | * First band (A): Red | + | |
| - | * Second band (B): Black | + | |
| - | * Exponent: Orange | + | |
| - | * Tolerance: Gold | + | |
| - | + | ||
| - | Resistance = (20+0)x10< | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * The relationship between voltage and current for a resistor is Ohm's Law: //v=iR//. Since a resistor only dissipates energy, the voltage and current for a resistor must always agree with the passive sign convention. | + | |
| - | * As noted in section 1.2, circuit elements can be either active or passive. Resistors are passive circuit elements. Passive elements can store or dissipate electrical energy provided to them by the circuit; they can subsequently return energy to the circuit which they have previously stored, but they cannot create energy. Resistors cannot store electrical energy, they can only dissipate energy by converting it to heat. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. The ideal voltage source shown in the circuit below delivers 18V to the resistor shown. What is the current //I// through the resistor? | + | |
| - | {{ : | + | |
| - | + | ||
| - | 2. The ideal current source shown in the circuit below delivers 4mA to the resistor shown. What is the voltage difference //V// across the resistor? | + | |
| - | {{ : | + | |
| - | + | ||
| - | 3. The ideal voltage source shown in the circuit below delivers 10V to the resistor shown. What is the current //I// in the direction shown? | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Kirchhoff' | + | |
| - | This section provides some basic definitions and background information for two important circuit analysis tools: Kirchhoff' | + | |
| - | + | ||
| - | We will use a // | + | |
| - | + | ||
| - | The lumped parameters approach toward modeling circuits is appropriate if the voltage and currents in the circui change slowly relative to the rate with which information can be transmitted through the circuit. Since information propagates in an electrical circuit at a rate comparable to the speed of light and circuit dimensions are relatively small, this modeling | + | |
| - | + | ||
| - | An alternate approach to circuit analysis is a // | + | |
| - | + | ||
| - | ==== Nodes ==== | + | |
| - | Identification of circuit nodes will be extremely important to the application of Kirchhoff' | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.7 ==== | + | |
| - | The circuit below has four nodes, as shown. A common error for beginning students is to identify points a, b, and c as being separate nodes, since they appear as separate points on the circuit diagram. However, these points are connected by perfect connectors (no circuit elements are between points a, b, and c) and thus the points are at the same voltage and are considered electrically to be at the same point. Likewise, points d, e, f, and g are at the same voltage potential and are considered to be the same node. Node 2 interconnects two circuit elements (a resistor and a source) and must be considered as a separate node. Likewise, node 4 interconnects two circuit elements and qualifies as a node. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Loops ==== | + | |
| - | A //loop// is any closed path through the circuit which encounters no node more than once. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.8 ==== | + | |
| - | There are six possible ways to loop through the circuit of the previous example. These loops are shown below. | + | |
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Kirchhoff' | + | |
| - | Kirchhoff' | + | |
| - | + | ||
| - | Kirchhoff' | + | |
| - | + | ||
| - | //**The algebraic sum of all currents | + | |
| - | + | ||
| - | A common alternate statement for KCL is: | + | |
| - | + | ||
| - | //**The sum of the currents entering any node equals the sum of the currents leaving the node.**// | + | |
| - | + | ||
| - | A general mathematical statement for Kirchhoff' | + | |
| - | + | ||
| - | $$ \sum_{k=l}^{N} i_k(t)=0 | + | |
| - | + | ||
| - | **Note**: Current directions (entering or leaving the node) are based on __assumed__ directions of currents in the circuit. As long as the assumed directions of the currents are consistent from node to node, the final result of the analysis will reflect the __actual__ current directions in the circuit. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.9 ==== | + | |
| - | In the figure below, the assumed directions of // | + | |
| - | {{ : | + | |
| - | + | ||
| - | If we (arbitrarily) choose a sign convention such that the currents entering the node are positive, then currents leaving the node are negative and KCL applied at this node results in: | + | |
| - | + | ||
| - | $$ i_1(t) + i_2(t) - i_3(t)=0 $$ | + | |
| - | + | ||
| - | If, on the other hand, we choose a sign convention that currents entering the node are negative, then currents leaving the node are positive and KCL applied at this node results in: | + | |
| - | + | ||
| - | $$ -i_1(t) - i_2(t) + i_3(t) = 0 $$ | + | |
| - | + | ||
| - | These two equations are the same; the second equation is simply the negative of the first equation. Both of the above equations are equivalent to the statement: | + | |
| - | + | ||
| - | $$ i_1(t) + i_2(t) = i_3(t) $$ | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.10 ==== | + | |
| - | Use KCL to determine the value of the current //i// in the figure below: | + | |
| - | {{ : | + | |
| - | + | ||
| - | Summing the currents entering the node results in: | + | |
| - | + | ||
| - | $$ 4A - (-1A) - 2A - i = 0 \Rightarrow 4A + 1A - 2A = 3A $$ | + | |
| - | + | ||
| - | And //i=3A//, __leaving__ the node. | + | |
| - | + | ||
| - | In the figure below, we have reversed our assumed direction of //i// in the above circuit: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Now, if we sum currents entering the node: | + | |
| - | + | ||
| - | $$ 4A - (-1A) - 2A - i = 0 \Rightarrow i = -4A - 1A + 2A = -3A $$ | + | |
| - | + | ||
| - | So now //i=-3A//, __entering__ the node. The negative sign corresponds to a change in direction, so we can interpret this result to a +3A current __leaving__ the node, which is consistent with out previous result. Thus, the assumed current direction has not affected our results. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | We can generalize Kirchhoff' | + | |
| - | + | ||
| - | //**The algebraic sum of all currents entering (or leaving) any enclosed surface is zero**//. | + | |
| - | + | ||
| - | Applying this statement to the circuit of Fig. 1 results in: | + | |
| - | + | ||
| - | $$ i_1 + i_2 + i_3 = 0 $$ | + | |
| - | + | ||
| - | {{ :learn: | + | |
| - | + | ||
| - | ==== Kirchhoff' | + | |
| - | Kirchhoff' | + | |
| - | + | ||
| - | Kirchhoff' | + | |
| - | + | ||
| - | //**The algebraic sum of all voltage differences around any closed loop is zero.**// | + | |
| - | + | ||
| - | An alternate statement of this law is: | + | |
| - | + | ||
| - | //**The sum of the voltage rises around a closed loop must equal the sum of the voltage drops around the loop.**// | + | |
| - | + | ||
| - | A general mathematical statement for Kirchhoff' | + | |
| - | + | ||
| - | $$ \sum_{k=l}^{N} V_k(t)=0 | + | |
| - | + | ||
| - | Where // | + | |
| - | + | ||
| - | **Note:** | + | |
| - | Voltage polarities are based on __assumed__ polarities of the voltage differences in the loop. As long as the assumed directions of the voltages are consistent from loop to loop, the final result of the analysis will reflect the __actual__ voltage polarities in the circuit. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.11 ==== | + | |
| - | In the figure below, the assumed (or previously known) polarities of the voltages v< | + | |
| - | + | ||
| - | Our sign convention for applying signs to the voltage polarities in our KVL equations will be as follows: when traversing the loop, if the positive terminal of a voltage difference is encountered before the negative terminal, the voltage difference will be interpreted as __positive__ in the KVL equation. If the negative terminal is encountered first, the voltage difference will be interpreted as __positive__ in the KVL equation. We use this sign convention for convenience; | + | |
| - | + | ||
| - | {{ :learn:courses: | + | |
| - | + | ||
| - | Applying KVL to the loop a-b-e-d-a, and using our sign convention as above results in: | + | |
| - | + | ||
| - | $$ v_1 - v_4 - v_6 - v_3 = 0 $$ | + | |
| - | + | ||
| - | The starting point of the loop and the direction that we loop in is abitrary; we could equivalently write the same loop equation as loop d-e-b-a-d, in which case out equation would become: | + | |
| - | + | ||
| - | $$ v_6 + v_4 - v_1 + v_3 = 0 $$ | + | |
| - | + | ||
| - | This equation is identical to the previous equation, the only difference is that the signs of all variables has changed and the variables appear in a different order in the equation. | + | |
| - | + | ||
| - | We now apply KVL to the loop b-c-e-b, which results in: | + | |
| - | + | ||
| - | $$ -v_2 +v_5 +v_4 = 0 $$ | + | |
| - | + | ||
| - | Finally, application of KVL to the loop a-b-c-e-d-a provides: | + | |
| - | + | ||
| - | $$ v_1 -v_2 + v_5 - v_6 - v_3 = 0 $$ | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Application Examples: Solving for Circuit Element Variables ==== | + | |
| - | Typically, when analyzing a circuit, we will need to determine voltages and/or currents in one or more elements in the circuit. In this chapter, we discuss use of the tools presented in previous chapters for circuit analysis. | + | |
| - | + | ||
| - | The complete solution of a circuit consists of determining the voltages and currents for __every__ elements in the circuit. A complete solution of a circuit can be obtained by: | + | |
| - | + | ||
| - | - Writing a voltage-current relationship for each element in the circuit (e.g. write Ohm's Law for the resistors). | + | |
| - | - Applying KCL at all but one of the nodes in the circuit. | + | |
| - | - Applying KVL for all but one of the loops in the circuit. | + | |
| - | + | ||
| - | This approach will typically result in a set of N equations in N unknowns, the unknowns consisting of the voltages and currents for each element in the circuit. Methods exist for defining a reduced set of equations or a complete analysis of a circuit; these approaches will be presented in later chapters. | + | |
| - | + | ||
| - | If KCL is written for __every__ node in the circuit and KVL written for __every__ loop in the circuit, the resulting set of equations will typically be over-determined and the resulting equations will, in general, not be independent. That is, there will be more than N equations in N unknowns and some of the equations will carry redundant information. | + | |
| - | + | ||
| - | Generally, we do not need to determine all the variables in a circuit. This often means that we can write fewer equations than those listed above. The equations to be written will, in these cases, be problem dependent and are often at the discretion of the person doing the analysis. | + | |
| - | + | ||
| - | Examples of using Ohm's Law, KVL, and KCL for circuit analysis are provided below. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.12 ==== | + | |
