Schaums outlines electric circuits 4th edition pdf download

Wicks, C. So depending on what exactly you are searching, you will be able to choose ebooks Joseph A. Access Fundamentals of Electric Circuits Edition solutions now. Actually, as a reader, you can get many lessons of life. Not Enough Time?

Textbook too Pricey? Calculate the value of a series resistance needed to measure 50 V of full scale. Everyday low prices and free delivery on eligible orders. 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. Joseph A Edminister Solutions. Below are Chegg supported textbooks by Joseph A Edminister. Select a textbook to see worked-out Solutions. This online book alsoprovides some example to explain the explanation clearly. Honda cr-v Honda crv service-manual. A 2-mF capacitor with zero initial charge is also connected in parallel with the inductor through an ideal diode such that the diode is reversed biased i.

Find the value of the inductor. The circuit will generally contain at least one voltage or current source. The arrangement of elements results in a new set of constraints between the currents and voltages. These new constraints and their corresponding equations, added to the current-voltage relationships of the individual elements, provide the solution of the network. The solution generally answers necessary questions about the operation of the device under conditions applied by a source of energy.

Some of the voltages will be sosurces, while others will result from current in passive elements creating a voltage, which is sometimes referred to as a voltage drop. The mesh current method of circuit analysis introduced in Section 4. Write the KVL equation for the circuit shown in Fig. The junction between two elements is called a simple node and no division of current results.

The junction of three or more elements is called a principal node, and here current division does take place. It may be stated alternatively that the sum of the currents entering a node is equal to the sum of the currents leaving that node.

The node voltage method of circuit analysis introduced in Section 4. The basis for the law is the conservation of electric charge. Write the KCL equation for the principal node shown in Fig. The voltages across the elements are v1 , v2 , and v3. The equivalent resistance of three resistors in series is Two of the resistors are What must be the ohmic resistance of the third resistor?

Repeat if C2 is Obtain the equivalent resistance of a two The concept extends beyond the set of resistors illustrated here and applies equally to impedances in series, as will be shown in Chapter 9. A voltage divider circuit of two resistors is designed with a total resistance of the two resistors equal to If the output voltage is 10 percent of the input voltage, obtain the values of the two resistors in the circuit. The ratio of the branch current i1 to the total current i illustrates the operation of the divider.

A current of Obtain the branch resistances. Solved Problems 3. However, KCL applies to the network taken as a single node.

This is in series with the resistor so that their sum is 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. 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. Hint: A zero voltage source corresponds to a short-circuited element and a zero current source corresponds to an open-circuited element.

Ans: 3. 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. 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. 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:c: is obtained from the three-mesh circuit shown in Fig.

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. When delivering its rated current of 40 A, the terminal voltage drops to V.

In the circuit of Fig. Move node B in Fig. Hint: 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. 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. The current in Rl is supplied by the op amp. Such a signal may be Fig. 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. 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! Several transient components are described below. Exponential Source The source starts at a constant initial value V0. After 15 ms, it starts decaying back to 1 V with a time constant of 2 ms. Write the data statement for the source and use Probe to plot the waveform. The pulse stays at 2 V for 11 ms. At t0 , the exponentially decaying sinusoidal component with frequency f , phase angle, starting amplitude V1 , and decay factor alpha is added to it.

The amplitude of the sine wave is 2 V and it decays to zero with a time constant of 10 ms. Sensitivity analysis is done using the. SENS statement. Fourier analysis is done using the. FOUR statement. These can be found in books or manuals for PSpice or Spice. The following summarizes the statements used in this chapter. AC hsweep typei hnumber of pointsi hstarting fi hending fi. DC hnamei hinitial valuei hfinal valuei hstep sizei.

LIB [hfile namei]. TF houtput variablei hinput sourcei. END Fig. We sweep Vs from 1 to 10 V. END equivalent of Fig. Find the complex magnitude of V2 for f varying from Hz to 10 kHz in 10 steps. We add to the netlist an. AC statement to sweep the frequency and obtain V 2 by any of the commands. PLOT, or. Find the complex magnitude of V2 for f varying from Hz to 10 kHz in steps.

