186 Inductance, Capacitance, and Mutual Inductance (Note that 5 V is the voltage on the capacitor at the end of the preceding interval.) Then, v = (10*7 - 12.5 X 10¥ - 10) V, p = t», = (62.5 X 10 12 / 3 - 7.5 x 10V + 2.5 X 10 5 / - 2) W, w = —Cv~, 2 0 V (V) t(fis) 12 A 9*3 ^),2 = (15.625 X 10 u r - 2.5 X l()r + 0.125 X 10 ( V For/ > 40^ts, - 2/ + 10~ 3 ) J. v = 10 V, p = vi = 0, w = —Cv 2 = 10/xJ. b) The excitation current and the resulting voltage, power, and energy are plotted in Fig. 6.12. c) Note that the power is always positive for the duration of the current pulse, which means that energy is continuously being stored in the capac- itor. When the current returns to zero, the stored energy is trapped because the ideal capacitor offers no means for dissipating energy. Thus a voltage remains on the capacitor after i returns to zero. 10 5 0 —n 10 1 20 l 30 i 40 1 50 I 60 t((XS) p(mw 500 400 300 200 100 0 ) ^1 10 1 20 1 30 40 1 50 J 60 tips) w( 10 8 6 4 2 0 nJ) 10 i 20 I 30 1 40 l 50 I 60 Figure 6.12 • The variables /', v, p, and to versus Example 6.5. t(fJ!S) t for ^/ASSESSMENT PROBLEMS Objective 2—Know and be able to use the equations for voltage, current, power, and energy in a capacitor 6.2 The voltage at the terminals of the 0.6 /xF capacitor shown in the figure is 0 for t < 0 and 4Qe -ism)t sin 30,000/ V for t > 0. Find (a) /'(0); (b) the power delivered to the capacitor at / = 77-/8O ms; and (c) the energy stored in the capacitor at t = TT/80 ms. 0.6 ^F NOTE: Also try Chapter Problems 6.16 and 6.17. Answer: (a) 0.72 A; (b) -649.2 mW; (c) 126.13 AtJ. 6.3 The current in the capacitor of Assessment Problem 6.2 is 0 for t < 0 and 3 cos 50,000/ A for t a 0. Find (a) v{t)\ (b) the maximum power delivered to the capacitor at any one instant of time; and (c) the maximum energy stored in the capacitor at any one instant of time. Answer: (a) 100 sin 50,000/ V, / > 0; (b)150W;(c)3mJ. 6.3 Series-Parallel Combinations of Inductance and Capacitance 187 6.3 Series-Parallel Combinations of Inductance and Capacitance Just as series-parallel combinations of resistors can be reduced to a single equivalent resistor, series-parallel combinations of inductors or capacitors can be reduced to a single inductor or capacitor. Figure 6.13 shows induc- tors in series. Here, the inductors are forced to carry the same current; thus we define only one current for the series combination. The voltage drops across the individual inductors are Figure 6.13 A Inductors in series. di di di „ = £,-, v 2 = L 1Jt , and v } = L, The voltage across the series connection is v = v\ + v 2 + v 3 = (L { + L 2 + L?)— at from which it should be apparent that the equivalent inductance of series- connected inductors is the sum of the individual inductances. For n induc- tors in series, L eQ = Li + L 2 + L 3 + ••• + L„. (6.19) < Combining inductors in series If the original inductors carry an initial current of i(f 0 ), the equivalent inductor carries the same initial current. Figure 6.14 shows the equivalent circuit for series inductors carrying an initial current. Inductors in parallel have the same terminal voltage. In the equivalent circuit, the current in each inductor is a function of the terminal voltage and the initial current in the inductor. For the three inductors in parallel shown in Fig. 6.15, the currents for the individual inductors are L, Ls v *0b) •^cu — L\+ L 2 + LT, i\=— I v dr + /JOO), '2 v dr + /2(?oX '2 At Figure 6.14 A An equivalent circuit for inductors in series carrying an initial current i(t () ). h -~T I vdT + i 3 (t {) ). ^3 At (6.20) v LiiliM L 2 i\i 2 (t n ) ^Jj'aft)) The current at the terminals of the three parallel inductors is the sum of Fi 9"re 6.15 A Three inductors in parallel, the inductor currents: i = ii + /'2 + /3. Substituting Eq. 6.20 into Eq. 6.21 yields (6.21) I — + — + — I I 1 : ; 188 Inductance, Capacitance, and Mutual Inductance Now we can interpret Eq. 6.22 in terms of a single inductor; that is, 1 f i = — I vdr + i(t {) ). (6.23) ^eq J la Comparing Eq. 6.23 with (6.22) yields 1111 — = - + — + — (6.24) ^eq -^1 *->2 JUT, ^c q L x L 2 L 3 i( tl) ) = i&o) + i 2 (t {) ) + i 3 (( () ). (6.25) *(%)! 