Schaum series network theory pdf

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    Schaum's outline of theory and problems of basic circuit analysis p. c.m. ( Schaum's outline series). Includes index. 1. Electric circuits. 2. Electric circuit analysis. Schaum's Outline of Theory and Problems of Electric Circuits . Network functions, frequency response, filters, series and parallel resonance, two-port networks. This Schaum's Outline gives you fully solved problems, extra practice on topics . Dr. Nahvi's areas of special interest and expertise include network theory.

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    Schaum Series Network Theory Pdf

    Use with these courses: [ΠΠ] Electric Circuits [ΠΠ] Circuit Theory. [ΠΠ] Electric Circuit Analysis. [ΠΠ] Electric SCHAUM'S OUTLINE SERIES. McGRAW-HILL. Theory and Problems of Schaum's Outline Series . used as a generic reference to a host of circuit simulators that use the SPICE2 solution. Theory and Problems of . Network functions, frequency response, filters, series and parallel resonance, . RLC Series Circuit; Series Resonance.

    Click Here for Terms of Use. Introduction 1. Four basic quantities and their SI units are listed in Table The other three basic quantities and corresponding SI units, not shown in the table, are temperature in degrees kelvin K , amount of substance in moles mol , and luminous intensity in candelas cd. All other units may be derived from the seven basic units. The electrical quantities and their symbols commonly used in electrical circuit analysis are listed in Table Two supplementary quantities are plane angle also called phase angle in electric circuit analysis and solid angle. Their corresponding SI units are the radian rad and steradian sr. The decimal multiples or submultiples of SI units should be used whenever possible. Work results when a force acts over a distance. Work and energy have the same units. Power is the rate at which work is done or the rate at which energy is changed from one form to another.

    At normal temperatures there is constant, random motion of these electrons. This electric potential is capable of doing work just as the mass in Fig.

    The potential energy mgh represents an ability to do work when the mass m is released. As the mass falls, it accelerates and this potential energy is converted to kinetic energy. In an electric circuit an energy of 9.

    The rate, in joules per second, at which energy is transferred is electric power in watts. In the following chapters time average power Pavg and a root-mean-square RMS value for the case where voltage and current are sinusoidal will be developed. Under these conditions at what rate is electric energy converted to heat? Solved Problems 1. Preface B. About the Authors 1. Introduction 2.

    Circuit Concepts 3. Circuit Laws 4. Analysis Methods 5. Amplifiers and Operational Amplifier Circuits 6. Waveforms and Signals 7. First-Order Circuits 8.

    Higher-Order Circuits and Complex Frequency 9. Sinusoidal Steady-State Circuit Analysis AC Power Supplementary Problems 8. Find the current transient, assuming zero initial charge on the capacitor.

    Find the transient voltage across the resistance. What are the scaled element values? In the circuits of Fig. At what! The response will also be sinusoidal. For a linear circuit, the assumption of a sinusoidal source represents no real restriction, since a source that can be described by a periodic function can be replaced by an equivalent combination Fourier series of sinusoids. This matter will be treated in Chapter In this chapter, the functions of v and i will be sines or cosines with the argument!

    The functions are sketched in Fig. Note that the current function i is to the right of v, and since the horizontal scale is! This illustrates that i lags v. This is a case of mixed units just as with! It is not mathematically correct but is the accepted practice in circuit analysis. If sketches are made of these responses, they will show that for a resistance R, v and i are in phase.

    Obtain the voltage v across the two circuit elements and sketch v and i. LI sin! LI cos! Consequently i leads v for a series RC circuit. A directed line segment, or phasor, such as that shown rotating in a counterclockwise direction at a constant angular velocity! The length of the phasor or its magnitude is the amplitude or maximum value of the cosine function.

    If a voltage or current is expressed as a sine, it will be changed to a cosine by subtracting from the phase. Consider the examples shown in Table Observe that the phasors, which are directed line segments and vectorial in nature, are indicated by boldface capitals, for example, V, I. The phase angle of the cosine function is the angle on the phasor. The frequency f Hz and! Obtain the voltages v and V, the phasor current I and sketch the phasor diagram.

