ELECTRIC CIRCUIT ANALYSIS DAVID E JOHNSON PDF

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Electric Circuit Analysis David E Johnson Pdf

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electric circuit analysis 4th johnson In our collection PDF Ebook is the best for you, supported Basic Electric Circuit Analysis 2nd Edition by David E Johnson, . JOHNSON JOHNNY ciofreedopadkin.ml electric circuit analysis pdf An electric motor is an electrical machine that converts electrical energy into. David E. Johnson's most popular book is Basic Electric Circuit Analysis. David David E, our (PDF) ELECTRIC CIRCUIT ANALYSIS DAVID E JOHNSON.

If the function graphed in Problem An element has constant current and voltage. Find the 14 battery and b the charge delivered to the battery in 2 h current entering the positive terminal and the charge be- hours.. CIs 1. What is the peak o 4 6 7 9 12 t s pet dissipated by the element and total energy supplied to each coulomb through the element?

How long does it take power supplied to the element? V for t Show that by definition of a passive element. If the function J t is the current in amperes entering 1. The net energy J t in joules absorbed by an ele- 2 1. If we early from 6 to 18 V as t varies from 0 to 10 min. Is this a passive or active element? When fully charged.

Find those values of p for which the 1. If the volt. A certain element with i and v that satisfy the pas. What should be the value of this voltage? What is the peak power we lutionaries. If the el. An element has a power rating of 1 kW. Justify your answer. Sketch the power 1. Ampere was born in Lyons. The electrical potential at a distance r away from a certain electrical charge is 20lr V r in meters. If the movement were made at uniform velocity and v satisfy the passive sign convention.

We wish to supply this element 5 J of ene. Determine if the rest of the circuit to which the The unit of electric current. These are III 2. A point of connection of two or more circuit elements.

Nothing more need be known to determine the resulting currents and. For the present circuit theory purposes. The element law for a resistor. Armed with these tools. Simplifications available through the use of equivalent circuits are then discovered in the context of Node 3 series-parallel and Thevenin-Norton equivalents. For mstance. We are now ready to discuss the all-important laws of Kirchhoff. We next introduce the basic relationship linking current and voltage in resistors.

Chapter Contents of elements constrains their possible currents and voltages. Kirchhoff's current law KCL states that For purposes of circuit analysis. These III 2. These laws are valid for circuits containing elements of all types: The fundamental concept of equivalent circuits is introduced.. The larger point within the node. Ohm's law. For circuits so defined. This impedes or resists the motion of the electrons.

Nodes are indicated in diagrams in two metals. We shall consider a resistor to be any device that exhibits solely a resistance. They are called Kirchhoff's current law and Kirchhoff's voltage law.

III 2. Whichever is used. Perfect conductors are idealized leads or wires that allow current to flow without impediment no charge accumulations or v9ltage drops along the leads. An example of a conductors exhibit properties that are characteristic of resistors. Materials that are commonly used in fabricating resistors include metallic alloys and carbon compounds and. The points marked a and b in Fig. In this chapter we first introduce two general laws governing the flow of current and the pattern of voltages in any circuit.

In other materials remember that a node is all the wire in direct contact with a given point. When currents flow in circuit with three nodes is shown in Fig. But this is a closed path.

The algebraic sum of voltage drops around any closed path is zero. Multiplying both sides of the KCL equation by Examining this equation. As with KCL. As an example of KCL. All minus signs disappear. In general. We now move to Kirchhoff's voltage law. For every charge that since i flows into the node. Summing the currents entering the node.

KVL is in fact an instance of same two equations.

The signs of By use of similar reasoning. Suppose that there is a single unknown voltage in a loop. This may be used to solve for a single unknown current at a node and in the traversal is in the opposite sense.

Traversing the loop in the plus-to-minus direction of the unknown voltage clockwise in Fig. Their choice sum of rises form. Traversing the loop clockwise beginning at point a. Multiplying each term in the sum of drops form of KVL stated above by -1 yields the Clearly. The electric fields produced in lumped. There is a common situation. If we hike along a path with uphill and downhill sections. All other things being equal. Had we chosen to move counterclockwise from b. The sum of rises form yields exactly these is invalid.

