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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. Author: ARDELLA HORGAN Language: English, Spanish, Portuguese Country: Canada Genre: Science & Research Pages: 243 Published (Last): 26.06.2016 ISBN: 907-3-16012-628-8 ePub File Size: 28.58 MB PDF File Size: 18.57 MB Distribution: Free* [*Regsitration Required] Downloads: 28616 Uploaded by: MONSERRATE

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.

Loop currents are chosen such that each source contains only one current. Consequently, a diagram should be examined before a solution is started and redrawn if necessary to show more clearly how the elements are interconnected. A capacitor bank is added, improving the power factor to 0. The future course and values of such signals can be predicted only in average and not precisely. See Figs.

This in turn is in parallel with the other resistor so that the overall equivalent resistance is Since this is in parallel with the mH inductance, the overall equivalent inductance is 5 mH. When it is a single resistor with an adjustable tap, it is called a potentiometer, or pot. Find Leq. This is similar to the case of current division in a two-branch parallel circuit where the other resistor has been replaced by R 0.

The loads are located at 4, 7, and 10 km from the generator and draw 50, 20, and A, respectively. The resistance of the line is 0. Find i and vAC. A zero voltage source corresponds to a short-circuited element and a zero current source corresponds to an open-circuited element. Find a the value of Rx and b the power delivered by the voltage source. Note that v changes proportionally with b. Compare with corresponding values obtained in Problem 3.

This produces a set of simultaneous equations which can be solved to obtain the currents. Other directions may be chosen for the branch currents and the answers will simply include the appropriate sign. It also results in more independent equations than either the mesh current or node voltage method requires.

They are sometimes referred to as loop currents. Each element and branch therefore will have an independent current. When a branch has two of the mesh currents, the actual current is given by their algebraic sum. The assigned mesh currents may have either clockwise or counterclockwise directions, although at the outset it is wise to assign to all of the mesh currents a clockwise direction.

In Example 4. The currents do not have to be restricted to the windows in order to result in a valid set of simultaneous equations, although that is the usual case with the mesh current method. For example, see Problem 4. In that problem they are called loop currents. Refer to Appendix B for an introduction to matrices and determinants.

Similarly, elements R22 and R33 are the sums of all resistances through which I2 and I3 , respectively, pass. Element R12 row 1, column 2 is the sum of all resistances through which mesh currents I1 and I2 pass.

Similarly, elements R21 , R23 , R13 , and R31 are the sums of the resistances common to the two mesh currents indicated by the subscripts, with the signs determined as described previously for R As a result, the resistance matrix is symmetric about the principal diagonal. The current matrix requires no explanation, since the elements are in a single column with subscripts 1, 2, 3,.

These are the unknowns in the mesh current method of network analysis. Element V1 in the voltage matrix is the sum of all source voltages driving mesh current I1. In other words, a voltage is positive if the source drives in the direction of the mesh current. The matrix equation arising from the mesh current method may be solved by various techniques. The unknown current I1 is obtained as the ratio of two determinants. The denominator determinant has the elements of resistance matrix. In the node voltage method, one of the principal nodes is selected as the reference and equations based on KCL are written at the other principal nodes.

At each of these other principal nodes, a voltage is assigned, where it is understood that this is a voltage with respect to the reference node. These voltages are the unknowns and, when determined by a suitable method, result in the network solution. KCL requires that the total current out of node 1 be zero: Indeed, the current in branch 1—2 is necessarily directed out of one node and into the other. The 1,1-element contains the sum of the reciprocals of all resistances connected to note 1; the 2,2-element contains the sum of the reciprocals of all resistances connected to node 2.

The 1,2- and 2,1-elements are each equal to the negative of the sum of the reciprocals of the resistances of all branches joining nodes 1 and 2. There is just one such branch in the present circuit. Both these terms are taken positive because they both drive a current into a node. Further discussion of the elements in the matrix representation of the node voltage equations is given in Chapter 9, where the networks are treated in the sinusoidal steady state. The circuit is redrawn in Fig.

With two principal nodes, only one equation is required. Current I3 in Example 4. Such a network is suggested in Fig. Since the only source is V1 , the equation for I1 is [see 7 of Example 4.

A voltage source applied to a passive network results in voltages between all nodes of the network. An external resistor connected between two nodes will draw current from the network and in general will reduce the voltage between those nodes.