| - | For the circuit below, determine // | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | We are free to arbitrarily choose wither the voltage polarity or the current direction in each element. Our choices are shown below: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Once the above voltage polarities and current directions are chosen, we must choose all other parameters in a way that satisfies the passive sign convention (current must enter the positive voltage polarity node). Our complete definition of all circuit parameters is shown below: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | We now apply the steps outlined above for an exhaustive circuit analysis. | + | |
| - | + | ||
| - | - Ohm's aw, applied for each resistor, results in: $ v_1=(1 \Omega )i_1 $; $v_3=(3 \Omega )i_3 $; $v_6 = 6 \Omega )i_6$ | + | |
| - | - KCL, applied at node a: $ i_1 + i_3 - i_6 = 0 $ | + | |
| - | - KVL, applied over any two of the three loops in the circuit: $ -12V + v_1 - v_3 = 0 $; $v_3 + v_6 = 0 $ | + | |
| - | + | ||
| - | The above provide six equatoins in six unknowns. Solving these for // | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 1.13 ==== | + | |
| - | Determine // | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | We choose voltages and currents as shown below. Since // | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | KVL around the single loop in the ciruit does not help us - the voltage across the current source is unknown, so inclusion of this parameter in a KVL equation simply introduces an additional unknown to go with the equation we write. KVL would, however, be useful if we wished to determine the voltage across the current source. | + | |
| - | + | ||
| - | KCL at node a tells us that i< | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL) govern the interactions between circuit elements. Governing equations for a circuit are created by applying KVL and KCL and applying the circuit element governing equations, such as Ohm’s Law. | + | |
| - | * Kirchhoff’s current law states that the sum of the currents entering or leaving a node must be zero. A node in a circuit is an point which has a unique voltage. | + | |
| - | * A node is a point of interconnection between two or more circuit elements. A circuit node has a particular voltage. Nodes can be “spread out” with perfect conductors. | + | |
| - | * Kirchhoff’s voltage law states that the sum of the voltage differences around any closed loop in a circuit must sum to zero. A loop in a circuit is any path which ends at the same point at which it starts. | + | |
| - | * A loop is a closed path through a circuit. Loops end at the same node at which they start, and typically are chosen so that no node is encountered more than once. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | - For the circuit below, determine: | + | |
| - | - The current through the 2Ω resistor | + | |
| - | - The current through the 1Ω resistor | + | |
| - | - The power (absorbed or generated) by the 4V power source | + | |
| - | + | ||
| - | {{ : | + | |
| <-- | <-- | ||
| - | ---- | ||
| - | + | -->Complete Chapters | |
| - | --> | + | Download |
| - | * {{ : | + | * {{ : |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | + | ||
| - | * {{ : | + | |
| - | * {{ : | + | |
| - | + | ||
| - | ===== 2. Introduction and Chapter Objectives ===== | + | |
| - | In [[https:// | + | |
| - | + | ||
| - | In the next few chapters, we will still apply Kirchhoff' | + | |
| - | + | ||
| - | In this chapter, we introduce analysis methods based on //circuit reduction// | + | |
| - | + | ||
| - | We begin our discussion of circuit reduction techniques by presenting two specific, but very useful, concepts: //Series// and // | + | |
| - | consequences of a voltage or current measurement. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== After Completing this Chapter, You Should be Able to: ==== | + | |
| - | * Identify series and parallel combinations of circuit elements | + | |
| - | * Determine the equivalent resistance of series resistor combinations | + | |
| - | * Determine the equivalent resistance of parallel resistor combinations | + | |
| - | * State voltage and current divider relationships from memory | + | |
| - | * Determine the equivalent resistance of electrical circuits consisting of series and parallel combinations of resistors | + | |
| - | * Sketch equivalent circuits for non-ideal voltage and current meters | + | |
| - | * Analyze circuits containing non-ideal voltage or current sources | + | |
| - | * Determine the effect of non-ideal meters on the parameter being measured | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 2.1 Series Circuit Elements and Voltage Division ===== | + | |
| - | There are a number of common circuit element combinations that are quite easily analyzed. These “special cases” are worth noting since many complicated circuits contain these circuit combinations as sub-circuits. Recognizing these sub-circuits and analyzing them appropriately can significantly simplify the analysis of a circuit. | + | |
| - | + | ||
| - | This chapter emphasizes two important circuit element combinations: | + | |
| - | + | ||
| - | ==== Series Connections ==== | + | |
| - | Circuit elements are said to be connected in //series// if all of the elements carry the same current. An example of two circuit elements connected in series is shown in Fig. 2.1. Applying KCL at node a and taking currents out of the node as positive we see that: | + | |
| - | + | ||
| - | $$-i_1+i_2=0$$ | + | |
| - | + | ||
| - | or | + | |
| - | + | ||
| - | $$i_1=i_2 | + | |
| - | + | ||
| - | Equation (2.1) is a direct outcome of the fact that the (single) node a in Fig. 2.1 interconnects only two elements - there are no other elements connected to this node through which curren can be diverted. This observation is so apparent (in many cases((If there is any doubt whether the elements are in series, apply KCL! Assuming elements are in series which are not in series can have disastrous consequences.))) that equation (2.1) is generally written by inspection for series elements such as those shown in Fig. 2.1 __without__ explicitly writing KCL. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | When resistors are connected in series, a simplification of the circuit is possible. Consider the resistive circuit shown in Fig. 2.2(a). Since the resistors are in series, they both carry the same current. Ohm's law gives: | + | |
| - | + | ||
| - | \begin{align*} | + | |
| - | v_1 = R_1i \\ | + | |
| - | v_2=R_2i \\ | + | |
| - | (Eq.2.2) | + | |
| - | \end{align*} | + | |
| - | + | ||
| - | Applying KVL around the loop: | + | |
| - | + | ||
| - | $$ -v+v_1+v_2=0 \Rightarrow v=v_1+v_2 | + | |
| - | + | ||
| - | Substituting equations (2.2) into equation (2.3) and solving for the current //i// results in: | + | |
| - | + | ||
| - | $$ i = \frac{v}{R_1+R_2} | + | |
| - | + | ||
| - | Now cinsider the circuit of Fig. 2.2(b). Application of Ohm's law to this circuit and solution for the current //i// gives: | + | |
| - | + | ||
| - | $$ i = \frac{v}{R_{eq}} | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Comparing equation (2.4) with equation (2.5), we can see that the circuits of Figs. 2.2(a) and 2.2(b) are indistinguishable if we select: | + | |
| - | + | ||
| - | $$R_{eq}=R_1+R_2 | + | |
| - | + | ||
| - | Figures 2.2(a) and 2.2(b) are called // | + | |
| - | + | ||
| - | This result can be generalized to a series combination of //N// resistances as follows: | + | |
| - | + | ||
| - | A series combination of //N// resistors R< | + | |
| - | + | ||
| - | ==== Voltage Division ==== | + | |
| - | Combining equations (2.2) with equation (2.4) results in the following expressions for // | + | |
| - | + | ||
| - | $$v_1= \frac{R_1}{R_1+R_2}v | + | |
| - | + | ||
| - | $$v_2= \frac{R_2}{R_1+R_2}v | + | |
| - | + | ||
| - | These results are commonly called //voltage divider// relationships, | + | |
| - | + | ||
| - | The above results can be generalized for a series combination of //N// resistance as follows: | + | |
| - | + | ||
| - | The voltage drop across any resistor in a series combination of N resistances is proportional to the total voltage drop across the combination of resistors. The constant of proportionality is the same as the ratio of the individual resistor value to the total resistance of the series combination. For example, the voltage drop of the // | + | |
| - | + | ||
| - | $$ v_k={R_k}{R_1+R_2+ \cdots +R_N}v | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ====Example 2.1 ==== | + | |
| - | For the circuit below, determine the voltage across the 5Ω resistor, //v//, the current supplied by the source, //i//, and the power supplied by the source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The voltage across the 5Ω resistor can be determined from our voltage divider relationship: | + | |
| - | + | ||
| - | $$ v = [\frac{5Ω}{5Ω + 15Ω + 10Ω}] \cdot 15V= \frac{5}{30} \cdot 15V = 2.5V $$ | + | |
| - | + | ||
| - | The current supplied by the source can be determined by dividing the total voltage by the equivalent resistance: | + | |
| - | + | ||
| - | $$i= \frac{15V}{R_{eq}}= \frac{15V}{5Ω+15Ω+10Ω}= \frac{15V}{30Ω}=0.5A$$ | + | |
| - | + | ||
| - | The power supplied by the source is the product of the source voltage and the source current: | + | |
| - | + | ||
| - | $$P=iv=(0.5A)(15V)=7.5W$$ | + | |
| - | + | ||
| - | We can double-check the consistency between the voltage //v// and the current //i// with Ohm's law. Applying Ohm's law to the 5Ω resistor, with a 0.5A current, results in $v=(5Ω)(0.5A)=2.5V$, | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * If only two elements connect at a single node, the two elements are in //series//. A more general definition, however, is that circuit elements in series all share the same current - this definition allows us to determine series combinations that contain more than two elements. Identification of series circuit elements allows us to simplify our analysis, since there is a reduction in the number of unknowns: there is only a single unknown current for all series elements. | + | |
| - | * A series combination of resistors can be replaced by a single // | + | |
| - | * If the total voltage difference across a set of series is known, the voltage differences across any individual resistor can be determined by the concept of //voltage division//. The term voltage division comes from the fact that the voltage drop across a series combination of resistors is divide among the individual resistors. The ratio between the voltage difference across a particular resistor and the total voltage difference is the same as the ratio between the resistance of that resistor and the total resistance of the combination. If // | + | |
| - | + | ||
| - | $$ \frac{v_k}{v_{TOT}} = \frac{R_k}{R_{TOT}} $$ | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. Determine the voltage V< | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 2.2 Parallel Circuit Elements and Current Division ===== | + | |
| - | Circuit elements are said to be connected in // | + | |
| - | + | ||
| - | $$ v_1=v_2 | + | |
| - | + | ||
| - | This result is so common that equation (2.10) is generally written by inspection fro parallel elements such as those shown in Fig. 2.3 __without__ explicitly writing KVL. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | We can simplify circuits, which consist of resistors connected in parallel. Consider the resistive circuit shown in Fig. 2.4(a). The ressitors are connected in parallel, so both resistors have a voltage difference of //v//. Ohm's law applied to each resistor results in: | + | |