    Using the methods of Example 9. Y and Z are complex numbers. The sign on the imaginary part may be positive or negative: When positive, X is called the inductive reactance, and when negative, X is called the capacitive reactance. When the admittance is written in Cartesian form, the real part is admittance G and the imaginary part is susceptance B.

    A positive sign on the susceptance indicates a capacitive susceptance, and a negative sign indicates an inductive susceptance. Find the equivalent impedance and admittance.

    Therefore, impedances combine exactly like resistances: Impedance Diagram In an impedance diagram, an impedance Z is represented by a point in the right half of the complex plane. Applying KVL, as in Section 4. For the network of Fig. It is easy to see that two loop currents might have the same direction in one impedance and opposite directions in another.

    Nevertheless, the preceding rules for writing the Z-matrix and the V-column have been formulated in such a way as to apply either to meshes or to loops. These rules are, of course, identical to those used in Section 4.

    Choosing meshes as in Fig. Setting up the matrix equation: Thus, for the single-source network of Fig. Y11 is the self-admittance of node 1, given by the sum of all admittances connected to node 1. Similarly, Y22 and Y33 are the self-admittances of nodes 2 and 3. Y12 , the coupling admittance between nodes 1 and 2, is given by minus the sum of all admittances connecting nodes 1 and 2.

    Similarly, for the other coupling admittances: The Y-matrix is therefore symmetric. On the right-hand side of the equation, the I-column is formed just as in Section 4.

    In words: If all sources have the same frequency, superposition is applied in the phasor domain. Otherwise, the circuit is solved for each source, and time-domain responses are added. Obtain the voltage vL. Obtain total voltage v and the angle by which i lags v. If the maximum voltage across the capacitance is 24 V, what is the maximum voltage across the series combination? Then, by the methods of Example 9. Determine the source frequency and the impedance Z. From the impedance diagram, Fig.

    Find the frequency for which the magnitude of the impedance is a twice that at f1 , b one-half that at f1.

    Determine the current which results when the resistance is reduced to a 30 percent, b 60 percent, of its former value. The frequency must also be known, but the phase angles of the voltages are not Fig. Determine R and XL. Find the applied voltage V. By voltage division in the two branches: The matrix equation can be written by inspection: In the netwrok of Fig. The network is redrawn in Fig. By the rule of Section 9.

    There are three principal nodes in the network. V0 is the node voltage V2 for the selection of nodes indicated in Fig. For the network shown in Fig. At terminals ab, Isc is the Norton current I 0. Make V 0 the voltage of a with respect to b. Find the phase angle by which the current leads the voltage. Determine the phase angle and whether the current leads or lags the total voltage.

    Find the resulting current i. Find the applied voltage V, if the voltage across Z1 is Obtain Rp and Lp in terms of Rs and Ls. Ls Fig. Obtain the circuit constants R and L.

    Find the source voltage V. For each choice, calculate the phasor voltage V. Assuming a source angular frequency! Find vA if! KCL at node A in the phasor domain. Apply 9. Find the current through the 4 H inductor. Find vA. AC Power If p is positive, energy is delivered to the circuit. If p is negative, energy is returned from the circuit to the source. Therefore, in the steady state and during each cycle, all of the energy received by an inductor or capacitor is returned. This is illustrated in Fig.

    During the rest of the cycle, the instan- CHAP. The plot of p vs. It depends on V, I, and the phase angle between them. This occurs when the load is purely resistive. The ratio of Pavg to Veff Ieff is called the power factor pf.

    Find the power factor. During the period of energy return, the power is negative. The power involved in this exchange is called reactive or quadrature power. The reactive power Q depends on V, I, and the phase angle between them. It is the product of the voltage and that component of the current which is out of phase with the voltage. This occurs for a purely resistive load, when V and I are in phase.