The output resistance is found by dividing the open-circuited voltage to the shortcircuited current at the desired node. The short-circuited current is found in Section 4. For example, a voltage source applied to a passive network results in an output current in that part of the network where a load resistance has been connected.

In such a case the network has an overall transfer resistance. Consider the passive network suggested in Fig. The mesh current equation for Is contains only one term, the one resulting from Vr in the numerator determinant: If a certain voltage in mesh r gives rise to a certain current in mesh s, then the same voltage in mesh s produces the same current in mesh r.

Consider now the more general situation of an n-mesh network containing a number of voltage sources. The solution for the current in mesh k can be rewritten in terms of input and transfer resistances [refer to 7 , 8 , and 9 of Example 4. A source far removed from mesh k will have a high transfer resistance into that mesh and will therefore contribute very little to Ik.

Source Vk , and others in meshes adjacent to mesh k, will provide the greater part of Ik. However, the equivalent resistance of series and parallel branches Sections 3. This method is tedious and usually requires the drawing of several additional circuits. Even so, the process of reducing CHAP.

The reduction begins with a scan of the network to pick out series and parallel combinations of resistors. This principle applies because of the linear relationship between current and voltage. With dependent sources, superposition can be used only when the control functions are external to the network containing the sources, so that the controls are unchanged as the sources act one at a time.

Voltage sources to be suppressed while a single source acts are replaced by short circuits; current sources are replaced by open circuits. Superposition cannot be directly applied to the computation of power, because power in an element is proportional to the square of the current or the square of the voltage, which is nonlinear.

As a further illustration of superposition consider equation 7 of Example 4. Note that the three terms on the right are added to result in current I1. If there are sources in each of the three meshes, then each term contributes to the current I1. Additionally, if only mesh 3 contains a source, V1 and V2 will be zero and I1 is fully determined by the third term. With the V source acting alone, the A current source is replaced by an open circuit, Fig.

The two resistances are the same, R 0. When terminals ab in Fig. If a short circuit is applied to the terminals, as suggested by the dashed line in Fig. From Fig. Now, if the circuits in b and c are equivalents of the same active network, they are equivalent to each other. In the present case, V 0 , R 0 , and I 0 were obtained independently. This is suggested in CHAP.

If this were attempted in the original circuit using, for example, network reduction, the task would be very tedious and time-consuming. Assuming that the network is linear, it can be reduced to an equivalent circuit as in Fig.

Solved Problems 4. With two principal nodes, only one equation is necessary. The elements in the matrix form of the equations are obtained by inspection, following the rules of Section 4. The circuit has been redrawn in Fig. By KCL, the net current out of node 1 must equal zero. The node voltage method will be used and the matrix form of the equations written by inspection.

The short-circuit current Is: Loop currents are chosen such that each source contains only one current. The circuit is redrawn with series resistors added [Fig. Obtain current I1 by expanding the numerator determinant about the column containing the voltage sources to show that each source supplies a current of 2.

Check the result by network reduction. Consider terminal a positive with respect to b. Consider a positive with respect to b. In the 2. If the source supplies W to the circuit, Ans: When delivering its rated current of 40 A, the terminal voltage drops to V. In the circuit of Fig. Node C: Contribution of the voltage source: Loop 3: Move node B in Fig. Use the linearity and superposition properties, along with the results of Problems 4. The input and output reference terminals are often connected together and form a common reference node.

For a better operation it is desired that Ri be high and Ro be low. Deviations from the above conditions can reduce the overall gain. Example 5. Thus, a tenfold increase in k produces only a 5. The common reference for inputs, output, and power supplies resides outside the op amp and is called the ground Fig.

The open-loop gain A is generally very high. In practice, Ri is large, Ro is small, and A ranges from to several millions. The model of Fig. Vcc is generally from 5 to 18 V. Find and sketch the open-loop output vo. Assume Fig. Therefore, the ideal op amp draws zero current at its inverting and noninverting inputs, and if it is not saturated these inputs are at the same voltage.

The noninverting terminal of the op amp is grounded see Fig. This circuit, called a summing circuit, is an extension of the inverting circuit. The input lines are set either at 0 or 1 V. Find vo in terms of v4 , v3 , v2 , v1 , given the following sets of inputs: With the inputs at 0 V low or 1 V high , the circuit converts the binary number represented by the input set fv4 ; v3 ; v2 ; v1 g to a negative voltage which, when measured in V, is equal to the base 10 representation of the input set.