| - | + | ||
| - | $$ i_1= \frac{v}{R_1} \\ | + | |
| - | i_2= \frac{v}{R_2} \\ | + | |
| - | (Eq. 2.11) $$ | + | |
| - | + | ||
| - | Applying KCL at node a: | + | |
| - | + | ||
| - | $$i=i_1 + i_2 (Eq. 2.12)$$ | + | |
| - | + | ||
| - | Substituting equations (2.11) into equation (2.12): | + | |
| - | + | ||
| - | $$ i = \left[ \frac{1}{R_1} + \frac{1}{R_2} \right] v (Eq. 2.13)$$ | + | |
| - | + | ||
| - | or | + | |
| - | + | ||
| - | $$v= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\cdot i (Eq. 2.14)$$ | + | |
| - | + | ||
| - | If we set $ R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$, | + | |
| - | + | ||
| - | We can generalize this result for //N// parallel resistances: | + | |
| - | + | ||
| - | A parallel combination of //N// resistors $R_1, R_2, \cdots, R_N$ can be replaced with a single equivalent resistance: | + | |
| - | + | ||
| - | $$ R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}+ \cdots \frac{1}{R_N}} | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | For the special case of two parallel resistances, | + | |
| - | + | ||
| - | $$R_{eq}= \frac{R_1R_2}{R_1+R_2} | + | |
| - | + | ||
| - | This alternative way to calculate $R_{eq}$ can be also used to calculate $R_{eq}$ for larger numbers of parallel resistors since any number of resistors could be combined two at a time. | + | |
| - | + | ||
| - | ==== Current Division ==== | + | |
| - | Substituting equation (2.14) into equations (2.11) results in: | + | |
| - | + | ||
| - | $$i_1 = \frac{1}{R_1}\cdot \frac{i}{ \frac{1}{R_1} + \frac{1}{R_2}} | + | |
| - | + | ||
| - | Simplifying: | + | |
| - | + | ||
| - | $$i_1=\frac{R_2}{R_1+R_2} | + | |
| - | + | ||
| - | Likewise, for the current $i_2$: | + | |
| - | + | ||
| - | $$i_2 = \frac{R_1}{R_1+R_2} | + | |
| - | + | ||
| - | Equations (2.18) and (2.19) are the current | + | |
| - | + | ||
| - | The above results can be generalized for a series combination of //N// resistances. By Ohm's law, $v = R_{eq} i$. Substituting our previous result for the equivalent resistance for a parallel combination of //N// resistors results in: | + | |
| - | + | ||
| - | $$v= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2} + \cdots \frac{1}{R_N}} \cdot i (Eq. 2.20)$$ | + | |
| - | + | ||
| - | Since the voltage difference across all resistors is the same, the current through the k< | + | |
| - | + | ||
| - | $$i_k=\frac{v}{R_k} | + | |
| - | + | ||
| - | Where $R_k$ is the resistance of the k< | + | |
| - | + | ||
| - | $$i_k= \frac{\frac{1}{R_k}}{\frac{1}{R_1}+\frac{1}{R_2}+ \cdots \frac{1}{R_N}} \cdot i (Eq. 2.22)$$ | + | |
| - | + | ||
| - | It is often more convenient to provide the generalized result of equation (2.20) in terms of the conductance of the individual resistors. Recall that the conductance is the reciprocal of the resistance, $G=\frac{1}{R}$. Thus, equation (2.22) can be re-expressed | + | |
| - | + | ||
| - | The Current through any resistor in a parallel combination of //N// resistances is proportional to the total current into the combination of resistors. The constant of proportionality is the same as the ratio of the conductance of the individual resistor value to the total conductance of the parallel combination. For example, the current through the // | + | |
| - | + | ||
| - | $$i_k=\frac{G_k}{G_1+G_2+ \cdots + G_N}i (Eq. 2.23)$$ | + | |
| - | + | ||
| - | Where //i// is the total current through the parallel combination of resistors. | + | |
| - | + | ||
| - | One final comment about notation: two parallel bars are commonly used as shorthand notation to indicate that two circuit elements are in parallel. For example, the notation $R_1 \parallel R_2$ indicates that the resistors $R_2$ and $R_2$ are in parallel. The notation $R_1 \parallel R_2$ is often used as shorthand notation for the __equivalent resistance__ of the parallel resistance combination, | + | |
| - | + | ||
| - | Double-checking results for parallel resistances: | + | |
| - | * The equivalent resistance for a parallel combination of //N// resistors will always be less than the smallest resistance in the combination. In fact, the equivalent resistance will always obey the following inequalities: | + | |
| - | + | ||
| - | $$\frac{R_{min}}{N} \leq R_{eq} \leq R_{min}$$ | + | |
| - | + | ||
| - | * Where $R_{min}$ is the smallest resistance value in the parallel combination. | + | |
| - | * In a parallel combination of resistances, | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * If several elements interconnect the same two nodes, the two elements are in // | + | |
| - | * A parallel combination of resistors can be replaced by a single // | + | |
| - | + | ||
| - | $$R_{eq} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\cdots\frac{1}{R_N}}$$ | + | |
| - | * If the total current through a set of parallel resistors is known, the current through any individual resistor can be determined by the concept of //current division//. The term current division comes form the fact that the current through a parallel combination of resistors is divided among the individual resistors. The ratio between the current through a particular resistor and the total current is the same as the ratio between the conductance of that resistor and the total conductance of the combination. If $i_k$ is the voltage across the k< | + | |
| - | + | ||
| - | $$\frac{v_k}{i_{TOT}}=\frac{G_k}{G_{TOT}}$$ | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. Determine the value of //I// in the circuit below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | 2. Determine the value of //R// in the circuit below which makes I=2mA. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 2.3 Circuit Reduction and Analysis ===== | + | |
| - | The previous results give us an ability to potentially simplify the analysis of some circuits. This simplification results if we can use //circuit reduction// techniques to convert a complicated circuit to a simpler, but equivalent, circuit which we can use to perform the necessary analysis. Circuit reduction is not always possible, but when it is applicable it can significantly simplify the analysis of a circuit. | + | |
| - | + | ||
| - | Circuit reduction relies upon identification of parallel and series combinations of circuit elements. The parallel and series elements are then combined into equivalent elements and the resulting //reduced// circuit is analyzed. The principles of circuit reduction are illustrated below in a series of examples. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.2 ==== | + | |
| - | Determine the equivalent resistance seen by the terminals of the resistive network shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The sequence of operations performed is illustrated below. The 6Ω and 3Ω resistances are combined in parallel to obtain an equivalent 2Ω resistance. This 2Ω resistance and the remaining 6Ω resistance are in series, these are combined into an equivalent 8Ω resistance. Finally, this 8Ω resistor and the 24Ω resistor are combined in parallel to obtain an equivalent 6Ω resistance. Thus, the equivalent resistance of the overall network is 6Ω. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.3 ==== | + | |
| - | In the circuit below, determine the power delivered by the source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | In order to determine power delivery, we need to determine the total current provided by the source to the rest of the circuit. We can determine current easily if we convert the resistor network to a single, equivalent, resistance. A set of steps for doing this are outlined below. | + | |
| - | + | ||
| - | Step 1: The four ohm and two ohm resistors, highlighted on the figure to the left below in blue, are in series. Series resistances add directly, so these can be replaced with a single six ohm resistor, as shown on the figure to the right below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Step 2: The tree ohm resistor and the two six ohm resistors are now all in parallel, as indicated on the figure to the left below. These resistances can be combined into a single equivalent resistor $R_{eq}=\frac{1}{\frac{1}{3}+\frac{1}{6}+\frac{1}{6}}=1.5\Omega$. The resulting equivalent circuit is shown to the right below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The current out of the source can now be readily determined from the figure to the right above. The voltage drop across the 1.5Ω resistor is 6V, so Ohm's law gives $i=\frac{6V}{1.5\Omega}=4A$. Thus, the power delivered by the source is $P=(4A)(6V)=24W$. Since the sign of the current relative to the current does __not__ agree with the passive sign convention, the power is __generated__ by the source. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.4 ==== | + | |
| - | For the circuit shown below, determine the voltage, $v_s$, across the 2A source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The two 1Ω resistors and the two 2Ω resistors are in series with one another, as indicated on the figure to the left below. These can be combined by simply adding the series resistances, | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The three remaining resistors are all in parallel (they all share the same nodes) so they can be combined using the relation $R_{eq}=\frac{1}{\frac{1}{2}+\frac{1}{4}+\frac{1}{4}}=1\Omega$. Note that it is not necessary to combine all three simultaneously, | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The voltage across the source can now be determined from Ohm's law: $v_s=(1\Omega)(2A)=2V$. The assumed polarity of the source voltage is correct. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.5: Wheatstone Bridge ==== | + | |
| - | A Wheatstone bridge circuit is shown below. The bridge is generally presented as shown in the figure to the left; we will generally use the equivalent circuit shown to the right. A Wheatstone bridge is commonly used to convert a variation in resistance to a variation in voltage. A constant supply voltage $V_s$ is applies to the circuit. The resistors in the circuit all have a nominal resistance of R; the variable resistor has a variation $\Delta R$ from this nominal value. The output voltage $v_{ab}$ indicates the variation $\Delta R$ in the variable resistor. The variable resistor in the network is often a transducer whose resistance varies dependent upon some external variable such as temperature. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | By voltage division, the voltages $v_b$ and $v_a$ (relative to ground) are: | + | |
| - | + | ||
| - | $$v_b=\frac{R+\Delta R}{2R+\Delta R}V_s$$ | + | |
| - | + | ||
| - | and | + | |
| - | + | ||
| - | $$v_a=Ri_2=\frac{V_s \cdot R}{2R}=\frac{V_s}{2}$$ | + | |
| - | + | ||
| - | The voltage $v_{ab}$ is then: | + | |
| - | + | ||
| - | $$v_{ab}=v_a-v_b= \left( \frac{1}{2} - \frac{R+ \Delta R}{2R + \Delta R} \right) V_s = \left( \frac{(2R+ \Delta R) - 2(R+ \Delta R)}{2(2R+ \Delta R)} \right) V_s = - \frac{\Delta R}{2(2R+ \Delta R)} \cdot V_s $$ | + | |
| - | + | ||
| - | For the case in which $\Delta R << 2R$, this simplifies to: | + | |
| - | + | ||
| - | $$v_{ab} \approx - \frac{V_s}{4R}\Delta R$$ | + | |
| - | + | ||
| - | And the output voltage is proportional to the change in resistance of the variable resistor. | + | |
| - | + | ||
| - | === Practical Applications: | + | |
