You may use the work for your own non-commercial and personal use; any other use of the work is strictly prohibited. McGraw-Hill has no responsibility for the content of any information obtained through the work.
The Laplace Transform Method 398
Fourier Method of Waveform Analysis 420
ELECTRICAL QUANTITIES AND SI UNITS
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). Electrical quantities and their symbols commonly used in electrical circuit analysis are listed in Table 1-2.
FORCE, WORK, AND POWER
ELECTRIC CHARGE AND CURRENT
Of greater importance in the analysis of electrical circuits is the current in metal conductors that occurs due to the movement of electrons occupying the outermost shell of the atomic structure. The electron charge is e¼ 1:6021019C, so that for a current of one ampere about 6:241018 electrons per second would have to pass through a fixed cross-section of the conductor.
ELECTRIC POTENTIAL
In copper, for example, the one electron in the outermost shell is only loosely bound to the central nucleus and moves freely from one atom to the next in the crystal structure. In an electrical circuit, an energy of 9.25mJ is required to transport 0.5mC from pointato pointb.
ENERGY AND ELECTRICAL POWER
CONSTANT AND VARIABLE FUNCTIONS
The ampere-hour rating is a measure of the energy the battery stores; consequently, the energy transferred for total discharge is the same whether it is transferred within 10 hours or 20 hours. Since power is the rate of energy transfer, the power for a 10-hour discharge is twice that in a 20-hour discharge.
PASSIVE AND ACTIVE ELEMENTS
SIGN CONVENTIONS
VOLTAGE-CURRENT RELATIONS
RESISTANCE
INDUCTANCE
CAPACITANCE
CIRCUIT DIAGRAMS
NONLINEAR RESISTORS
Find the total energy delivered to the capacitor and verify that the energy delivered equals the energy stored in the capacitor. Find the released energy W and the capacitance of the cloud to ground C if the voltage before the discharge is (a) 100 MV; (b) 500 MV.
INTRODUCTION
KIRCHHOFF’S VOLTAGE LAW
KIRCHHOFF’S CURRENT LAW
CIRCUIT ELEMENTS IN SERIES
Starting from the lower left corner of the circuit, for the current direction as shown, we have vaþv1þvbþv2þv3¼0. Note: When two capacitors in series differ by a large amount, the equivalent capacitance is essentially equal to the value of the smaller of the two.
CIRCUIT ELEMENTS IN PARALLEL
The equivalent resistance of two resistors in parallel is given by the product over the sum. For several parallel capacitors,Ceq¼C1þC2þ , which are of the same shape as resistors in series.
VOLTAGE DIVISION
CURRENT DIVISION
Since this is in parallel with the 10-mH inductance, the total equivalent inductance is 5 mH. 3.6 Express the total capacitance of the three capacitors in Fig. For C2 and C3 in parallel,Ceq¼C2þC3. The resistor Rin in parallel with the 250 resistor has a corresponding resistance Req¼ 250ð106Þ. 3.8 Find all branch currents in the network shown in fig.
THE BRANCH CURRENT METHOD
THE MESH CURRENT METHOD
MATRICES AND DETERMINANTS
A voltage is counted positive in the sum if I1 passes from the þ terminal of the source; otherwise it is considered negative. In other words, a voltage is positive if the source moves downstream of the grid.
THE NODE VOLTAGE METHOD
Further discussion of the elements in the matrix representation of the node voltage equations is given in Chapter 9, where sinusoidal steady-state networks are considered. Assuming that all flows are directed from the top node and the bottom node is the reference.
INPUT AND OUTPUT RESISTANCES
The 1,1 element contains the sum of the reciprocal values of all resistors connected to node1; 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 in all branches that connects nodes 1 and 2. The output resistance is found by dividing the open circuit voltage by the short circuit current at the desired node.
TRANSFER RESISTANCE
NETWORK REDUCTION
The reduction begins with a scan of the network to select series and parallel combinations of resistors. EXAMPLE 4.6 Obtain the total power delivered by the 60-V source and the power absorbed in each resistor in the network of Fig.
SUPERPOSITION
THE´VENIN’S AND NORTON’S THEOREMS
The utility of The'venin and Norton equivalent circuits is evident when an active network is to be examined under a number of load conditions, each represented by a resistance. If this were attempted in the original circuit using e.g. network reduction, the task would be very tedious and time consuming.
MAXIMUM POWER TRANSFER THEOREM
Apply KVL around the mesh on the upper part of the circuit to find current I coming from the voltage source, then find VA and VC. 4-48 removes the 2-A current source and sets the other two sources to zero, reducing the circuit to a source-free resistive circuit.