The supposition yielding the same result.

At the upper-left node. We would imply a violation of KCL at one or more nodes in the region. At the upper-right node.

Applying generalized KCL to the circuit of Fig. In the circuit for Exercise 2. Find il and i2 for the circuit shown. Repeating for the left loop. For this reason.

Ohm's law is currents and environmental conditions. Mathematically Treated. In nonlinear resistors. In he published the results in a paper titled "The Galvanic Chain. Since sets of linear equations are far easier to solve than sets containing even a single nonlinear equation.

The constant of proportionality is the resistance value In reality. Many in Section 1. Answer Since R is constant. It is ironic. In all likelihood you will 2. Sketch a loop satisfying the equation VI. A l-kQ resistor pulled out of a lab drawer will reliably draw 1 rnA of current when placed across where R 2: Ohm's law states that the voltage across a resistor is directly proportional to the current flowing through the resistor.

If we choose to state this law using variables. Resistors with different i-v laws are called nonlinear resistors. Implied by the choice to concentrate on linear elements is the result that symbol for the resistor. For R v of any resistor is affected by conditions such as temperature. Had he not failed to publish his findings. To avoid the minus sign. By you should find a I-MV power supply with which to excite it.

In this book our interest will be exclusively on linear elements. An example is an ordinary 7V incandescent lamp. The circuit symbol for the resistor is shown in Fig.

Since v and i together satisfy the passive sign convention. They may be open-circuited by removing posItive net energy. Thus a resistor does indeed satisfy the definition of a paSSIve element. Another umt of conductance.

A subcircuit containing any number of in- reSIstance.. The important concepts of short circuit and open circuit may be defined in terms of A subcircuit is any part of a circuit. A long cable carries 3 A of current while dissipating 72 W.

Two points may be short-circuited by joining them with a wire. The symbol for the mho is an inverted o: We wish to determine the current i and the power p absorbed by the l-kQ resistor in Fig.

Note that is constant over the half-hour. Answer a 0.

Find a the conductance. The terminal voltage of a lO-kQ resistor is 5 V. What Werner and William Siemens. For resistors the instantaneous power is all current pathways between them. In hght of 2. The voltage across and current into these terminals are through It. Examples include two-terminal elements. Resistors dissipate. Such a beneficial trade is possible only when the original and voltage replacement subcircuits are equivalent to one another in a specific sense to be defined.

Find a the current i and b the power delivered to the resistors CombInIng 2. For instance. Two two-terminal subcircuits are said to be equivalent if they have the same terminal law. Its pI ace as mIg. For these simplest of subcircuits. Let's find the terminal laws for the two subcircuits of Fig.

VSI is are the source func. The element law for the independent voltage source of Fig. The behavior of a two-terminal subcircuit. The element law for a resistor is Ohm's law. The terminal laws for ideal voltage and current sources are specified in terms of only one terminal variable.

The terminal law is what is needed to understand the action of elements and i lt 5Q i lt 5Q subcircuits. The independence of the terminal a b voltage v from the current i flowing through the source is what characterizes this ele- ment as an ideal voltage source. Since the terminal laws are identical. The source functions Vs and is might in fact vary with time.

All other elements. SIllce R] and R3 context. The opposite choice of reference 2. While the currents and voltages external to equivalent subcircuits will not be which we define as one that can be completely described by a smgle equatIOn. Consider first the circuit of Fig. We will proceed by replacing more complex subcircuits by simpler ones.

By KCL. One type. The notion of equivalent circuits will prove very helpful in simplifying circuit problems. The test for equivalence will always be the identity of the terminal Now that the laws of Kirchhoff and Ohm and the notion of equivalent circuits have been laws of the two subcircuits. We could have assigned its reference direction counterclockwise had we preferred. Wherever this is the case. So chains of series elements may be formed of any Finally. Find a value for R for which the two circuits of Exercise 2.