| - | A number of common sensors result in a resistance variation resulting from some external influence. // | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * In a circuit, which contains obvious series and/or parallel combinations of resistors, analysis can be simplified by combining these resistances into equivalent resistances. The reduction in the overall number of resistances reduces the number of unknowns in the circuit, with a corresponding reduction in the number of governing equations. Reducing the number of equations and unknowns typically simplifies the analysis of the circuit. | + | |
| - | * Not all circuits are reducible. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | - For the circuit shown, determine: | + | |
| - | - $R_{eq}$ (the equivalent resistance seen by the source) | + | |
| - | - The currents $I_1$ and $I_2$ | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 2.4 Non-ideal Power Supplies ===== | + | |
| - | In section 1.2, we discussed ideal power sources. In that section, an ideal voltage supply was characterized as providing a specified | + | |
| - | + | ||
| - | In this section, we present simple models for voltage and current sources which incorporate more realistic assumptions as to the behavior of these devices. | + | |
| - | + | ||
| - | ==== Non-ideal Voltage Sources ==== | + | |
| - | An ideal voltage source was defined in section 1.2 as providing a specified voltage, regardless of the current flow out of the device. For example, an __ideal__ 12V battery will provide 12V across its terminals, regardless of the load connected to the terminals. A real 12V battery, however, provides 12V across its terminals only when its terminals are open-circuited. As we draw current from the terminals, the battery will provide less than 12V - the voltage will decrease as more and more current is drawn from the battery. The real battery thus appears to have an internal voltage drop which increases with increased current. | + | |
| - | + | ||
| - | We will model a real or // | + | |
| - | + | ||
| - | $$V=V_s-i \cdot R_s (Eq. 2.24)$$ | + | |
| - | + | ||
| - | Equation (2.24) indicates that the voltage delivered by our non-ideal voltage source model decreases as the current out of the voltage source increases, which agrees with expectations. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.6 ==== | + | |
| - | Consider the case in which we connect a resistive load to the non-ideal voltage source. The figure below provides a schematic of the overall system; $R_L$ is the load resistance, $V_L$ is the voltage delivered to the load, and $i_L$ is the current delivered to the load. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | In the case above, the current delivered to the load is $i=\frac{V_s}{R_s+R_L}$ and the load voltage is $V_L=V_s \frac{R_L}{R_s+R_L}$. Thus, if the load resistance is infinite (the load is an open circuit), $V_L=V_S$, but the power supply delivers no current and hence no power to the load. If the load resistance is zero (the load is a short circuit), $V_L=0$ and the power supply delivers current $i_L=\frac{V_s}{R_s}$ to the load; the power delivered to the load, however, is still zero. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.7: Charging a Battery ==== | + | |
| - | We have a " | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | If we attempt to analyze this circuit by applying KVL around the loop, we obtain 12V=4V. This is obviously incorrect and we cannot proceed with our analysis - our model disagrees with reality! | + | |
| - | + | ||
| - | To resolve this issue, we will include the internal resistance of the batteries. Assuming a 3Ω internal resistance in each battery, we obtain the following model for the system: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Applying KVL around the loop, and using Ohm's law to write the voltages across the battery internal resistances in terms of the current between the batteries results in: | + | |
| - | + | ||
| - | $$-12V+(3\Omega)i+(3\Omega)i+4V=0$$ | + | |
| - | + | ||
| - | Which can be solved for the current //i// to obtain: | + | |
| - | + | ||
| - | $$i= \frac{12V-4V}{6\Omega}=1.33A$$ | + | |
| - | + | ||
| - | Notice that as the voltage of the " | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Non-ideal Current Sources ==== | + | |
| - | An ideal current source was defined in section 1.2 as providing a specified current, regardless of the voltage difference across the device. This model suffers from the same basic drawback as our ideal voltage source model - the model can deliver infinite power, which is inconsistent with the capabilities of a real current source. | + | |
| - | + | ||
| - | We will use the circuit shown schematically in Fig. 2.6 to model a non-ideal current source. The non-ideal model consists of an ideal current source, $i_s$, placed in parallel with an internal resistance, $R_s$. The source delivers a voltage //V// and current //i//. The output current is given by: | + | |
| - | + | ||
| - | $$i=i_S-\frac{V}{R_S} | + | |
| - | + | ||
| - | Equation (2.25) shows that the current delivered by the source decreases as the delivered voltage increases. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 2.8 ==== | + | |
| - | Consider the case in which we connect a resistive load to the non-ideal current source. The figure below provides a schematic of the overall system; $R_L$ is the load resistance, $V_L$ is the voltage delivered to the load, and $i_L$ is the current delivered to the load. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | In the case above, the current delivered to the load can be determined from a current divider relation as $i_l=i_s \cdot \frac{R_s}{R_s+R_L}$ and the load voltage, by Ohm's law, is $V_L=i_L R_L = i_s \frac{R_S R_L}{R_s + R_L}$. If the load resistance is zero (the load is a short circuit), $i_L=i_s$, but the power supply delivers no voltage and hence no power to the load. In the case of infinite load resistance (the load is open circuit), $i_L = 0$. In this case, we can neglect $R_s$ in the denominator of the load voltage equation to obtain $V_L \approx i_s \frac{R_S R_L}{R_L}$ so that $V_L \approx i_s R_S$. Since the current is zero, however, the power delivered to the load is still zero. | + | |
| - | + | ||
| - | If we explicitly calculate the power delivered to the load, we obtain $V_L = i_{s}^{2} \frac{R_S R_L}{R_s + R_L} \cdot \frac{R_s}{R_s+R_L}$. A plot of the power delivered to the load as a function of the load resistance is shown below; a logarithmic scale is used on the horizontal axis to make the plot more readable. As expected, the power is zero for high and low load resistances. The peak of the curve occurs when the load resistance is equal to the source resistance, $R_L=R_s$. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * In many cases, power supplies can be modeled as ideal power supplies, as presented in section 1.2. However, in some cases representation as a power supply as ideal results in unacceptable errors. For example, ideal power supplies can deliver infinite power, which is obviously unrealistic. | + | |
| - | * In this section, we present a simple model for a non-ideal power supply. | + | |
| - | * Our non-ideal voltage source consists of an ideal voltage source in series with a resistance which is internal to the power supply. | + | |
| - | * Our non-ideal current supply consists of an ideal current source in parallel with a resistance which is internal to the power supply. | + | |
| - | * Voltage and current divider formulas allow us to easily quantify the effects of the internal resistances of the non-ideal power supplies. Our analysis indicates that the non-ideal effects are negligible, as long as the resistance of the load is large relative to the internal resistance of the power supply. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. A voltage source with an internal resistance of 2Ω as shown below is used ot apply power to a 3Ω resisotr. What voltage would you measure across the 3Ω resistor? | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | 2. The voltage source of exercise 1 above is used to apply power to a 2kΩ resistor. What voltage would you measure across the 2kΩ resistor? | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 2.5 Practical Voltage and Current Measurement ===== | + | |
| - | The process of measuring a physical parameter will almost invariably change the parameter being measures. This effect is both undesirable and, in general, unavoidable. One goal of any measurement is to effect the parameter being measured __as little as possible__. | + | |
| - | + | ||
| - | The above statement is true of voltage and current measurements. An __ideal__ voltmeter, connected in parallel with some circuit element, will measure the voltage across the element without affecting the current flowing through the element. Unfortunately, | + | |
| - | + | ||
| - | In this section, we introduce some effects of measuring voltages and currents with practical meters. | + | |
| - | + | ||
| - | ==== Voltmeter and Ammeter Models ==== | + | |
| - | We will model both voltmeters and ammeters as having some internal resistance and a method for displaying the measured voltage difference or current. Fig. 2.7 shows schematic representations of voltmeters and ammeters. | + | |
| - | + | ||
| - | The ammeter in Fig. 2.7(a) has an internal resistance $R_M$; the current through the ammeter is $i_A$ and the voltage difference across the ammeter is $V_M$. The ammeter' | + | |
| - | + | ||
| - | The voltmeter in Fig. 2.7(b) is also represented as having an internal resistance $R_M$; the current through the meter is $i_v$ and the voltage difference across the meter is $V_v$. The current through the voltmeter should be as small as possible - the voltmeter should have an extremely high internal resistance. | + | |
| - | + | ||
| - | The effects of non-zero ammeter voltages and non-zero voltmeter currents are explored in more detail in the following subsections. | + | |
| - | + | ||
| - | ==== Voltage Measurement ==== | + | |
| - | Consider the circuit shown in Fig. 2.8(a). A current source, $i_s$, provides current to a circuit element with resistance, R. We want to measure the voltage drop, V, across the circuit element. We do this by attaching a voltmeter across the circuit element as shown in Fig. 2.8(b). | + | |
| - | + | ||
| - | In Fig. 2.8(b) the voltmeter resistance is in parallel to the circuit element we wish to measure the voltage across and the combination of the circuit element and the voltmeter becomes a current divider. The current through the resistor R then becomes: | + | |
| - | + | ||
| - | $$i=i_s \frac{R_M}{R+R_M} | + | |
| - | + | ||
| - | The voltage across the resistor R is then, by Ohm's law: | + | |
| - | + | ||
| - | $$V=i_s \frac{R \cdot R_M}{R+R_M} | + | |
| - | + | ||
| - | If $R_M >>R$, this expression simplifies to: | + | |
| - | + | ||
| - | $$V \approx i_s \frac{R \cdot R_M}{R_M} = R \cdot i_s (Eq. 2.28)$$ | + | |
| - | + | ||
| - | And negligible error is introduces into the measurement - the measured voltage is approximately the same as the voltage without the voltmeter. If, however, the voltmeter resistance is comparable to the resistance R, the simplification of equation (2.28) is not appropriate and significant changes are made to the system by the presence of the voltmeter. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Current Measurement ==== | + | |