AMPLIFIER MODEL
EXAMPLE 5.1 A practical voltage source with an internal resistance Rsis connected to the input of a voltage amplifier with input resistance Rias in Fig. in charge.
FEEDBACK IN AMPLIFIER CIRCUITS
The open loop gain is reduced by the factor Ri=ðRiþRsÞ. 5-3 a practical voltage source VS with internal resistance Rs feeds a load R1 through an amplifier with input and output resistances RiandRo, respectively. v1¼ Ri RiþRsvs Likewise is the output voltage. -4 and express it as a function of the ratiob¼R1=ðR1þR2Þ. We know that from the amplifier. a) Findv2=vs as a function of the open loop gain k. a) Figures 5-4 and 5-5 differ only in the polarity of the dependent voltage source.
OPERATIONAL AMPLIFIERS
The common reference for the input, output and power supplies lies outside the operational amplifier and is called the ground (Fig. 5-6).
ANALYSIS OF CIRCUITS CONTAINING IDEAL OP AMPS
INVERTING CIRCUIT
SUMMING CIRCUIT
With the inputs at 0 V (low) or 1 V (high), the circuit converts the binary number represented by the input string fv4;v3;v2;v1g to a negative voltage that, measured in V, is equal to the base 10 representation of the input string.
NONINVERTING CIRCUIT
The input resistance of the circuit is infinite because the op amp draws no current.
VOLTAGE FOLLOWER
DIFFERENTIAL AND DIFFERENCE AMPLIFIERS
CIRCUITS CONTAINING SEVERAL OP AMPS
The current in the 5-k input resistor of the first op-amp provides the output of the second op-amp through the 40-k feedback resistor.
INTEGRATOR AND DIFFERENTIATOR CIRCUITS Integrator
The desired initial state, vo, of the integration can be provided by a reset switch as shown in Fig. By placing an inductor instead of a resistor in the feedback path of an inverting amplifier, the derivative of the input signal is produced at the output.
ANALOG COMPUTERS
LOW-PASS FILTER
COMPARATOR
Replace the circuit to the left of the node including vs, R1 and Ri with its The'venin equivalent. 5.16 (a) Find the The'venin equivalent of the circuit to the left of the A-Bin nodes Fig. a) Thevenin equivalent of the circuit in Fig.
INTRODUCTION
PERIODIC FUNCTIONS
However, as will be seen in Chapter 17, they can be represented by a sum of sinusoids.
SINUSOIDAL FUNCTIONS A sinusoidal voltage vðtÞ is given by
TIME SHIFT AND PHASE SHIFT
COMBINATIONS OF PERIODIC FUNCTIONS
THE AVERAGE AND EFFECTIVE (RMS) VALUES
Þand (b) the value of a constant current source Idc which can produce the same voltage across the above capacitor att¼5kms when applied att>0. CompareIdcwithhiðtÞi, the mean vaniðtÞin Fig.
NONPERIODIC FUNCTIONS
THE UNIT STEP FUNCTION
THE UNIT IMPULSE FUNCTION
In all previous cases, the charge accumulated on the capacitor at the end of the transient period is Q¼.
THE EXPONENTIAL FUNCTION
EXAMPLE 6.21 (a) Show that the rate of change with respect to time of an exponential function v¼Aestis at any time is proportional to the value of the function at that time. Find the proportionality coefficient. a) The rate of change of a function is equal to the derivative of the function, which for the given exponential function is.
DAMPED SINUSOIDS
RANDOM SIGNALS
Determine its average and effective values for a period of 10 s. Therefore, the mean of vðtÞ can be approximated as. The equation is vðtÞ ¼Aet=2þB. 6.21 The voltage ¼V0eajtjfora>0 is connected through a parallel combination of a resistor and a capacitor as shown in Fig. a) Find the current C,iR, andi¼iCþiR.
INTRODUCTION
CAPACITOR DISCHARGE IN A RESISTOR
The voltage and current of the capacitor are exponential with initial values of V0 and V0=R respectively. EXAMPLE 7.2 A 5 mF capacitor with an initial voltage of 4 V is connected to a parallel combination of a 3- and a 6-k resistor (Fig. 7-2).
ESTABLISHING A DC VOLTAGE ACROSS A CAPACITOR
EXAMPLE 7.3 A 4-mF capacitor with an initial voltage of vð0Þ ¼2 V is connected to a 12-V battery through a resistor R¼5 katt¼0. And so the voltage increases exponentially from an initial value of 2 V to a final value of 12 V, with a time constant of 20 ms, as shown in Fig.