We next make the voltage variable assignments V]. R2 and R3 are III senes. Find an equivalent to the two-tenninal circuit shown in Exer. Only where it is thought necessary to emphasize the incompleteness of a part of a circuit will the term subcircuit be used. Find the tenninal laws for the two circuits sketched. I amperes clockwise is the same as.

I amperes counterclockwIse.

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Rand R respectively. All that and KVL applied around the loop yield a single equation in i that descnbes the CircUlt is guaranteed by their equivalence is that they can be freely interchanged in any cir. Note that the voltage reference directions have been chosen 2. Nonadjacent elements are III senes If they are no way to tell which of two equivalents is in the circuit. III t. It is never necessary somehow to guess the nght reference direction in order to arrive at a correct answer.

Since all elements in the loop are III senes. Let us determine the loop current. Comparing this result with 2. V and Note that these voltages are in proportion to their resistances.

Less formally. This equivalent resistor can be freely substituted in any circuit without changing any current or voltage external to Although we did not explicitly use the voltage-divider principle. Then 38 Chapter 2 Resistive Circuits Section 2. This follows from the voltage-divider principle and the Rs fact that series elements carry a common current. Larger the voltage reference direction.

Examining Fig. As another example. Continuing our analysis of Fig. Since RI and R2 form a voltage divider. Thus a positive vi product means resistances have larger voltage drops in a voltage divider. This is the principle of i together violate the passive sign convention for the source the voltage division: Replacing the three series resistors Generalizing in the obvious way.

Since VI. This is also the terminal law for the two-terminal circuit. Either the given element laws are incorrect or KCL is violated. Note that we could have assigned pendent of their voltages. If the given source functions reference direction for an equivalent source function.

Yet we know this cannot happen. It is not necessary to guess the best tions.. This is a case of inconsistent math- Since the three voltage sources are in series.

We next consider the series connection of N voltage sources as in Fig. The "give" would be on the part of the element law. Inconsistent equations cannot be solved. Comparing the terminal law Finally. From Fig. This is an illustration of Tellegen' s theorem. But what if the series current source functions we are given are not all equal? With reference directions as Once a mathematical inconsistency is discovered in a problem statement.

More concisely. The terminal law for the shaded circuit in Fig. Reduce the circuit to two equivalent elements and find i and the the instantaneous power dissipated by the V source. By KVL. Applying KCL at the upper node. Two elements are connected in parallel if together they form a loop containing no other elements.

Generalizing in the obvious way for N resistors connected in parallel. By denoting the conductance of loop circuits lead naturally to series interconnections and voltage dividers. Find R in the circuit shown. Answer 2. Just as single. Answer -6 A. SOV cos 2tV Designate the common voltage of these elements v and define resistive currents to satisfy the passive sign convention relative to v.

Note that the currents are in the predicted proportions 4: The equivalent parallel conductance is or the sum The equivalent resistance of two resistors in parallel is the product of their resistances divided by the sum. Then the currents will be. Given their resistances. Smaller resistances larger conductances have larger current flows in a current divider. Note by 2. Note that the same caution holds with this as with the product-by-sum rule: This is the principle 18mA n n n n v of current division: In this case 2.

The terminal laws are identical if Finally. Continuing the example. Since the two current sources are in parallel. Note the minus sign in Ohm's law here. If the given source functions are not all equal.

P2 and P3 are found to be 8 mW and P4 is 4 mW. The two series resistors are equivalent to a single Q dissipated by is given by the vi product. A set of voltage sources in parallel must have add. Then noting that v and that each re- sistive current in Fig. Turning to the source. Determine the equivalent resistance of the circuit shown.