    Note that, while P is always nonnegative, Q can assume positive values for an inductive load where the current lags the voltage or negative values for a capacitive load where the current leads the voltage. It is also customary to specify Q by it magnitude and load type. We use the notation Veff and Ieff to include the phase angles. S is discussed in Section V2 sin 2! In such cases, the inductor and the capacitor will exchange some energy with each other, bypassing the ac source.

    This reduces the reactive power delivered by the source to the LC combination and consequently improves the power factor. See Sections Substituting the values of pR , pL , and pC found in Examples The three scalar quantities S, P, and Q may be represented geometrically as the hypotenuse, horizontal and vertical legs, respectively, of a right triangle CHAP. The power triangle is simply the triangle of the 2 as shown in Fig.

    Power triangles for an inductive load impedance Z scaled by the factor Ieff and a capacitive load are shown in Figs. In summary, Complex Power: Ieff Referring to Fig. In the example shown in Fig. In such diagrams, some of the triangles may degenerate into straight-line segments if the corresponding R or X is zero.

    If the power data for the individual branches are not important, the network may be replaced by its equivalent admittance, and this used directly to compute ST. However, this approach is more time-consuming. Each individual load tends to be either pure resistance, with unity power factor, or resistance and inductive reactance, with a lagging power factor.

    Schaum's Outline of Theory and Problems of Electric Circuits

    All of the loads are parallel-connected, and the equivalent impedance results in a lagging current and a corresponding inductive quadrature power Q. To improve the power factor, capacitors, in three-phase banks, are connected to the system either on the primary or secondary side of the main transformer, such that the combination of the plant load and the capacitor banks presents a load to the serving utility which is nearer to unit power factor.

    The decrease, VA, amounts to 4. Consequently, an improvement in the power factor, with its corresponding reduction in kVA, releases some of this generation and transmission capability so that it can be used to serve other customers.

    This is the reason behind the rate structures which, in one way or another, make it more costly for an industrial customer to operate with a lower power factor.

    An economic study comparing the cost of the capacitor bank to the savings realized is frequently made. A capacitor is added in parallel such that the power factor is improved to 0. Find the reduction in current drawn from the generator. Before improvement: Find Q4 and the resulting S. In Example Then, tan Addition of the compensating load Q4 reduces the reactive power from This reduces the apparent power S from The current is proportionally reduced.

    Table Load If a common period T may be found for the sources i. The preceding result may be generalized to the case of any n number of sinusoidal sources operating simultaneously on a network.

    If the n sinusoids form harmonics of a fundamental frequency, then superposition of powers applies. Let v1 by itself produce i1. Likewise, let v2 produce i2. Note that in addition to P1 and P2 , we need to take into account a third term hi1 i2 i which, depending on!

    The average power depends on the duration of averaging. Solved Problems Find and plot a v across the series RC combination and b the instantaneous power p entering RC. It is the same result obtained in Example The power exchanged between the source and the circuit during one cycle also agrees with the result obtained in Example Find P in each of the three cases. Determine the complete power information. Since the angle on the equivalent admittance is the negative of the angle on the equivalent impedance, its cosine also gives the power factor.

    Now, for a change in power factor to 0. Obtain the power triangles for the branches and combine them into the total power triangle. Calculate the average power P and the reactive power Q for each load. Find the branch powers P1 and P2. How many kvar must these capacitors furnish, and what is the resulting percent reduction in apparent power?

    What percent reduction in current results? What percent of the transformer rating does this load represent? How many kW in additional load may be added at unity power factor before the transformer exceeds its rated kVA? Give complete input power information. The power factor is therefore variable, ranging from values near 0.

    Supplementary Problems Find the complex power and the power factor. Obtain the complete power triangle. Obtain the power triangles for the branches and combine them to obtain the total power triangle. Obtain the total apparent power and the overall power factor. Determine the parallel capacitance in microfarads necessary to improve the power factor to a 0. What further reduction was achieved in part b? How many kW do these units take, if the new overall power factor is 0. A capacitor bank is added, improving the power factor to 0.