| - | Consider the circuit shown in Fig. 2.9(a). A voltage source, $V_s$, provides power to a circuit element with resistance, R. We want to measure the current, $i$, through the circuit element. We do this by attaching an ammeter in series with the circuit element as shown in Fig. 2.9(b). | + | |
| - | + | ||
| - | In Fig. 2.9(b) the series combination of the ammeter resistance and the circuit element whose current we wish to measure creates a voltage divider. KVL around the single circuit loop provides: | + | |
| - | + | ||
| - | $$V_s=i(R_M+R) | + | |
| - | + | ||
| - | Solving for the current results in: | + | |
| - | + | ||
| - | $$i= \frac{V_s}{R_M+R} | + | |
| - | + | ||
| - | If $R_M << R$, this simplifies to: | + | |
| - | + | ||
| - | $$i \approx \frac{V_s}{R} | + | |
| - | + | ||
| - | And the measured current is a good approximation to current in the circuit of Fig. 2.9(a). However, if the ammeter resistance is not small compared to the resistance R, the approximation of equation (2.31) is not appropriate and the measured current is no longer representative of the circuit' | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Caution ==== | + | |
| - | Incorrect connections of ammeters or voltmeters can cause damage to the meter. For example, consider the connection of an ammeter in __parallel__ with a relatively large resistance, as shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | In this configuration the ammeter current, $i_M= \frac{V_S}{R_M}$. Since the ammeter resistance is typically very small, this can result in high currents being provided to the ammeter. This, in turn, may result in excessive power being provided to the ammeter and resulting damage to the device. | + | |
| - | + | ||
| - | Ammeters are generally intended to be connected in __series__ with the circuit element(s) whose current is being measured. Voltmeters are generally intended to be connected in __parallel__ with the circuit element(s) whose voltage is being measured. Alternate connections can result in damage to the meter. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * Measurement of voltage and/or current in a circuit will always result in some effect on the circuit' | + | |
| - | * In this section, we present simple models for voltmeters and ammeters (voltage and current measurement devices, respectively). | + | |
| - | * Our non-ideal voltmeter consists of an ideal voltmeter (which had infinite resistance, and thus draws no current from the circuit) in parallel with a resistance which is internal to the voltmeter. This model replicates the finite current which is necessarily drawn from the circuit by a real voltmeter. | + | |
| - | * Our non-ideal ammeter consists of an ideal ammeter (which has zero resistance, and thus introduces no voltage drop in the circuit) in series with a resistance which is internal to the ammeter. This resistance allows us to model the finite voltage drop which is introduced into the circuit by a real current measurement. | + | |
| - | * Voltage and current divider formulas allow us to easily quantify the effects of the internal resistances of voltage and current meters. Our analysis indicates that the non-ideal effects are negligible, as long as: | + | |
| - | * The resistance of the voltmeter is large relative to the resistance across which the voltage measurement is being made. | + | |
| - | * The resistance of the ammeter is small compared to the overall circuit resistance. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | - A voltmeter with an internal resistance of 10MΩ is used to measure the voltage $v_{ab}$ in the circuit below. What is the measured voltage? What voltage measurement would you expect from an ideal voltmeter? | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| <-- | <-- | ||
| - | -->Real Analog: Chapter 3# | + | -->Complete Chapters Solutions# |
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| - | ====== 3. Introduction and Chapter Objectives ====== | + | |
| - | In Chapters [[https:// | + | |
| - | * Ohm's Law | + | |
| - | * Kirchhoff' | + | |
| - | * Circuit reduction | + | |
| - | + | ||
| - | Circuit reduction, it should be noted, is not fundamentally different from direct application of Ohm's and Kirchhoff' | + | |
| - | + | ||
| - | In Chapter 1, we saw that direct application of Ohm's law and Kirchhoff' | + | |
| - | + | ||
| - | In cases where circuit reduction is not feasible, approaches are still available to reduce the total number of unknowns in the system. //Nodal analysis// and //mesh analysis// are two of these. Nodal and mesh analysis approaches still rely upon application of Ohm's law and Kirchhoff' | + | |
| - | + | ||
| - | Since nodal and mesh analysis approaches are fairly closely related, section 3.1 introduces the basic ideas and terminology associated with __both__ approaches. Section 3.2 provides details of nodal analysis, and mesh analysis is presented in section 3.3. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== After Completing this Chapter, You Should be Able to: ==== | + | |
| - | * Use nodal analysis techniques to analyze electrical circuits | + | |
| - | * Use mesh analysis techniques to analyze electrical circuits | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 3.1 Introduction and Terminology ===== | + | |
| - | As noted in the introduction, | + | |
| - | + | ||
| - | Consider the circuit shown in Fig. 3.1(a). The circuit nodes are labeled in Fig. 3.1(a), for later convenience. The circuit is not readily analyzed by circuit reduction methods. If the exhaustive approach toward applying KCL and KVL is taken, the circuit has 10 unknowns (the voltages and currents of each of the five resistors), as shown in Fig. 3.1(b). Ten circuit equations must be written to solve for the ten unknowns. Nodal analysis and mesh analysis provide approaches for defining a __reduced__ number of unknowns and solving for these unknowns. If desired, any other desired circuit parameters can subsequently be determined from the reduced set of unknowns. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | In nodal analysis, the unknowns will be //node voltages//. Node voltages, in this context, are the // | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Basic Idea ==== | + | |
| - | In nodal analysis, Kirchhoff' | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | In mesh analysis, the unknowns will be //mesh currents//. Mesh currents are defined only for //planar circuits//; planar circuits are circuits which can be drawn in a single plane such that no elements overlap one another. When a circuit is drawn in a single plane, the circuit will be divided into a number of distinct areas; the boundary of each area is a //mesh// of the circuit. A //mesh current// is the current flowing around a mesh of the circuit. The circuit of Fig. 3.1 has three meshes: | + | |
| - | - The mesh bounded by $V_s$, node a, and node d | + | |
| - | - The mesh bounded by node a, node c, and node b | + | |
| - | - The mesh bounded by node b, node c, and node d | + | |
| - | + | ||
| - | These three meshes are illustrated schematically in Fig. 3.2. Thus, in a mesh analysis of the circuit of Fig. 3.1, three equations must be solved in three unknowns (the mesh currents). Any other desired circuit parameters can be determined from the mesh currents. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Basic Idea ==== | + | |
| - | In mesh analysis, Kirchhoff' | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * In nodal analysis: | + | |
| - | * Unknowns in the analysis are called the //node voltages// | + | |
| - | * Node voltages are the voltages at the __independent__ nodes in the circuit | + | |
| - | * Two nodes connected by a voltage source are __not independent__. The voltage source constrains the voltages at the nodes relative to one another. A node which is not independent is also called dependent. | + | |
| - | * In mesh analysis: | + | |
| - | * Unknowns in the analysis are called //mesh currents//. | + | |
| - | * Mesh currents are defined as flowing through the circuit elements which form the perimeter of the circuit meshes. A mesh is any enclosed, non-overlapping region in the circuit (when the circuit schematic is drawn on a piece of paper). | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. The circuit below has three nodes, A, B, and C. Which two nodes are dependent? Why? | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | 2. Identify meshes in the circuit below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 3.2 Nodal Analysis ===== | + | |
| - | As noted in section 3.1, in nodal analysis we will define a set of node voltages and use Ohm's law to write Kirchhoff' | + | |
| - | + | ||
| - | The steps used in nodal analysis are provided below. The steps are illustrated in terms of the circuit of Fig. 3.3. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 1: Define Reference Voltage ==== | + | |
| - | One node will be arbitrarily as a //reference node// or //datum node//. The voltages of all other nodes in the circuit will be defined to be __relative__ to the voltage of this node. Thus, for convenience, | + | |
| - | + | ||
| - | For our example circuit, we will choose node d as our reference node and define the voltage at this node to be 0V, as shown in Fig. 3.4. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 2: Determine Independent Nodes ==== | + | |
| - | We now define the voltages at the // | + | |
| - | * " | + | |
| - | * " | + | |
| - | + | ||
| - | After removal of the sources, the remaining nodes (with the exception of the reference node) are defined as // | + | |
| - | + | ||
| - | For our example circuit of Fig. 1, removal of the voltage source (replacing it with a short circuit) results in nodes remaining only at nodes b and c. This is illustrated in Fig. 3.5. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 3: Replace Sources in the Circuit and Identify Constrained Voltages ==== | + | |
| - | With the independent voltages defined as in Step 2, replace the sources and define the voltages at the dependent nodes in terms of the independent voltages and the known voltage difference. | + | |
| - | + | ||
| - | For our example, the voltage at node a can be written as a known voltage $V_s$ above the reference voltage, as shown in Fig. 3.6. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 4: Applying KCL at Independent Nodes ==== | + | |
| - | Define currents and write Kirchhoff' | + | |
| - | + | ||
| - | Node b: | + | |
| - | + | ||
| - | $$-i_1+i_3+i_4=0 | + | |
| - | + | ||
| - | Node c: | + | |
| - | + | ||
| - | $$-i_2-i_3+i_5=0 | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 5: Use Ohm's Law to Write the Equations From Step 4 in Terms of Voltages ==== | + | |
| - | The currents defined in Step 4 can be written in terms of the node voltages defined previously. For example, from Fig. 3.7: $i_1= \frac{V_S-V-b}{R_1}$, | + | |
| - | + | ||
| - | $$-\frac{V_s-V_b}{R_1} + \frac{V_b-V_c}{R_3} + \frac{V_b-0}{R_4} = 0 $$ | + | |
| - | + | ||
| - | So the KCL equation for node b becomes: | + | |
| - | + | ||
| - | $$ \left( \frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4} \right) V_b - \frac{1}{R_3} V_c = \frac{1}{R_1}V_S | + | |
| - | + | ||
| - | Likewise, the KCL equation for node c can be written as: | + | |
| - | + | ||