3 ^cq Figure 6.16 shows the equivalent circuit for the three parallel inductors in Fig. 6.15. The results expressed in Eqs. 6.24 and 6.25 can be extended to Figure 6.16 • An equivalent circuit for three inductors n inductors m parallel: in parallel. Ill 1 Combining inductors in parallel • = — + — + + — (6.26) L eq L\ L 2 L n Equivalent inductance initial current • /(r 0 ) = /^) + j 2 (t 0 ) + • • + i n (t 0 ). (6.27) Capacitors connected in series can be reduced to a single equivalent capacitor. The reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. If each capacitor carries its own initial voltage, the initial voltage on the equivalent capacitor is the algebraic sum of the initial voltages on the individual capacitors. Figure 6.17 and the following equations summarize these observations: 111 1 Combining capacitors in series • = 1 h • • • -\ , (6.28) Equivalent capacitance initial voltage • v(t 0 ) = vi(t 0 ) + v 2 (t Q ) + • • • + v n (t 0 ). (6.29) We leave the derivation of the equivalent circuit for series-connected capacitors as an exercise. (See Problem 6.32.) The equivalent capacitance of capacitors connected in parallel is sim- ply the sum of the capacitances of the individual capacitors, as Fig. 6.18 and the following equation show: Combining capacitors in parallel • C eq = C { + C 2 + • •• + C n . (6.30) Capacitors connected in parallel must carry the same voltage. Therefore, if there is an initial voltage across the original parallel capacitors, this same initial voltage appears across the equivalent capacitance C eq . The deriva- tion of the equivalent circuit for parallel capacitors is left as an exercise. (See Problem 6.33.) We say more about series-parallel equivalent circuits of inductors and capacitors in Chapter 7, where we interpret results based on their use. 6.4 Mutual Inductance 189 + v i Ci? C 2 T- Cj (a) + ^Wi(fe) + =: «20») + ~v„(ta) + V 1 c c d Wq y + ^v(t u ) 1 + L 4- + «i('o) + 1¾¾)) + (b) + V e —^- c,;fc o 4) c„ (a) C + «„('<)) c, cq c, + c? +- + c„ Figure 6.17 • An equivalent circuit for capacitors connected in series, (a) The series capacitors, (b) The equivalent circuit. (b) Figure 6.18 • An equivalent circuit for capacitors connected in parallel, (a) Capacitors in parallel, (b) The equivalent circuit. /ASSESSMENT PROBLEMS Objective 3—Be able to combine inductors or capacitors in series and in parallel to form a single equivalent inductor 6.4 The initial values of i[ and / 2 in the circuit shown are + 3 A and -5 A, respectively. The voltage at the terminals of the parallel induc- tors for t > 0 is -30e" 5 ' mV. a) If the parallel inductors are replaced by a single inductor, what is its inductance? b) What is the initial current and its reference direction in the equivalent inductor? c) Use the equivalent inductor to find /(f). d) Find / t (f) and / 2 (f). Verify that the solutions for *']_(*), / 2 (f), and /(f) satisfy Kirchhoff s current law. I'W Answer: (a) 48 mH; (b) 2 A, up; (c) 0.125e~ 5 ' - 2.125 A, f > 0; (d)/!(f) = O.le -5 ' + 2.9 A, t > 0, kit) = 0.025e~ 5 ' - 5.025 A, f ; 0. 6.5 The current at the terminals of the two capaci- tors shown is 240e~ 1() '/xA for f > 0. The initial values of v^ and v 2 are -10 V and -5 V, respectively. Calculate the total energy trapped in the capacitors as f —* oo. (Hint: Don't com- bine the capacitors in series—find the energy trapped in each, and then add.) + v, 2/x¥ /i(0H60mH / 2 (0H240mH Answer: 20 /xJ. NOTE: Also try Chapter Problems 6.21, 6.25, 6.26, and 6.31. 