THE SOURCE-FREE RL CIRCUIT
After the battery is disconnected, at t>0, the circuit will be as in Fig.
ESTABLISHING A DC CURRENT IN AN INDUCTOR
THE EXPONENTIAL FUNCTION REVISITED
The tangent to the exponential curve att¼0þ can be used to estimate the time constant. In this case, any pair of data points, perhaps read from the instruments, can be used to find the transient equation.
COMPLEX FIRST-ORDER RL AND RC CIRCUITS
The capacitor discharges its initial voltage V0 in an exponential form with a time constant¼RC¼349×103××9106×¼0:034 s.
DC STEADY STATE IN INDUCTORS AND CAPACITORS
TRANSITIONS AT SWITCHING TIME
RESPONSE OF FIRST-ORDER CIRCUITS TO A PULSE
IMPULSE RESPONSE OF RC AND RL CIRCUITS
By taking their derivatives and scaling them down by 1/9, we find the unit impulse responses.
SUMMARY OF STEP AND IMPULSE RESPONSES IN RC AND RL CIRCUITS Responses of RL and RC circuits to step and impulse inputs are summarized in Table 7-1. Some of
RESPONSE OF RC AND RL CIRCUITS TO SUDDEN EXPONENTIAL EXCITATIONS Consider the first-order differential equation which is derived from an RL combination in series with
The natural response ihðtÞ is the solution of RiþLðdi=dtÞ ¼0; i.e., the case with a zero forcing function. If the forcing function has the same exponent as that of the natural responseðs¼ R=LÞ, the forced response must beipðtÞ ¼I0teRt=L.
SUMMARY OF FORCED RESPONSE IN FIRST-ORDER CIRCUITS Consider the following differential equation
FIRST-ORDER ACTIVE CIRCUITS
The unit step response of the circuit is the time derivative of the preceding solution. When the rest of the circuit is viewed from the terminals of the inductance, there is equivalent resistance Req¼5þ5ð10Þ.
INTRODUCTION
SERIES RLC CIRCUIT
The signs of A1 and A2 are fixed by the polarity of the initial voltage on the capacitor and its relationship to the assumed positive direction of the current. Once again, the polarity is a matter of choosing the direction of the current relative to the polarity of the starting voltage on the capacitor.
PARALLEL RLC CIRCUIT
When < !0, s1 and s2 in the solution of the differential equation presented in the preceding is complex conjugation1 ¼þjands2¼j, which is now given by. The function0:667e1000t, dashed in the graph, provides an envelope within which the sine function is bounded.
0 A solution is of the form
TWO-MESH CIRCUIT
The solution of (7) in the state of dynamic equilibrium is obviously i1ð1Þ ¼V=R1; the transient solution will be determined by the roots s1 and s2. Also, depending on the values of the four circuit constants, the transient process can be over- or under-damped, which is oscillation.
COMPLEX FREQUENCY
If ¼0, there is no damping and the result is a cosine function with maximum values of Vm (not shown). A driving voltage of the form v¼Vmest applied to a passive network will result in branch currents and voltages across the elements, each having the same time dependence; e.g. Iaej est and Vbejest.
NETWORK FUNCTION AND POLE-ZERO PLOTS
In linear circuits consisting of lumped elements, the network function HðsÞis arational function of and can be written in the following general form. Þthe poles of HðsÞ, take on particular importance when HðsÞis is interpreted as the ratio of the response (in one part of this domain network) to the excitation (in another part of the network).
THE FORCED RESPONSE
In Chapter 12, where series and parallel resonance are discussed, it will be seen that the parallelLC circuit is resonant at !¼1= ffiffiffiffiffiffiffi. However, it is clear that the expression within the constant factor k can be written from the known poles and zeros of HðsÞ in the pole-zero graph.
THE NATURAL RESPONSE
The result means that in the time domain iðtÞ ¼0:248vðtÞ so that both voltage and current become infinite with respect to the function e1t. The poles are again at 2 Np/s and 6 Np/s, which is the same result as in Example 8.10.
MAGNITUDE AND FREQUENCY SCALING Magnitude Scaling
HIGHER-ORDER ACTIVE CIRCUITS
For sustained oscillations, the roots of the characteristic equation in Example 8.14 should be imaginary numbers. a) Get the current transient if the capacitor has no initial charge. First apply an excitation, IiðsÞ, and obtain the natural frequency from the denominator of the network function.