| - | $$-\frac{1}{R_3} V_b + \left( \frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5} \right) V_c = \frac{1}{R_2} V_S (Eq. 3.4)$$ | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Double-checking Results ==== | + | |
| - | If the circuit being analyzed contains only __independent sources__, and the sign convention used in KCL equations is the same as used above (__currents leaving nodes are assumed positive__), | + | |
| - | * The term multiplying the voltage at that node will be the sum of the conductances connected to that node. For the example above, the term multiplying $V_b$ in the equation for node b is $\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_4}$ while the term multiplying $V_c$ nin the equation for node c is $\frac{1}{R_3} + \frac{1}{R_2} + \frac{1}{R_5}$. | + | |
| - | * The term multiplying the voltages adjacent to the node will be the negative of the conductance connecting the two nodes. For the example above, the term multiplying $V_c$ in the equation for node b is $-\frac{1}{R_3}$, | + | |
| - | + | ||
| - | If the circuit contains dependent sources, or a different sign convention is used when writing the KCL equations, the resulting equations will not necessarily have the above form. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Step 6: Solve the System of Equations Resulting From Step 5 ==== | + | |
| - | Step 5 will always result in N equations in N unknowns, where N is the number of independent nodes identified in Step 2. These equations can be solved for the independent voltages. Any other desired circuit parameters can be determined from these voltages. | + | |
| - | + | ||
| - | The example below illustrates the above approach. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.1 ==== | + | |
| - | Find the voltage V for the circuit shown below: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | //Steps 1, 2, and 3//: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | //Steps 4 and 5//: Writing KCL at nodes A and B and converting currents to voltages using Ohm's law results in the following two equations: | + | |
| - | + | ||
| - | Node A: | + | |
| - | + | ||
| - | $$\frac{V_A - 6}{2} + \frac{V_A - V_B}{2} + \frac{V_A - V_B}{1} = 0 \Rightarrow \left( \frac{1}{2} + \frac{1}{2} + \frac{1}{1} \right) V_A - V_B = 3 \Rightarrow 2V_A - V_B = 3$$ | + | |
| - | + | ||
| - | Node B: | + | |
| - | + | ||
| - | $$\frac{V_B-V_A}{1} + \frac{V_B-0}{0.5} + 16 = 0 \Rightarrow \left( \frac{1}{1} + \frac{1}{0.5} \right) V_B - V_A = -16 \Rightarrow 3V_B - V_A = 16 $$ | + | |
| - | + | ||
| - | //Step 6//: Solving the above equations results in $V_A=5V$ and $V_B=7V$. The voltage $V$ is: | + | |
| - | + | ||
| - | $$V=V_A - V_B = -2V$$ | + | |
| - | + | ||
| - | Several comments should be made relative to the above example: | + | |
| - | 1. Steps 4 and 5 (applying KCL at each independent node and using Ohm's law to write these equations in terms of voltages) have been combined into a single step. This approach is fairly common, and can provide a significant savings in time. | + | |
| - | + | ||
| - | 2. There may be a perceived inconsistency between the two node equations, in the assumption of positive current direction in the 1Ω resistor. In the equation for node A, the current is apparently assumed to be positive from node A to node B, as shown below: | + | |
| - | {{ : | + | |
| - | + | ||
| - | This leads to the corresponding term in the equation for node A becoming: $\frac{V_A - V_B}{1}$. In the equation for node B, however, the positive current direction appears to be from node B to node A, as shown below: | + | |
| - | {{ : | + | |
| - | + | ||
| - | This definition leads to the corresponding term in the equation for node B becoming: $\frac{V_B - V_A}{1}$/ | + | |
| - | + | ||
| - | The above inconsistency in sign is, however, insignificant. Suppose that we had assumed (consistently with the equation for node A) that the direction of positive current for the node B equation is from Node A to B. Then, the corresponding term in the equation for node B would have been: $-\frac{V_A - V_B}{1}$ (note that a negative sign has been applied to this term to accommodate our assumption that __currents flowing into nodes are negative__). This is equal to $\frac{V_B - V_A}{1}$, which is exactly what our original result was. | + | |
| - | + | ||
| - | 3. The current source appears __directly__ in the nodal equations. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Note ==== | + | |
| - | When we write nodal equations in these chapters, we will generally assume that any unknown currents are flowing away from the node for which we are writing the equation, __regardless of any previous assumptions we have made for the direction of that current__. The signs will work out, as long as we are consistent in our sign convention __between assumed voltage polarity and current direction__ and our sign convention relative to __positive currents flowing out of nodes__. | + | |
| - | + | ||
| - | The sign applied to currents induced by current sources must be consistent with the current direction assigned by the source. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Supernodes ==== | + | |
| - | In the previous examples, we identified __dependent__ nodes and determined __constrained__ voltages. Kirchhoff' | + | |
| - | + | ||
| - | Example: for the circuit below, determine the voltage difference, V, across the 2mA source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 1: define Reference Node ==== | + | |
| - | Choose reference node (somewhat arbitrarily) as shown below; label the reference node voltage, $V_R$, as zero volts. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 2: Define Independent Nodes ==== | + | |
| - | Short circuit voltage sources, open circuit current sources as shown below and identify independent nodes/ | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 3: Replace Sources and Label any Known Voltages ==== | + | |
| - | The known voltages are written in terms of node voltages identified above. There is some ambiguity in this step. For example, either of the representations below will work equally well - either side of the voltage source can be chosen as the node voltage, and the voltage on the other side of the source written in terms of this node voltage. Make sure, however, that the correct __polarity__ of the voltage source is preserved. In our example, the left side of the source has a potential that is three volts higher than the potential of the right side of the source. This fact is represented correctly by both of the choices below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 4: Apply KCL at the Independent Nodes ==== | + | |
| - | It is this step that sometimes causes confusion among readers, particularly when voltage sources are present in the circuit. Conceptually, | + | |
| - | + | ||
| - | For our example, we will arbitrarily choose the circuit to the left above to illustrate this approach. The supernode is chosen to include the voltage source and both nodes to which it is connected, as shown below. We define two currents leaving the supernode, $i_1$ and $i_2$, as shown. KCL, applied at the supernode, results in: | + | |
| - | + | ||
| - | $$-2mA + i_1 + i_2 = 0$$ | + | |
| - | + | ||
| - | As before, currents leaving the node are assumed to be positive. This approach allows us to account for the current flowing through the voltage source without ever explicitly solving for it. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 5: Use Ohm's Law to Write the KCL Equations in Terms of Voltages ==== | + | |
| - | For the single KCL equation written above, this results in: | + | |
| - | + | ||
| - | $$-2mA + \frac{V_A - 0}{3k \Omega} + \frac{(V_A - 3) - 0}{6k \Omega} = 0$$ | + | |
| - | + | ||
| - | ==== Step 6: Solve the System of Equations to Determine the Nodal Voltages ==== | + | |
| - | Solution of the equation above results in $V_A = 5V$. Thus, the voltage difference across the current source is $V=5V$. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Alternate Approach: Constraint Equations ==== | + | |
| - | The use of supernodes can be convenient, but is not a necessity. An alternate approach, for those who do not wish to identify supernodes, is to restrain separate nodes on either side of the voltage source and then write a constraint equation relating these voltages. Thus, in cases where the reader does not recognize a supernode, the analysis can proceed correctly. We now revisit the previous example, but use constraint equations rather than the previous supernode technique. | + | |
| - | + | ||
| - | In this approach, Steps 2 and 3 (identification of independent nodes) are not necessary. One simply writes Kirchhoff' | + | |
| - | + | ||
| - | Example (revisited): | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Choice of a reference voltage proceeds as previously. However, now we will not concern ourselves too much with identification of independent nodes. Instead, we will just make sure we account for voltages and currents everywhere in the circuit. For our circuit, this results in the node voltages and currents shown below. Notice that we have now identified two unknown voltages ($V_A$ and $V_B$) and three unknown currents, one of which ($i_3$) is the current through the voltage source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Now we write KCL at each of the identified nodes, making sure to account for the current through the voltage source. This results in the following equations (assuming currents leaving the node are positive): | + | |
| - | + | ||
| - | Node A: $-2mA+i_1+i_3=0$ | + | |
| - | + | ||
| - | Node B: $-i_3+i_2=0$ | + | |
| - | + | ||
| - | Using Ohm's law to convert the currents $i_1$ and $i_2$ to voltages results in: | + | |
| - | + | ||
| - | Node A: $-2mA+ \frac{V_A - 0}{3k \Omega} + i_3 = 0$ | + | |
| - | + | ||
| - | Node B: $-i_3 + \frac{V_B - 0}{6k \Omega} = 0$ | + | |
| - | + | ||
| - | Notice that we cannot, by inspection, determine anything about the current $i_3$ from the voltages; __the voltage-current relationship for an ideal source is not known__. | + | |
| - | + | ||
| - | The two equations above have three unknowns - we cannot solve for the node voltages from them without a third equation. This third equation is the __constraint__ equation due to the presence of the voltage source. For our circuit, the voltage source causes a direct relationship between $V_A$ and $V_B$: | + | |
| - | + | ||
| - | $$V_B=V_A-3$$ | + | |
| - | + | ||
| - | These thee equations (the two KCL equations, written in terms of the node voltages and the constraint equation) constitute three equations in three unknowns. Solving these for the node voltage $V_A$ results in $V_A=5V$, so the voltage across the current source is $V=5V$. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | The example below uses the concept of a supernode to write the governing KCL equations. In the example below, Steps 1, 2, and 3 have been condensed into a single process, as have Steps 4 and 5. It is suggested that the reader re-do the example below using constraint equations. Note again that the current sources appear __directly__ in the KCL equations. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.2 ==== | + | |