6.4 Mutual Inductance The magnetic field we considered in our study of inductors in Section 6.1 was restricted to a single circuit. We said that inductance is the parameter that relates a voltage to a time-varying current in the same circuit; thus, inductance is more precisely referred to as self-inductance. We now consider the situation in which two circuits are linked by a magnetic field. In this case, the voltage induced in the second circuit can be related to the time-varying current in the first circuit by a parameter 190 Inductance, Capacitance, and Mutual Inductance Figure 6.19 • Two magnetically coupled coils. Figure 6.20 • Coil currents i { and i 2 used to describe the circuit shown in Fig. 6.19. Figure 6.21 A The circuit of Fig. 6.20 with dots added to the coils indicating the polarity of the mutually induced voltages. known as mutual inductance. The circuit shown in Fig. 6.19 represents two magnetically coupled coils. The self-inductances of the two coils arc labeled L] and L 2 , and the mutual inductance is labeled M. The double- headed arrow adjacent to M indicates the pair of coils with this value of mutual inductance.This notation is needed particularly in circuits contain- ing more than one pair of magnetically coupled coils. The easiest way to analyze circuits containing mutual inductance is to use mesh currents.The problem is to write the circuit equations that describe the circuit in terms of the coil currents. First, choose the refer- ence direction for each coil current. Figure 6.20 shows arbitrarily selected reference currents. After choosing the reference directions for /, and / 2 , sum the voltages around each closed path. Because of the mutual inductance M, there will be two voltages across each coil, namely, a self-induced voltage and a mutually induced voltage. The self- induced voltage is the product of the self-inductance of the coil and the first derivative of the current in that coil. The mutually induced voltage is the product of the mutual inductance of the coils and the first deriva- tive of the current in the other coil. Consider the coil on the left in Fig. 6.20 whose self-inductance has the value L\. The self-induced voltage across this coil is L x (di x fdt) and the mutually induced voltage across this coil is M(di 2 /dt). But what about the polarities of these two voltages? Using the passive sign convention, the self-induced voltage is a voltage drop in the direction of the current producing the voltage. But the polarity of the mutually induced voltage depends on the way the coils are wound in relation to the reference direction of coil currents. In general, showing the details of mutually coupled windings is very cumbersome. Instead, we keep track of the polarities by a method known as the dot convention, in which a dot is placed on one terminal of each winding, as shown in Fig. 6.21. These dots carry the sign information and allow us to draw the coils schematically rather than showing how they wrap around a core structure. The rule for using the dot convention to determine the polarity of mutually induced voltage can be summarized as follows: Dot convention for mutually coupled coils • When the reference direction for a current enters the dotted termi- nal of a coil, the reference polarity of the voltage that it induces in the other coil is positive at its dotted terminal. Or, stated alternatively. Dot convention for mutually coupled coils (alternate) • When the reference direction for a current leaves the dotted termi- nal of a coil, the reference polarity of the voltage that it induces in the other coil is negative at its dotted terminal. For the most part, dot markings will be provided for you in the circuit diagrams in this text. The important skill is to be able to write the appro- priate circuit equations given your understanding of mutual inductance and the dot convention. Figuring out where to place the polarity dots if they are not given may be possible by examining the physical configura- tion of an actual circuit or by testing it in the laboratory. We will discuss these procedures after we discuss the use of dot markings. In Fig. 6.21, the dot convention rule indicates that the reference polar- ity for the voltage induced in coil 1 by the current i 2 is negative at the dot- ted terminal of coil l.This voltage (Mdi 2 /dt) is a voltage rise with respect to /]_. The voltage induced in coil 2 by the current /| is Mdi\jdt, and its ref- erence polarity is positive at the dotted terminal of coil 2. This voltage is a voltage rise in the direction of i 2 . Figure 6.22 shows the self- and mutually induced voltages across coils 1 and 2 along with their polarity marks. Figure 6.22 • The self- and mutually induced voltages appearing across the coils shown in Fig. 6.21. 