The circuit is a digital-to-analog converter. The inverting terminal is connected to the output through R2 and also to the ground through R1 see Fig. The output v2 follows the input v1. Therefore, vs reaches the load with no reduction caused by the load current.

## Solved Problems in Electric Circuits Schaums

The current in Rl is supplied by the op amp. Such a signal may be Fig. Node B: Apply KCL to the currents leaving node B. The current is drawn from vs is, therefore, zero. For further discussion see Section 5. Find v2. The inverting node is at zero voltage, and the sum of currents arriving at it is zero.

To obtain the input-output relationship, apply KCL to currents arriving at the inverting node: Use the summer-integrator op amp 1 in Fig. Supply inputs to op amp 1 through the following connections. The complete circuit is shown in Fig. Following the steps used in Example 5. The plot of gain versus frequency is called a frequency response. The leaky integrator of Fig.

Find jv2 j for! By repeating the procedure of Example 5. Table Frequency Response of the Low-pass Filter! However, in their present form they are not the binary numbers representing input amplitudes.

Yet, by using a coder we could transform the above sequences into the binary numbers corresponding to the values of analog inputs. Table 5. Compare the results. From 5 in Example 5.

The circuit of Fig. However, the circuit of Fig. Table k 1 10 1 5. Discuss the effects of Ri and k on the overall gain. Note that 38 is identical with Then use 5 to derive Refer to Figs. Find v1 and v2.

Find the sequence of outputs corresponding to values of vi from 0 to 1 V in steps of 0. Show that vC , i1 , v2 in Problem 5. The current source i1 delivers no power as the voltage vAB across it is zero. It is desired that v1 follow the signal with a constant gain of regardless of the value of Rl.

Design a current-to-voltage converter to accomplish this task. The transducer should feed Rl indirectly through an op amp. Design 1: However, a resistor of such a large magnitude is expensive and Design 2: The design of Fig.

## Schaum's Outline of Electric Circuits, Sixth Edition

Find il and compare with part a. The op amp circuit converts the practical current source to an ideal current source. See Figure c. Using results of Problem 5. The two frontal op amps are voltage followers. However, it employs two small and large resistors which are rather out of ordinary range. Show that in the circuit of Fig. The current also goes through the two R3 resistors, creating voltage drops iR3 across them. The op amp circuit is a negative impedance converter.

Supplementary Problems 5. Find the maximum value of R2 before the op amp is saturated. Write KCL at node B and solve for v2. Design an op amp circuit to obtain v2 from vs. However, as will be seen in Chapter 17, they may be represented by a sum of sinusoids.

This type of function will be developed in the following sections. The angular velocity! Since cos! In summary, for sinusoidal functions we have! Therefore, for a given phase shift the higher is the frequency, the smaller is the required time shift. Thus a phase shift proportional to! The output follows the input with no distortion. The output is a distorted form of the input. Otherwise, the sum is not a periodic function. Trigonometric Identities The trigonometric identities in Table are useful in the study of circuit analysis. Find Vdc such that the average power during the period remains the same.

Every 5 ms the above amount is added to the capacitor voltage. Examples are discussed in the following sections. Express vAB using the step function. The magnitude A is sometimes called the strength of the impulse. Find the limit of integral I in 33 when T approaches zero. S is the area of the resulting trapezoid.

It decays with time if the real part of s is negative and grows if the real part of s is positive. We will discuss exponentials eat in which the constant a is a real number. This observation provides a convenient approximate approach to plotting the exponential function as described in Example 6. Draw the tangent line AB. Find the current i in the capacitor. Graphs of v and i are shown in Figs. Sketch i as a function of time. For example, the values of a sinusoidal waveform, such as the line voltage, can be determined for all times if its amplitude, frequency, and phase are known.

Such signals are called deterministic. These are called random signals. Random signals can carry information and should not be mistaken with noise, which normally corrupts the information contents of the signal. The voltage recorded at the terminals of a microphone due to speech utterance and the signals picked up by an antenna tuned to a radio or TV station are examples of random signals.

The future course and values of such signals can be predicted only in average and not precisely. Other examples of random signals are the binary waveforms in digital computers, image intensities over the area of a picture, and the speech or music which modulates the amplitude of carrier waves in an AM system. It can change its sign at 1-ms intervals. The sign change is not known a priori, but it has an equal chance for positive or negative values. Therefore, if measured for a long time, it spends an equal amount of time at the 0.