| - | For the circuit below, find the power generated or absorbed by the 2V source __and__ the power generated or absorbed by the 2A source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | //Steps 1, 2, and 3//: we choose our reference node (arbitrarily) as shown below. Shorting voltage sources and open-circuiting current sources identifies three independent node voltages (labeled below as $V_A$, $V_B$, and $V_C$) and one dependent node, with voltage labeled below as $V_A-2$. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | //Steps 4 and 5//: Writing KCL at nodes A, B, and C and converting the currents to voltages using Ohm's law results in the equations below. Note that we have (essentially) assumed that all unknown currents at a node are flowing out of the node, consistent with our node 2 for example 1 above. | + | |
| - | + | ||
| - | Node A: | + | |
| - | + | ||
| - | $$2A+ \frac{V_A - 0V}{4 \Omega} + \frac{( V_A-2V) -V_B}{4 \Omega} + \frac{(V_A - 2V) - V_C}{8\Omega} = 0 \Rightarrow 5V_A - 2V_B - V_C = -10$$ | + | |
| - | + | ||
| - | Node B: | + | |
| - | + | ||
| - | $$\frac{V_B - 0V}{3\Omega} + \frac{(V_A - 2V)}{4\Omega} + 3A = 0 \Rightarrow 7V_B - 3V_A = 42$$ | + | |
| - | + | ||
| - | Node C: | + | |
| - | + | ||
| - | $$\frac{V_C - (V_A - 2)}{8\Omega} - 3A = 0 \Rightarrow V_C - V_A = 22$$ | + | |
| - | + | ||
| - | Step 6: Solving the above results in $V_A = 0V$, $V_B = -6V$, and $V_C=22V$. Thus, the voltage difference across the 2A source is zero volts, and the __2A source delivers no power__. KCL at node A indicates that the current through the 2V source is 2A, and the __2V source generates 4W__. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Dependent Sources ==== | + | |
| - | In the presence of dependent sources, nodal analysis proceeds approximately as outlined above. The main difference is the presence of additional equations describing the dependent source. As before, we will discuss the treatment of dependent sources in the context of examples. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.3 ==== | + | |
| - | Write the nodal equations for the circuit below. The dependent source is a voltage controlled voltage source. $I_S$ is an independent current source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | As always, the choice of reference node is arbitrary. To determine independent voltages, dependent voltage sources are short-circuited in the same way as independent voltage sources. Thus, the circuit below has two independent nodes; the dependent voltage source and the nodes on either side of it form a supernode. The reference voltage, independent voltages, supernode, and resulting dependent voltage are shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | We now, as previously, write KCL for each independent node, taking into account the dependent voltage resulting from the presence of the supernode: | + | |
| - | + | ||
| - | $$\frac{(V_A-2V_X)-0}{R_3} + \frac{V_A-0}{R_4} + \frac{V_A-V_B}{R_2}=0$$ | + | |
| - | + | ||
| - | $$\frac{V_B-V_A}{R_2} + \frac{V_B-(V_A+2V_X)}{R_1} -I_S=0$$ | + | |
| - | + | ||
| - | The above equations result in a system with two equations and three unknowns: $V_A$, $V_B$, and $V_X$ ($I_S$ is a known current). We now write any equations governing the dependent sources. Writing the controlling voltage in terms of the independent voltages results in: | + | |
| - | + | ||
| - | $$V_X=V_A-V_B$$ | + | |
| - | + | ||
| - | We now have three equations in three unknowns, which can be solved to determine the independent voltages, $V_A$ and $V_B$. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.4 ==== | + | |
| - | Write the nodal equations for the circuit below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | The reference node, independent voltages and dependent voltages ae shown on the figure below. A supernode, consisting of the 4V source and the nodes on either side of it, exists but is not shown explicitly on the figure. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Applying KCL for each independent node results in: | + | |
| - | + | ||
| - | $$\frac{(V_A+4V)-3V}{2\Omega} + \frac{(V_A + 4V)-0}{4\Omega} + \frac{V_A-0}{5\Omega} + \frac{V_A-V_B}{3\Omega} = 0$$ | + | |
| - | + | ||
| - | $$\frac{V_B-V_A}{3\Omega}-3I_X=0$$ | + | |
| - | + | ||
| - | This consists of two equations with three unknowns. The equation governing the dependent current source provides the third equation. Writing the controlling current in terms of independent voltages results in: | + | |
| - | + | ||
| - | $$I_X= \frac{V_A-0}{5\Omega}$$ | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * Basic steps in nodal analysis are: | + | |
| - | * Define a reference node. All node voltages will be relative to this reference voltage. | + | |
| - | * Identify independent nodes. This can be done by short-circuiting voltage sources, open-circuiting current sources, and identifying the remaining nodes in the circuit. The voltages at these nodes are the node voltages. | + | |
| - | * Determine dependent voltages. This can be done by replacing the sources in the circuit schematic, and writing voltage constraints introduced by voltage sources. | + | |
| - | * Use Ohm's law to write KCL at each independent node, in terms of the node voltages. This will result in N equations in N unknowns, where N is the number of node voltages. Independent " | + | |
| - | * Solve the equations of step 4 to determine the node voltages. | + | |
| - | * Use the node voltages to determine any other esired voltages/ | + | |
| - | * Modifications to the above approach are allowed. For example, it is not necessary to define supernodes in step 4 above. Once can define unknown voltages at either terminal of a voltage source and write KCL at each of these nodes. However, the unknown current through the voltage source must be accounted for when writing KCL - this introduces an additional unknown into the governing equations. This added unknown requires an additional equation. This equation is obtained by explicitly writing a constraint equation relating the voltages at the two terminals of the voltage source. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises === | + | |
| - | 1. Use nodal analysis to write a set of equations from which you can find $I_1$, the current through the $12\Omega$ resistor. Do not solve the equations. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | 2. Use nodal analysis to find the current I flowing through the $10\Omega$ resistor in the circuit below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ===== 3.3: Mesh Analysis ==== | + | |
| - | In mesh analysis, we will define a set of mesh currents and use Ohm's law to write Kirchhoff' | + | |
| - | + | ||
| - | Mesh analysis is appropriate for //planar circuits//. Planar circuits can be drawn in a single plane((Essentially, | + | |
| - | + | ||
| - | The steps used in mesh analysis are provided below. The steps are illustrated in terms of the circuit of Fig. 3.8. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 1: Define Mesh Currents ==== | + | |
| - | In order to identify our mesh loops, we will turn off all sources, much like what we did in nodal analysis. To do this, we: | + | |
| - | * Short-circuit all voltage sources. | + | |
| - | * Open-circuit all current sources. | + | |
| - | + | ||
| - | Once the sources have been turned off, the circuit can be divided into a number of __non-overlapping__ areas, each of which is completely enclosed by circuit elements. The circuit elements bounding each of these areas form the meshes of our circuit. The mesh currents flow around these meshes. Our example circuit has two meshes after removal of the sources, the resulting mesh currents are as shown in Fig. 3.9. | + | |
| - | + | ||
| - | === Note === | + | |
| - | We will always choose our mesh currents as flowing __clockwise__ around the meshes. This assumption is not fundamental to the application of mesh analysis, but it will result in a special form for the resulting equations which will later allow us to do some checking of our results. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 2: Replace Sources and Identify Constrained Loops ==== | + | |
| - | The presence of current sources in our circuit will result in the removal of some meshes during Step 1. We must now account for these meshes in our analysis by returning the sources to the circuit and identifying // | + | |
| - | + | ||
| - | We have two rules for constrained loops: | + | |
| - | - Each current must have __one and only one__ constrained loop passing through it. | + | |
| - | - The __direction and magnitude__ of the constrained loop current must agree with the direction and magnitude of the source current. | + | |
| - | + | ||
| - | For our example circuit, we choose our constrained loop as shown below. It should be noted that constrained loops can, if desired, cross our mesh loops - we have, however, chosen the constrained loop so that is does not overlap any of our mesh loops. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ==== Step 3: Write KVL Around the Mesh Loops ==== | + | |
| - | We will apply Kirchhoff' | + | |
| - | + | ||
| - | Note that more than one mesh current may pass through a circuit element. When determining voltage drops across individual elements, the __contributions from all mesh currents passing through that element must be included in the voltage drop__. | + | |
| - | + | ||
| - | When we write KVL for a given mesh loop, we will base our sign convention for the voltage drops on the direction of the mesh current for that loop. | + | |
| - | + | ||
| - | For example, when we write KVL for the mesh current $i_1$ in our example, we choose voltage polarities for resistors $R_1$ and $R_4$ as shown in the figure below - these polarities agree with the passive sign convention for voltages __relative to the direction of the mesh current $i_1$__. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Using the above sign conventions, | + | |
| - | + | ||
| - | $$R_4 \left( i_2-i_1 \right) + R_2i_2 + R_5 \left( i_2 + I_S \right) = 0 $$ | + | |
| - | + | ||
| - | Please note that the currents $i_2$ and $I_s$ are in the same direction in the resistor $R_5$, resulting in a summation of these currents in the term corresponding to the voltage drop across the resistor $R_5$. | + | |
| - | + | ||
| - | === Notes: === | + | |
| - | - Assumed sign conventions on voltage drops for a particular mesh loop are based on the assumed direction of that loop's mesh current. | + | |
| - | - The current passing through an element is the __algebraic sum__ of all mesh and constraint currents passing through that element. This algebraic sum of currents is used to determine the voltage drop of the element. | + | |
| - | + | ||
| - | ==== Step 4: Solve the System of Equations to Determine the Mesh Currents of the Circuit ==== | + | |
| - | Step 3 will always result in N equations in N unknowns, where N is the number of mesh currents identified in Step 1. These equations can be solved for the mesh currents. Any other desired circuit parameters can be determined from the mesh currents. | + | |
| - | + | ||
| - | The example below illustrates the above approach. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.5 ==== | + | |