6.4 Mutual Inductance 191 Now let's look at the sum of the voltages around each closed loop. In Eqs. 6.31 and 6.32, voltage rises in the reference direction of a current are negative: di\ di 2 dii du ioRo + L 2 -f - M- 1 dt dt 0. (6.31) (6.32) The Procedure for Determining Dot Markings We shift now to two methods of determining dot markings. The first assumes that we know the physical arrangement of the two coils and the mode of each winding in a magnetically coupled circuit. The following six steps, applied here to Fig. 6.23, determine a set of dot markings: a) Arbitrarily select one terminal—say, the D terminal—of one coil and mark it with a dot. b) Assign a current into the dotted terminal and label it / D . c) Use the right-hand rule 1 to determine the direction of the magnetic field established by / D inside the coupled coils and label this field <j6 D . d) Arbitrarily pick one terminal of the second coil—say, terminal A—and assign a current into this terminal, showing the current as / A . e) Use the right-hand rule to determine the direction of the flux estab- lished by / A inside the coupled coils and label this flux <£ A . f) Compare the directions of the two fluxes <£ D and <£ A . If the fluxes have the same reference direction, place a dot on the terminal of the second coil where the test current (/ A ) enters. (In Fig. 6.23, the fluxes <£ D and c/> A have the same reference direction, and therefore a dot goes on terminal A.) If the fluxes have different reference direc- tions, place a dot on the terminal of the second coil where the test current leaves. The relative polarities of magnetically coupled coils can also be deter- mined experimentally.This capability is important because in some situations, determining how the coils are wound on the core is impossible. One experi- mental method is to connect a dc voltage source, a resistor, a switch, and a dc voltmeter to the pair of coils, as shown in Fig. 6.24. The shaded box covering the coils implies that physical inspection of the coils is not possible. The resis- tor R limits the magnitude of the current supplied by the dc voltage source. The coil terminal connected to the positive terminal of the dc source via the switch and limiting resistor receives a polarity mark, as shown in Fig. 6.24. When the switch is closed, the voltmeter deflection is observed. If the momentary deflection is upscale, the coil terminal connected to the positive terminal of the voltmeter receives the polarity mark. If the e P 2) __ Arbitrarily dotted D terminal (Stepl) Figure 6.23 • A set of coils showing a method for determining a set of dot markings. R + Switch dc voltmeter Figure 6.24 • An experimental setup for determining polarity marks. See discussion of Faraday's law on page 193. 192 Inductance, Capacitance, and Mutual Inductance deflection is downscale, the coil terminal connected to the negative termi- nal of the voltmeter receives the polarity mark. Example 6.6 shows how to use the dot markings to formulate a set of circuit equations in a circuit containing magnetically coupled coils. Example 6.6 Finding Mesh-Current Equations for a Circuit with Magnetically Coupled Coils a) Write a set of mesh-current equations that describe the circuit in Fig. 6.25 in terms of the currents /j and i 2 . b) Verify that if there is no energy stored in the cir- cuit at t = 0 and if L = 16 - 16tf _5t A, the solu- tions for i\ and i 2 are b) To check the validity of /j and i 2 , we begin by testing the initial and final values of i L and i 2 . We know by hypothesis that /^(0) = i 2 (0) = 0. From the given solutions we have /j(0) = 4 + 64 - 68 = 0, 4; /,=4 + 64e _: * - 6&T" A, / 2 (0) = 1 - 52 + 51 = 0. / 2 = 1 - 52«?"