END analysis of Fig. We model the resistor as a single-parameter resistor element with a single-parameter R and change the value of its parameter R from 2 to 10 in steps of 2. We use the. AC command to sweep the frequency from Hz to 3 kHz in steps. TRAN and. The values of the resistor R are changed by using. MODEL and. We also use.

Compared with the open-loop circuit of Fig. Pulse-Step Vs R C. During the transition period of 0 The current source is a 1 mA square pulse which lasts Model the resistor as a single-parameter resistor element with a single parameter R and change the value of R from 0. This is because pulse width is a multiple of the period of natural oscillations of the circuit.

The Laplace transform method described in this chapter may be viewed as generalizing the concept of the s-domain to a mathematical formulation which would include not only exponential excitations but also excitations of many other forms. This permits going back in the other direction, from the s-domain to the time Copyright , , , by The McGraw-Hill Companies, Inc. Here again, the integration need not actually be performed unless it is a question of adding to existing tables of transform pairs.

It should be remarked that taking the direct Laplace transform of a physical quantity introduces an extra time unit in the result. The exponential decay function, which appeared often in the transients of Chapter 7, is another time function which is readily transformed. The time-domain circuit is shown in Fig. This equation is transformed, term by term, into the s-domain equation ii. A circuit can be drawn in the s-domain, as shown in Fig. In this section, we extend the use of the complex frequency to transform an RLC circuit, containing sources and initial conditions, from the time domain to the s-domain.

The s-domain circuit is shown in Fig. More importantly, we introduce Laplace transform models of R, L, and C elements which, contrary to generalized impedances, incorporate initial conditions.

The input-output relationship is therefore derived directly in the transform domain. What is the relationship between the complex frequency and the Laplace transform models? A short answer is that the generalized impedance is the special case of the Laplace transform model i.

Apply this result to Problem Find the resulting current. Assume the direction of the current as shown in the diagram. Write the sdomain equations in matrix form and construct the corresponding circuit.

In the s-domain the 0. Referring to Fig. Find the resulting current i. Then, in the steady state, both capacitors are charged to 50 V and the current is zero. Supplementary Problems Find the resulting current using the Laplace transform method.

Find the current in the intervals 0 t 0. Find0 the current in the intervals 0 t 0. Find the resulting current transient if the charge is a of the 3same polarity as that deposited by the source, and b of the opposite polarity. Ans: mC, opposite polarity to that deposited by source Find the transient current in the intervals 0 t 0. Write the timedomain equations, transform them5 into the corresponding s-domain equations, and obtain the currents i1 5 and i2.

The switch is closed at an instant when the voltage is increasing at its maximum rate. Find the resulting mesh currents, with directions as shown in the diagram.

Two alternate forms of the trigonometric series result. Replace the sine and cosine terms in 1 by their complex exponential equivalents. Other waveforms will have only cosine terms; and sometimes only odd harmonics are present in the series, whether the series contains sine, cosine, or both types of terms.

Knowledge of such symmetry results in reduced calculations in determining the Fourier series. The sum or product of two or more even functions is an even function, and with the addition of a constant the even nature of the function is still preserved. They are symmetrical with respect to the vertical axis, as indicated by the construction in Fig.

The sum of two or more odd functions is an odd function, but the addition of a constant removes the odd nature of the function. The product of two odd functions is an even function.

The waveforms shown in Fig. They are symmetrical with respect to the origin, as indicated by the construction in Fig. Two waveforms with half-wave symmetry are shown in Fig. When the type of symmetry of a waveform is established, the following conclusions are reached. If the waveform is even, all terms of its Fourier series are cosine terms, including a constant if the waveform has a nonzero average value. If the waveform is odd, the series contains only sine terms. The wave may be odd only after its average value is subtracted, in which case its Fourier representation will simply contain that constant and a series of sine terms.