Answer 3 Q Find v and i I in Fig. Thus the 36 mW produced by the source is divided among the resis. If it is to our advantage. They will prove to be of great utilIty m First. Note that these additional 16V 4Q lA series simplifications were not possible before the Thevenin-Norton 2. The following example shows how it may be to our advantage to do so. How many of each are needed? In this section we develop a pair of equivalents. Section 2. Replacing the original by its two-element equivalent The voltage Vo in Fig.

Rr and the intercept is Vr. This is a simple circuit of the 2. To do so. Choosing the Norton form. It would be more convenient to determine the equivalents directly from the circuit diagram.

By Norton equivalents. The resulting Thevenin form. Since no current can flow if the terminals of the circuit are open circuited. Comparing the terminal law for the Thevenin form. If Instead of open circuiting the terminals. In the previous examples. We first simplify the has the desIred effect of simplifying the circuit. This follows from the linearity of Kirchhoff's laws and the element 2Q 4V laws and the fact that combinations of linear equations remain linear.

This will always be the case with resistive circuits of interest here. Use of Eq. Since a 0. To find ise. Then the Note that killing a source means setting its source function to zero.

Consider the circuit of Fig. V v 24V two leftmost resistors are in parallel they form a loop containing no voltage source is simply a short circuit. The resistance looking into its resistors. To find V oe. Returning to 2. Combining these two results. After killing internal sources. This follows from the fact that in the absence of internal sources. Practical sources. For now we content ourselves with modeling capability to represent practical sources by their Thevenin or Norton equivalent being aware of the power of Thevenin and Norton equivalents to simplify circuit analysis circuits.

The effect of the source resistance Rs in the practical voltage source model of Answer -1 V Fig. As RL converges toward zero. We will use this universal or voltage in any circuit are developed in Chapter 4. One clear sign that these ideal sources are not physically realizable is examining 2.

For large values of RL the source is lightly loaded little current is being drawn. An ideal current source with RL Q across its terminals is similarly required to that this equivalent is in series with. Since the voltage across the resistor Rs serves to limit the maximum or short-circuit current the source may supply.

Practical sources connected to load resistors are shown in Fig. The presence of a resistor in both circuits limits their current. Find the Thevenin equivalent of the circuit to the right of line cd.

If the circuit contains many loops and nodes. Repeat Exercise 2. This can be seen by infinity as the load resistance RL goes to zero. How then are we to model practical. In Section 2. Use the results of Exercises 2. This produce infinite power i 2RL as RL goes to infinity. VL decreases from its ideal value. The source resistance considering what happens as RL goes to zero. The power dissipated by the load is In Chapter 1 an independent voltage source was defined as an ideal element whose terminal voltage is equal to its source function.

Systematic methods for finding any current equivalent circuits for all linear two-terminal subcircuits. Noting that RT and 29 Q form a voltage divider. It is '. Vs accurately reflects the behavior of practical sources.. The Thevenin or Norton equivalent for a given source may found by measuring any. P goes to zero. Find the Thevenin and Norton equivalents of the C cell.

This is 1. In this case 1. The last measurement is often difficult to Practical voltage "'" arrange when it is a source that is being modeled. The comparisons Maximum current will flow if the device is short-circuited The source resistance Rs serves to limit the open-circuit voltage the The Thevenin equivalent of the C cell is shown in Fig. For instance P goes to infinity with RL Vs and is in Fig.

The power to the load is.

105213053-Electric-Circuit-Analysis.pdf

When a Q resistor is placed across it. The carbon film device consists of carbon powder that is deposited on an combine RSH with the original current source in a Norton source insulating substrate. Referring to Fig. By current division. Their characteristics include a internally across their source resistance. Multicolored model. The most common type of discrete resistor found in electrical circuits is the carbon load R L.

Then granules. Maximum power dissipation Silvera -2 Yellow 4 will occur when the load is an open circuit.. A typical resistor of this type is shown in Fig. This power is not available to supply to other nominal resistance value.

The new source model. Determine an external resistance RSH that can be placed in composition or carbon film resistor. Sources feel warm to the We next tum to practical resistors. The composition type is made of hot-pressed carbon "shunt.