* + 51e" 4 ' A. 50 4H 8H 20 n 1 16 H f i 2 \ 60 H Figure 6.25 • The circuit for Example 6.6. Now we observe that as t approaches infinity the source current (/<,) approaches a constant value of 16 A, and therefore the magnetically coupled coils behave as short circuits. Hence at t = oo the circuit reduces to that shown in Fig. 6.26. From Fig. 6.26 we see that at t = oo the three resistors are in parallel across the 16 A source. The equivalent resistance is 3.75 fl and thus the voltage across the 16 A current source is 60 V. It follows that z,(oo) = — + — = 4A, u } 20 60 Solution a) Summing the voltages around the i x mesh yields 4-^ + 8-¾ - i 2 ) + 20(/, - i 2 ) + 5(i t - i g ) = 0. , 2( oo)=|| = lA. These values agree with the final values pre- dicted by the solutions for i x and i 2 . Finally we check the solutions by seeing if they satisfy the differential equations derived in (a). We will leave this final check to the reader via Problem 6.37. The i 2 mesh equation is 20(/ 2 - h) + 60/ 2 + 16-7-(¾ - **) ~ 8-r = 0. 1 dt ~ gJ dt Note that the voltage across the 4 H coil due to the current (i g - i 2 ), that is, 8d(i g - i 2 )/dt, is a voltage drop in the direction of i x . The voltage induced in the 16 Ff coil by the current / t , that is, 8di\/dt, is a voltage rise in the direction of i 2 . 16A 20 n -VAr- :60O Figure 6.26 • The circuit of Example 6.6 when t — oo. 6.5 A Closer Look at Mutual Inductance 193 •ASSESSMENT PROBLEM Objective 4—Use the dot convention to write mesh-current equations for mutually coupled coils 6.6 a) Write a set of mesh-current equations for the circuit in Example 6.6 if the dot on the 4 H inductor is at the right-hand terminal, the reference direction of/^, is reversed, and the 60 ft resistor is increased to 780 ft. b) Verify that if there is no energy stored in the circuit at t = 0,andifi ff = 1.96 - 1.96e~* A, the solutions to the differential equations derived in (a) of this Assessment Problem are Answer: (a) A{dijdt) + 25i x + 8(di 2 /dt) - 20i 2 = -5i g - $(dig/dt) and %{di<Jdt) - 20¾ + 16(di 2 /dt) + 800/ 2 = -16(dig/dt); (b) verification. -0.4 - -0.01 11.6e" 4 ' + 12e _5 'A, ,-4/ 0.99<T" + e~™ A. -St NOTE: Also try Chapter Problem 6.39. 6.5 A Closer Look at Mutual Inductance In order to fully explain the circuit parameter mutual inductance, and to examine the limitations and assumptions made in the qualitative discussion presented in Section 6.4, we begin with a more quantitative description of self-inductance than was previously provided. A Review of Self-Inductance The concept of inductance can be traced to Michael Faraday, who did pio- neering work in this area in the early 1800s. Faraday postulated that a magnetic field consists of lines of force surrounding the current-carrying conductor. Visualize these lines of force as energy-storing elastic bands that close on themselves. As the current increases and decreases, the elas- tic bands (that is, the lines of force) spread and collapse about the conduc- tor. The voltage induced in the conductor is proportional to the number of lines that collapse into, or cut, the conductor. This image of induced volt- age is expressed by what is called Faraday's law; that is, dX_ dt ' (6.33) where A is referred to as the flux linkage and is measured in weber-turns. How do we get from Faraday's law to the definition of inductance pre- sented in Section 6.1? We can begin to draw this connection using Fig. 6.27 as a reference. The lines threading the N turns and labeled 4> represent the magnetic lines of force that make up the magnetic field. The strength of the mag- netic field depends on the strength of the current, and the spatial orienta- tion of the field depends on the direction of the current. The right-hand rule relates the orientation of the field to the direction of the current: When the fingers of the right hand are wrapped around the coil so that the fingers point in the direction of the current, the thumb points in the direc- tion of that portion of the magnetic field inside the coil. The flux linkage is the product of the magnetic field (<£), measured in webers (Wb), and the number of turns linked by the field (N): N turns Figure 6.27 • Representation of a magnetic field link- ing an vV-turn coil. A = N<f>. (6.34) 194 Inductance, Capacitance, and Mutual Inductance The magnitude of the flux, ¢, is related to the magnitude of the coil current by the relationship $ = SWi, (6.35) where N is the number of turns on the coil, and SP is the permeance of the space occupied by the flux. Permeance