    After improvement, what percent of rated kVA is the transformer carrying? How many kVA in additional load at 0. Find the power factor of the kVA load, if the overall power factor is 0. Synchronous motors totaling kVA are added and operated at a leading power factor.

    If the overall power factor is then 0. Compute the complex power. Find the impedance of the circuit and its equivalent circuit made of two series elements. Show the balance sheet for the average andpreactive powers between the source and R, L, and C. For this reason, polyphase systems are used.

    3000 Solved Problems in Electric Circuits Schaums

    Another advantage is having more than one voltage value on the lines. This chapter deals mainly with three-phase circuits which are the industry standard. However, examples of two-phase circuits will also be presented. The system feeds two identical loads [Fig.

    Find currents, voltages, the instantaneous and average powers delivered. Thus, This is realized by positioning three coils at electrical angle separations on the same rotor. Normally, the amplitudes of the three phases are also equal. The generator is then balanced. Voltage polarities reverse for each change of pole.

    Voltage B is electrical degrees later than A, and C is later. This is referred to as the ABC sequence. Changing the direction of rotation would result in. The phasor diagram for the voltage is shown in Fig.

    If the neutral point is carried along with the lines, it is a three-phase, four-wire system. If line impedances must be considered, then the current direction through, for example, line aA would be IaA , and the phasor line voltage drop VaA.

    In this chapter, an angle of zero will always be associated with the phasor voltage of line B with respect to line C: All ABC-sequence voltages are shown in Fig. The currents in the impedances are referred to either as phase currents or load currents, and the three will be equal in magnitude and mutually displaced in phase by The line currents will also be equal in magnitude and displaced from one another by ; by convention, they are given a direction from the source to the load.

    All current directions are shown in Fig. The line-to-line voltages and all currents are shown on the phasor diagram, Fig. Note particularly the balanced currents. Note also that Determine the line currents and draw the voltage-current phasor diagram. Note that with one line current calculated, the other two can be obtained through the symmetry of the phasor diagram.

    All three line currents return through the neutral. Therefore, the neutral current is the negative sum of the line currents: In actual power circuits, it must not be physically removed, since it carries the small unbalance of the currents, carries short-circuit or fault currents for operation of protective devices, and prevents overvoltages on the phases of the load.

    Since the computation in Example Looking into any two terminals, the two connections will be equivalent if corresponding input, output, and transfer impedances are equal.

    The criteria for equivalence are as follows: In many cases, for instance, in power calculations, only the common magnitude, IL , of the three line currents is needed.

    This may be obtained from the single-line equivalent, Fig. The currents will be unequal and will not have the symmetry of the balanced case. The voltage-current phasor diagram is shown in Fig. The line currents are unequal and the currents on the phasor diagram have no symmetry.

    VAN See Problem for an alternate method. It may be written in terms of line voltage VL and line current IL.

    The line voltage VL in industrial systems is always known. If the load is balanced, the total power can then be computed from the line current and power factor. Thus, in Fig. A meter will attempt to go downscale if the phase angle between the voltage and current exceeds In this event, the current-coil connections can be reversed and the upscale meter reading treated as negative in the sum.

    The same reasoning establishes the analogous result for a Y-connected load. Figure corresponds to Fig. The two voltage sources are out of phase. Find a the line currents, voltages, and their phase angles, and b the instantaneous and average powers delivered by the generator.

    Schaum's Outline of Theory and Problems of Electric Circuits - PDF Free Download

    Obtain the line currents and draw the voltage-current phasor diagram. The line currents are: Obtain the currents and draw the voltage-current phasor diagram. With balanced Y-loads, the neutral conductor carries no current. See the phasor diagram, Fig.

    Obtain the line and phase currents by the single-line equivalent method. Obtain the total power. Obtain the line currents and the total power. To obtain the currents a sequence must be assumed; let it be ABC. Then, using Fig. Obtain the line and neutral currents.

    Obtain the voltages across the impedances and the displacement neutral voltage VON. However, the mesh currents I1 and I2 suggested in Fig. From the results of Problem Obtain the load impedance, if the line voltage is Obtain the line-to-line voltage at the load. The single-line equivalent circuit is shown in Fig. The wire size and total length control the resistance in Zc , while the enclosing conduit material e.