Resistors are manufactured from a variety of touch when they are operating because they always drop some of the power they produce materials and are available in many sizes and values. Then the Norton equivalent of the left of ab in Fig. Integrated circuits are rapidly superceding discrete element circuits for most low.

For resistance values less than 10 n. Values of carbon resistors range from 2. This device is constructed by suspending an electrical coil between the poles of a permanent magnet. Low-temperature-coefficient wire 2. An integrated-circuit resistor has and has a terminal current of zero.

In the real world of physical design and use of electric circuits. The resistance is then adjusted by etching or grinding a pattern through the film. These resistors are made by vacuum deposition of a thin metal layer on a low-thermal-expansion substrate.

A rnA fuse is located in the wire connecting a power supply Therefore. A dc current passing through the coil causes a rotation of the coil.

A pointer is attached to the coil so that the rotation. Find the resistance range of carbon resistors having color bands of a brown. In contrast. An automobile battery in a poor state of charge has an open-circuit must be gold or silver. In this circuit.

Answer W. Practical doping. The resistive layer is very thin. An ideal which active and passive elements are fabricated by diffusion and deposition processes. Determine its Thevenin and Norton parameters VT. Answer a to Q.

Determine the power delivered to the load and the power being dissipated Another resistor type that is commonly used in applications requiring a high power internally in the source resistance under these conditions. This device consists of a metallic wire. It consists of the same doped semiconductor material used to form transistors ammeters. An integrated circuit is a single monolithic chip of semiconductor such as current. An ideal ohmmeter measures the resistance connected the typical structure shown in Fig.

Accuracy and stability for these resistors approach that of wire-wound types. The most common type of ammeter consists of a mechanical movement known as Resistive layer a d' Arsonval meter. They can be made to measure with precision unavailable using the d' Arsonval meter.

Scales may then be laid out on the face of the instrument to correspond with the resistance values Rx given in the preceding equation for each available Rs. Answer a Infinite. Determine a b Rp in Fig. RM is usually a few ohms. It is obvious that the Combining the last two equations. Applying KVL. In the YOM. Suppose that we wish. We must not allow a current greater than Ips to flow through the device without risking damage to the delicate device.

A circuit to accomplish this is shown in QfV. From current division we see that A simple ohmmeter circuit employing a d' Arsonval meter for measuring an un- known resistance Rx is shown in Fig. Electronic volt-ohm-ammeters employ active circuits. Solving for Rp. The d' Arsonval meter is an ammeter suitable for measuring dc currents not greater The current sensitivity of a voltmeter. If we put N identical R-Q resistors in that to a good approximation. Resistive circuits are not without important practical applications.

Each of 1 1! What voltage would each meter design of Exercise 2. Why are the two measurements different? They must be in parallel once again. This typically occurs when the single resistor is not of an available resistance. As with most design problems. Series-parallel transformations may be used to reduce a set of interconnected re. These elements will be introduced in the next several chapters.

For our first design example. We cannot have two or more in series. In this section we present three design examples using resistive circuits.

Electric Circuit Analysis

In order that the circuit dissipate 2 W of power. We could have begun connected between terminals a and a'. VB the battery voltage. But in this case the component count advantage would be lost.

We begin the design by noting that the circuit shown in favor the series alternative. We wish to deliver 10 rnA of current to larger batteries. Our goal is to design a battery charger circuit that will deliver con- trolled levels of dc current to charge batteries of different sizes. We could work circuit. This may seem to save resistors. Since 1 around this problem by substituting a pair of parallel 6.

We need to determine if this resistor array satisfies the power R L. The total current 1 passing through the circuit divides equally among the 10 resistors in Rei. At any terminal voltage with resistors in parallel.

Battery We will determine the Norton source. Thus we have a solution. Connecting the source and load as shown in Fig. If desired. Now the the examples above. This is the case with 68 Chapter 2 Resistive Circuits Section 2. This can easily be remedied by "tuning" this One of many possible solutions first -cut try. I will roughly scale by 10 as desired. The load current ing current of 10 rnA.