is a quantity that describes the magnetic properties of this space, and as such, a detailed discussion of per- meance is outside the scope of this text. Here, we need only observe that, when the space containing the flux is made up of magnetic materials (such as iron, nickel, and cobalt), the permeance varies with the flux, giving a nonlinear relationship between 4> and i. But when the space containing the flux is comprised of nonmagnetic materials, the permeance is constant, giving a linear relationship between cj> and L Note from Eq. 6.35 that the flux is also proportional to the number of turns on the coil. Here, we assume that the core material—the space containing the flux— is nonmagnetic. Then, substituting Eqs. 6.34 and 6.35 into Eq. 6.33 yields v = dX d(N4>) dt dt d<b d, dt dt dt di dt' (6.36) which shows that self-inductance is proportional to the square of the num- ber of turns on the coil. We make use of this observation later. The polarity of the induced voltage in the circuit in Fig. 6.27 reflects the reaction of the field to the current creating the field. For example, when i is increasing, di/dt is positive and v is positive. Thus energy is required to establish the magnetic field. The product vi gives the rate at which energy is stored in the field. When the field collapses, di/dt is negative, and again the polarity of the induced voltage is in opposition to the change. As the field collapses about the coil, energy is returned to the circuit. Keeping in mind this further insight into the concept of self-inductance, we now turn back to mutual inductance. Figure 6.28 A Two magnetically coupled coils. The Concept of Mutual Inductance Figure 6.28 shows two magnetically coupled coils. You should verify that the dot markings on the two coils agree with the direction of the coil wind- ings and currents shown. The number of turns on each coil are A^ and N 2 , respectively. Coil 1 is energized by a time-varying current source that establishes the current i\ in the A 7 ] turns. Coil 2 is not energized and is open. The coils are wound on a nonmagnetic core. The flux produced by the current i\ can be divided into two components, labeled <f> n and 0 2J . The flux component 4> X \ is the flux produced by i\ that links only the Ny turns.The component <£ 21 is the flux produced by f] that links the N 2 turns and the Ny turns. The first digit in the subscript to the flux gives the coil number, and the second digit refers to the coil current. Thus </> n is a flux linking coil 1 and produced by a current in coil 1, whereas cf) 2 \ is a flux link- ing coil 2 and produced by a current in coil 1. The total flux linking coil 1 is <^>|, the sum of </> n and 0 2 i : 01 = 0n + <t>2\- ( 6 -37) The flux <f> { and its components 4>n and 4>z\ are related to the coil current ii as follows: </>! = ?i > 1 N 1 /i, (6.38) 011 = 9 u Nii u (6.39) ^21 = ^2lWi, (6.40) where flf^ is the permeance of the space occupied by the flux 01,27*11 is the permeance of the space occupied by the flux 0 U , and P? 21 i s tne permeance of the space occupied by the flux 02i- Substituting Eqs. 6.38,6.39, and 6.40 into Eq. 6.37 yields the relationship between the permeance of the space occupied by the total flux 4> x and the permeances of the spaces occupied by its components 4>\\ and 0 2I : PJ>, = 0>ii + 9*21- ( 6 -4l) We use Faraday's law to derive expressions for ?; t and v 2 : rfA, d(NM d , di\ » di\ di\ = Aft*,, + 9> 2 ,)^ = NP^ = L,^, (6.42) and dk 2 d(N 2 (f>2\) iT d ,™ A , . x = N 2 N^ 2l -^. (6.43) The coefficient of d/ t /d7 in Eq. 6.42 is the self-inductance of coil 1. The coefficient of di x jdt in Eq. 6.43 is the mutual inductance between coils 1 and 2. Thus M 21 = N 2 Ni^2\- (6.44) The subscript on M specifies an inductance that relates the voltage induced in coil 2 to the current in coil 1. The coefficient of mutual inductance gives V 2 = M 21-^- (6.45) Note that the dot convention is used to assign the polarity reference to v 2 in Fig. 6.28. For the coupled coils in Fig. 6.28, exciting coil 2 from a time-varying cur- rent source (i 2 ) and leaving coil 1 open produces the circuit arrangement