| - | In the circuit below, determine the voltage drop, $V$, across the 3Ω resistor. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Removing the sources results in a single mesh loop with mesh current $i_1$, as shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Replacing the sources and defining one constrained loop per source results in the loop definitions shown below (note that each constrained loop goes through only one source and that the amplitude and direction of the constrained currents agrees with source). | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Applying KVL around the loop $i_1$ and using Ohm's law to write voltage drops in terms of currents: | + | |
| - | + | ||
| - | $$-3V+7\Omega \left( i_1 + 1A \right) + 3\Omega \left( i_1 + 1A + 3A \right) + 4\Omega \left( i_1 + 3A \right) = 0 \Rightarrow i_1 = -2A$$ | + | |
| - | + | ||
| - | Thus, the current $i_1$ is 2A, in the opposite direction to that shown. The voltage across the $3\Omega$ resistor is $V = 3\Omega \left( i_1 + 3A + 1A \right) = 3\Omega \left( -2A +3A +1A \right) = 3 \left( 2A \right) = 6V$. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Alternate Approach to Constraint Loops: Constraint Equations ==== | + | |
| - | In the above examples, the presence of current sources resulted in a reduced number of meshes. Constraint loops were then used to account for current sources. An alternate approach, in which we retain additional mesh currents and then apply // | + | |
| - | + | ||
| - | Example: determine the voltage, $V$, in the circuit below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Define three mesh currents for each of the three meshes in the above circuit and define unknown voltages $V_1$ abd $V_3$ across the two current sources as shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Applying KVL around the three mesh loops results in three equations with five unknowns: | + | |
| - | + | ||
| - | $$V_1 + 3\Omega \cdot \left(i_1 - i_3 \right) + 7\Omega \cdot \left( i_1 - i_2 \right) = 0$$ | + | |
| - | + | ||
| - | $$-3V + 7\Omega \cdot \left(i_2 - i_1 \right) + V_3 = 0$$ | + | |
| - | + | ||
| - | $$-V_3 + 3\Omega \cdot \left(i_3 - i_1 \right) + 4\Omega \cdot i_3 = 0$$ | + | |
| - | + | ||
| - | Two additional //constrain equations// are necessary. These can be determined by the requirement that the algebraic sum of the mesh currents passing through a current source must equal the current provided by the source. Thus, we obtain: | + | |
| - | + | ||
| - | $$-i_2 + i_3 = 3A$$ | + | |
| - | + | ||
| - | $$-i_1 = 1A$$ | + | |
| - | + | ||
| - | Solving the five simultaneous equations above results in the same answer determined previously. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Clarification: | + | |
| - | Previously, it was claimed that the choice of constraint loops is somewhat arbitrary. The requirements are that each source has only one constraint loop passing through it, and that the magnitude and direction of the constrained loop current be consistent with the source. Since constraint loops __can__ overlap other mesh loops without invalidating the mesh analysis approach, the choice of constraint loops is not unique. The examples below illustrate the effect of different choices of constraint loops on the analysis of a particular circuit. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ====Example 3.6: Version 1 ==== | + | |
| - | Using mesh analysis, determine the current, $i$, through the 4Ω resistor. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | === //Step 1//: Define mesh loops === | + | |
| - | + | ||
| - | Replacing the two current sources with open circuits and the two voltage sources with short circuits results in a single mesh current, $i_1$, as shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | === //Step 2//: Constrained loops, version 1 === | + | |
| - | Initially, we choose the constrained loops shown below. Note that each loop passes through only one source and has the magnitude and direction imposed by the source. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ===//Step 3//: Write KVL around the mesh loops === | + | |
| - | Our example has only one mesh current, so only one KVL equation is required. This equation is: | + | |
| - | + | ||
| - | $$-8V + 2\Omega \left( i_1 + 1A - 2A \right) + 4\Omega \left( i_1 - 2A \right) + 10V + 6\Omega \left( i_1 \right) = 0$$ | + | |
| - | + | ||
| - | === //Step 4//: Solve the system of equations to determine the mesh currents of the circuit === | + | |
| - | Solving the above equation results in $i_1 = 0.667A$. The current through the $4\Omega$ resistor is then, accounting for the 2A constrained loop passing through the resistor, $i = i_1 -2A = -1.333A$. | + | |
| - | + | ||
| - | ==== Example 3.6: Version 2 ==== | + | |
| - | In this version, we choose an alternate set of constraint loops. The alternate set of loops is shown below; all constraint loops still pass through only one current source, and retain the magnitude and direction of the source current. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Now, writing KVL for the single mesh results in: | + | |
| - | + | ||
| - | $$-8V + 2\Omega \left( i_1 + 1A \right) + 4\Omega \cdot i_1 + 10V + 6\Omega \left( i_1 +2A \right) = 0$$ | + | |
| - | + | ||
| - | Solving for the mesh current results in $i_1 = -1.333A$; note that this result is __different__ than previously. However, we determine the current through the $4\Omega$ resistor as $i=i_1 = -1.333A$, which __is__ the same result as previously. | + | |
| - | + | ||
| - | ===Note=== | + | |
| - | Choice of alternate constrained loops may change the values obtained for the mesh currents. The currents through the circuit elements, however, do not vary with choice of constrained loops. | + | |
| - | + | ||
| - | ==== Example 3.6: Version 3 ==== | + | |
| - | In this version, we choose yet another set of constrained loops. These loops are shown below. Again, each loop passes through one current source and retains that source' | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | KVL around the mesh loop results in: | + | |
| - | + | ||
| - | $$-8V + 2\Omega \cdot i_1 + 4\Omega \left( i_1 - 1A \right) + 10V + 6\Omega \left( i_1 - 1A + 2A \right) = 0$$ | + | |
| - | + | ||
| - | Which results in $i_1 =-0.333A$. Again, this is different from the result from our first two approaches. However, the current through the $4\Omega$ resistor is $i=i_1 - 1A = -1.333A$, which __is__ the same result as previously. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Dependent Sources ==== | + | |
| - | As with nodal analysis, the presence of dependent sources does not significantly alter the overall mesh analysis approach. The primary difference is simply the addition of the additional equations necessary to describe the dependent sources. We discuss the analysis with dependent sources in the context of the following examples. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.7 ==== | + | |
| - | Determine the voltage V in the circuit below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Shorting both of the voltage sources in the circuit above results in two mesh circuits. These are shown in the figure below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Writing KVL around the two mesh loops results in: | + | |
| - | + | ||
| - | $$-2V + 2\Omega \cdot i_1 + 3\Omega \left( i_1-i_2 \right) = 0 $$ | + | |
| - | + | ||
| - | $$2I_X + 3\Omega \left( i_2 - i_1 \right) + 4\Omega \cdot i_2 = 0$$ | + | |
| - | + | ||
| - | we have two equations and three unknowns. We need an additional equation to solve the system of equations. The third equation is obtained by writing the dependent source' | + | |
| - | + | ||
| - | $$I_X = i_1$$ | + | |
| - | + | ||
| - | The above three equations can be solved to obtain $i_1 = 0.4375 \Omega$ and $i_2 = 0.0625 \Omega$. The desired voltage $V=4i_2 = 0.25V$. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Example 3.8 ==== | + | |
| - | Write mesh equations for the circuit shown below. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Mesh loops and constraint loops are identified as shown below: | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | Writing KVL for the two mesh loops results in: | + | |
| - | + | ||
| - | $$4\Omega \cdot i_1 + 2\Omega \left(i_1 - 3V_X \right) + 12V = 0$$ | + | |
| - | + | ||
| - | $$-12V + 3\Omega \left(i_2-3V_X \right) + 5\Omega \cdot i_2 = 0$$ | + | |
| - | + | ||
| - | Writing the controlling voltage $V_X$ in terms of the mesh currents results in: | + | |
| - | + | ||
| - | $$V_X = 5\Omega \cdot i_2$$ | + | |
| - | + | ||
| - | The above consist of three equations in three unknowns, which can be solved to determine the mesh currents. Any other desired circuit parameters can be determined from the mesh currents. | + | |
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Section Summary ==== | + | |
| - | * Basic steps in mesh analysis are: | + | |
| - | * Identify mesh currents. This can be done by short-circuiting voltage sources, open-circuiting current sources, and identifying the enclosed, non-overlapping regions in the circuit. The perimeters of these areas are the circuit meshes. The mesh currents flow around the circuit meshes. | + | |
| - | * Determine constrained loops. The approach in Step 1 will ensure that no mesh currents will pass through the current sources. The current source currents can be accounted for by defining constrained loops. Constrained loops are defined as loop currents which pass through the current sources. Constrained loops are identified by replacing the sources in the circuit schematic, and defining mesh currents which pass through the current sources; these mesh currents form the constrained loops and must match both the magnitude and direction of the current in the current sources. | + | |
| - | * Use Ohm's law to write KVL around each mesh loop, in terms of the mesh currents. This results in N equations in N unknowns, where N is the number of mesh currents. Keep in mind that the voltage difference across each element must correspond to the voltage difference induced by __all__ the mesh currents which pass through that element. | + | |
| - | * Solve the equations of Step 4 to determine the mesh currents. | + | |
| - | * Use the mesh currents to determine any other desired voltages/ | + | |
| - | * The constrained loops in Step 2 above are not unique. Their only requirement is that they must account for the currents through the current sources. | + | |
| - | * Modifications to the above approach are allowed. For example, it is not necessary to define constrained loops in Step 3 above. One can define (unknown) mesh currents which pass through the current sources and write KVL for these additional mesh currents. However, the unknown voltage across the current source must be accounted for when writing KVL - this introduces an additional unknown into the governing equations. This added unknown requires an additional equation which is obtained by explicitly writing a constraint equation equating algebraic sum of the mesh currents passing through a current source to the current provided by the source. | + | |
| - | + | ||
| - | + | ||
| - | ---- | + | |
| - | + | ||
| - | ==== Exercises ==== | + | |
| - | 1. Use mesh analysis to write a set of equations from which you can find $I_1$, the current through the $12\Omega$ resistor. Do not solve the equations. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | 2. Use mesh analysis to find the current $I$ flowing through the $10\Omega$ resistor in the circuit below. Compare your result to your solution to exercise 2 of section 3.2. | + | |
| - | + | ||
| - | {{ : | + | |
| - | + | ||
| - | ---- | + | |
| <-- | <-- | ||
| - | --> | ||