During the s period, there are 10, intervals, each of 1-ms duration, which on average are equally divided between the 0. Solved Problems 6. Specify the phase angle in degrees. If not, determine the additional needed information.

Find its period and frequency. Express it in the form Fig. The signal is a cosine function with amplitude B added to a constant value A. Find its period, Vmax , and the times when v attains its maximum value. Find and plot the voltages across R, L, and RL. The plot of the resistor voltage vR has the same shape as that of the current [see Fig.

In this example the average power and the peak power are 25 W and W, respectively. Find the frequency, and the average, maximum, and minimum values of p.

The capacitor is initially uncharged. Find the voltage across the capacitor. Find the voltage across the parallel RC combination. During the pulse, iR remains negligible because v cannot exceed 1 V and iR remains under 1 mA. Draw the curve as shown. Figures b — e show the plots of v, iC , iR , and i, respectively, for the given data. The resistor current needed to sustain the exponential voltage across it is supplied by the capacitor.

Find the voltage across RL. Therefore, v1 is not periodic. This period is called the transient. After the transient has passed, the circuit is said to be in the steady state. The complementary function corresponds to the transient, and the particular solution to the steady state.

We will also present and solve important issues relating to natural, force, step, and impulse responses, along with the dc steady state and the switching behavior of inductors and capacitors. When a conducting path R is provided, the stored charge travels through the capacitor from one plate to the other, establishing a current i. Thus, the capacitor voltage v is gradually reduced to zero, at which time the current also becomes zero.

In the RC circuit of Fig. See Figs. The circuit is shown in Fig. Following the analysis of Example 7. The inductor current is then constant and its voltage is zero.

Using the results of Example 7. The solution follows. The current in the inductor and the voltage across it are given by 9 and 10 and plotted in Fig. For the RC circuit of Section 7. It may also be said that the function has undergone Thus, referring to Fig. This reduces the complex circuit to a simple RC or RL circuit which may be solved according to the methods described in the previous sections.

The initial value of the inductor current is zero. At this time, capacitor voltage Fig. After passage of 10 time constants the transient component equals to 0. At the dc steady state of RLC circuits, assuming no sustained oscillations exist in the circuit, all currents and voltages in the circuit are constants.

When the voltage across a capacitor is constant, the current through it is zero. All capacitors, therefore, appear as open circuits in the dc steady state. Similarly, when the current through an inductor is constant, the voltage across it is zero. All inductors therefore appear as short circuits in the dc steady state. The dc steady-state behavior presented in the preceding paragraph is valid for circuits containing any number of inductors, capacitors, and dc sources.

When the steady state is reached, the circuit will be as shown in Fig. By inspection of Fig. The time constant of the circuit of Fig. A jump in the capacitor voltage requires an impulse current. Similarly, a jump in the inductor current requires an impulse voltage.

If no such impulses can be present, the capacitor voltages and the inductor currents remain continuous. Therefore, the post-switching conditions of L and C can be derived from their pre-switching conditions. Find i and v for all times. The derivation applies to RC or RL circuits where the input can be a current or a voltage.

As an example, we use the series RC circuit in Fig. For convenience, time will be expressed in ms, voltages in V, and currents in mA. For 0 0: Impulse response is a useful tool in analysis and synthesis of circuits.

## 3000 Solved Problems in Electric Circuits Schaums

It may be derived in several ways: A unit impulse may be considered the derivative of a unit step. Thus CHAP. The responses of the circuit to a step of amplitude 9 were already found in Example 7. Some of the entries in this table have been derived in the previous sections. The remaining entries will be derived in the solved problems. Following an argument similar to that of Section 7. Several examples were given in the previous sections. By weighted linear combination of the entries in Table and their time delay, the forced response to new functions may be deduced. The usual methods of analysis are then applied to the circuit as illustrated in the following examples.

The op amp in the circuit of Fig. Find its unit-step response and compare with the answer in Example 7. Let vA be the voltage of the node connecting R1 and C.

Find the relationship between v2 and v1 in the circuit of Fig. Let node D be the reference node. KCL at node B: Show that the relationship between v2 and v1 in the circuit of Fig. Obtain the current and Fig. The switch in the RL circuit shown in Fig. Obtain vR and vL with polarities as indicated. And, since CHAP. Thus, as in Problem 7.