    As a general rule, a phasor diagram should be constructed for every polyphase problem. Obtain the line currents. CBA; Obtain the four line currents. Obtain the three line currents. Obtain the line currents, using the single-line equivalent method. Construct a phasor diagram similar to Fig. Obtain the phase power of each load. After using the single-line equivalent method to obtain the total line current, compute the total power, and compare with the sum of the phase powers.

    Find the readings of two wattmeters used to measure the total average power. Find the impedance of the balanced Y-connected load. The lines between the system and the load have impedances 2: Find the linevoltage magnitude at the load. Find the power absorbed by the load if the three wires connecting the source to the load have resistances of 0.

    Power is halved. This response is a function of the frequency. We have already seen that a sinusoid can be represented by a phasor which shows its magnitude and phase. It is a real function of j! For example, if a current source is connected across the network of Fig. Conversely, if a voltage source is applied to the input and Fig. For the two-port network of Fig. Both Hv1 and Hz1 are real constants, independent of frequency, since no reactive elements are present. If the network contains either an inductance or a capacitance, then Hv1 and Hz1 will be complex and will vary with frequency.

    If jHv1 j decreases as Fig. Four two-element circuits are shown in Fig. In a similar manner, the frequency response of the output-to-input voltage ratio can be obtained. This transfer function approaches unity at high frequency, where the output voltage is the same as the input.

    L versus!. For the open-circuit condition CHAP. The voltage transfer function Hv1 approaches zero at high frequencies and unity at! Consequently, any stray shunting capacitance, in parallel with C2 , serves to reduce the response of the circuit. In most cases, 0 The other transfer CHAP. This useful method is illustrated in the following example. The network passes low-frequency sinusoids and rejects, or attenuates, the high-frequency sinusoids. The pole-zero plot is shown in Fig.

    Therefore, for any value of s, the network function may be expressed in terms of three vectors A, B, and C as follows: Figure b shows the magnitude plot. Therefore, the three network functions are low-pass with the same half-power frequency of! Those containing additional dependent sources are called active. The circuit in Fig. The frequency response of the bandpass function is k2! The half-power frequencies are at! This behavior is also called resonance see Sections When the quality factor is high,!

    Again, from! C CHAP. The capacitive reactance, inversely proportional to! Consequently, the net reactance at frequencies below! At frequencies above! The points where the response is 0. The bandwidth is the width between these two frequencies: See Section The half-power frequencies can be expressed in terms of the circuit elements, or in terms of!

    A practical inductor, in which both resistance and inductance are present, is modeled in Fig. It is usually applied at resonance, in which case it has the equivalent forms! Complex series-parallel networks may have several high-impedance resonant frequencies! L is plotted magnitude only in Fig.

    Half-power frequencies! A reasonable model for the practical tank is shown in Fig. This becomes evident when the expression for! Given Rs , Ls , and the operating frequency! Lp Fig. There are times when the RC circuit in either form should be converted to the other form see Fig. Again, the equivalence depends on the operating frequency.

    This same information can be presented in a single plot: In this section we shall discuss locus diagrams for the input impedance or the input admittance; in some cases the variable will not be! For the series RL circuit, Fig. Either Fig. Note the points on the locus corresponding to! To illustrate the addition, the points corresponding to frequencies! Higher Qind corresponds to lower values of R. It is seen from Fig. Consider C variable without limit, and R1 , R2 , L, and!

    This may occur for two values of the real, positive parameter C [the case illustrated in Fig. Here we summarize the method see also Section 8. Dividing inductor and capacitor values in a circuit by a factor k will scale-up the!

    For example, a 1-mH inductor operating at 1 kHz has the same impedance as a 1-mH inductor operating at 1 MHz. This is called frequency scaling and is a useful property of linear circuits.

    By using a 1-nF capacitor,! Obtain the voltage Fig. The load resistance, 20 k , reduced the ratio from 0.

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