The final design is shown in Fig. If the resulting element values do not satisfy the specifications and constraints. The process of design begins with a complete statement of design specifications and constraints. Circuit analysis methods are then used to determine the performance of the dc proposed circuit.

The voltages supplied need not have a common ground terminal. In the examples above. This will be the case for most of the design examples in subsequent chapters. In very simple designs it may be possible to solve for solution values of the circuit elements exactly rather than through repeated cut-and-try cycles. But as the designs get more complex.

Even if a solution has been found. The minimum current is then increased to Specify Norton forms. Sketch the circuit. Others are introduced in later chapters. Each type of element has its characteristic element or i-v law. Examples include c In sum-entering-equals-sum-Ieaving form. Write KVL around each of the three loops.

These may be found by determining the open-circuit voltage. Write KVL for this loop: Equivalent subcircuits b In sum-leaving-equals zero form. Write KCL at each node. Write KCL for this node: While KCL. The force between a fixed and variable magnetic field moves a pointer. There are two interconnection laws. KCL equations for a certain three-node circuit are: Node 1: In this chapter we introduce two elements.

Determine the power supplied by. Find the energy used by a toaster with a resistance Find the equivalent resistance looking in terminals c -d if ter- 4Q 3Q of 12 n. A circuit has 4 nodes. By what factor would we increase the source func- tion in the circuit of Problem 2. Open means no other connections.

Find v for each subcircuit. Find all currents and voltages. In a given circuit. Assign names and reference directions. In each case. Rbd 2. Find VI. Find the terminal law for this circuit. Assign voltage reference directions for all five ele. If a toaster is dissipating W equivalent resistance looking in terminals a -b if terminals at a voltage of V. With what signs?

Determine 2. Find ii. Design an. RI and R2. Ofl 2. Problem 2. Start by taking GI to be 1 S. Use a series of Thevenin-Norton and series-parallel vider are scaled by the same factor a. A current divider is to be constructed using 2.

Find the terminal law of the new subcircuit and its Norton equivalent. If all resistors in a voltage divider are scaled by v the same factor a. If the conductances of all resistors in a current di.

All this circuit to Q? The voltages across them? The power? Find i and the total power dissipated by this circuit equivalent resistance of the divider. A V source and two resistors.

What must R be to set the equivalent resistance of 2. A V source in series with a resistive load R car. Find i and the power delivered to the 4-Q resistor. Start by taking RI to be 1 Q. Find the 2. Find the Thevenin equivalent circuit. Ten identical copies of a subcircuit N with terminal 2. Find the Thevenin equiv- resistors must be integer multiples of 1 kQ.

If a resistor RI is added in series with the source and load. Find the Thevenin equivalent of this circuit by first 2. When a load resistor is connected between 1 and 1'. Use any number of 2. Find a similar relation involving resistors and current sources. Find i and v. Using any number of 10 0.

Use a Thevenin-Norton transformation to source supplying 25 rnA of short-circuit current and 15 rnA help. Find the Thevenin equivalent of this subcircuit. Find the Norton equivalent to the left of ab. Find v. Design Problems a c. Design a resistive circuit limiting a Sketch the circuit.

Give an example. Find the internal resistors as possible. Find the Thevenin equivalent of everything except 2. Design a dc power supply circuit with four pairs ImA of load terminals marked ' through '.

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Data Papers.As RL converges toward zero. This follows from the linearity of Kirchhoff's laws and the element 2Q 4V laws and the fact that combinations of linear equations remain linear. Of those in the latter category, special thanks to Profs. The emphasis on analysis perhaps merits some comment.

Finally, just as several new examples and exercises time. If 10 J of work is required to move C of negative Note that this model shows that vo.. Use a series of Thevenin-Norton and series-parallel vider are scaled by the same factor a. The most Sufficient space has been allocated for judicious repetition, reinforcement and learn- challenging problems have been set off as a separate group.

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