Consequently, as t! Then, by Sections 6. Using the two-point method of Section The voltage across the inductance is CHAP. Hence, as in Problem 7. Obtain the expression for the transient voltage vR.

The two parallel capacitors have an equivalent capacitance of 3 mF. Then this capacitance is in series with the 6 mF, so that the overall equivalent capacitance is 2 mF. Obtain the current i and capacitor voltage vC , for Fig.

Furthermore, as t! Obtain the currents i1 CHAP. The complementary function is the general solution of 27 when the right-hand side is replaced by zero: The capacitor voltage is continuous. Calculate the time in which the transient voltage across the resistor drops from 40 to 10 volts. Find the time at which the voltage across the resistor is zero, reversing polarity. Find t 0 such that the current is constant at 0.

Obtain the current for a 0 0: Assuming that RC 7. In this chapter, several examples of second-order circuits will be presented. This will then be followed by more direct methods of analysis, including complex frequency and pole-zero plots.

In order to visualize the three possibilities, a second-order mechanical system is shown in Fig. The mass M is suspended by a spring with a constant k. A damping device D is attached to the mass M. Obtain the current transient. As in Example 8. Once again the polarity is a matter of the choice of direction for the current with respect to the polarity of the initial voltage on the capacitor. The reader is encouraged to examine the results, selecting several values for t, and comparing the currents.

However, a step function voltage applied to the circuit will initiate the same transient response. Here it is the frequency of the damped oscillation. It is referred to as the damped radian frequency. In fact, it is merely a curiosity, since it is a set of circuit constants whose response, while damped, is on the verge of oscillation. The following result is a second-order equation for i1 , of the types treated in Sections 8. Once the expression for i1 is known, that for i2 follows from 4.

There will be a damping factor that insures the transient will ultimately die out. Also, depending on the values of the four circuit constants, the transient can be overdamped or underdamped, which is oscillatory. We begin by expressing the exponential function in the equivalent cosine and sine form: In Table , several functions are given with corresponding values of s for the expression Aest. Consequently, only the magnitudes of currents and voltages and the phase angles need be determined this will also be the case in sinusoidal circuit analysis in Chapter 9.

We are thus led to consider the network in the s-domain see Fig. Obtain the current i by an s-domain analysis. Obtain the current by an s-domain analysis. Therefore, only the magnitude I and phase angle need be determined. Consequently, a voltage function at this frequency results in a current of zero. Figure shows the poles and zeros of Example 8. Draw the vectors from the poles and zeros to the test point and compute the lengths and angles see Fig.

See Problem 8. However, the natural frequencies, which characterize the transient response, are easily obtained. They are the poles of the network function. Obtain the natural response when a Fig. We say that the network has been magnitude-scaled by a factor Km. That is, the new network has the same impedance at complex frequency Kf s as the old had at s. We say that the network has been frequency-scaled by a factor Kf. A convenient tool for developing the equations is the complex frequency s and generalized impedance in the s-domain as used throughout Sections 8.

Again, we assume ideal op amps see Section 7. The method is illustrated in the following examples. At terminal A: At terminal B: In Example 8. Solved Problems 8. Obtain the current transient, assuming zero initial charge on the capacitor. Everything remains the same as in Problem 8. Obtain the expression for the voltage across the network. See columns 2 and 3 of the table.

See column 3 of the table. At very high frequencies the capacitance acts like a short circuit across the RL branch. The response to the excitation is the input current. The conductor at yy 0 in Fig.

Obtain the magnitude-scaling factor Km and the element values which will limit the current to 89 mA maximum value. The scaled element values are as follows: Scale the network with Fig. Before scaling, 1! Any other value of the impedance is likewise attained, after scaling, at a frequency times that at which it was attained before scaling. 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.

## Schaum's Outline of Theory and Problems of Electric Circuits

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.

All of the loads are parallel-connected, and the equivalent impedance results in a lagging current and a corresponding inductive quadrature power Q. Schaum's is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easy-to-follow, topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills.

This Schaum's Outline gives you fully solved problems, extra practice on topics such as amplifiers and operational amplifier circuits, waveforms and signals, AC power, etc, and support for all the major textbooks for electric circuits courses. Edminister Abstract: Full details.

Table of Contents A. 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