Electric Circuits, 9th Edition P23 doc

196 Inductance, Capacitance, and Mutual Inductance shown in Fig. 6.29. Again, the polarity reference assigned to Vy is based on the dot convention. 1 (T^H +I >l jLL> i /*TN The total flux linking coil 2 is Figure 6.29 • The magnetically coupled coils of Fig. 6.28, with coil 2 excited and coil 1 open. <t>2 = 4>22 + <Ai2- (6.46) The flux 4> 2 and its components (f> 2 2 and <pn are related to the coil current h as follows: (f) 2 = 8^2^2*2» (f> 22 = <3> 22 N 2 i 2 , 012 = ^12^2/2- (6.47) (6.48) (6.49) Tlie voltages v 2 and Vj are (6.50) (6.51) The coefficient of mutual inductance that relates the voltage induced in coil 1 to the time-varying current in coil 2 is the coefficient of di 2 /dt in Eq. 6.51: M u = NyN^u- (6.52) For nonmagnetic materials, the permeances 0^ 2 an d ^21 are equal, and therefore Myi = M 2 \ = M. (6.53) Hence for linear circuits with just two magnetically coupled coils, attach- ing subscripts to the coefficient of mutual inductance is not necessary. Mutual Inductance in Terms of Self-Inductance The value of mutual inductance is a function of the self-inductances. We derive this relationship as follows. From Eqs. 6.42 and 6.50, L 2 = Np 2 , (6.54) (6.55) respectively. From Eqs. 6.54 and 6.55, — \j2\rZc L X L 2 = NiNfr&i (6.56) 6.5 A Closer Look at Mutual Inductance 197 We now use Eq. 6.41 and the corresponding expression for ?P 2 to write L,L 2 = NM(®n + ^21)(¾ + ^12). (6.57) But for a linear system, 0> 2 i = ^12, so Eq. 6.57 becomes =«•('*&)('* a) Replacing the two terms involving permeances by a single constant expresses Eq. 6.58 in a more meaningful form: *-(••&)(••&} Substituting Eq. 6.59 into Eq. 6.58 yields M 2 = k 2 L } L 2 or M = A:VLjL 2 , (6.60) ^ Relating self-inductances and mutual inductance using coupling coefficient where the constant k is called the coefficient of coupling. According to Eq. 6.59, l/k 2 must be greater than 1, which means that k must be less than 1. In fact, the coefficient of coupling must lie between 0 and 1, or 0 < k < 1. (6.61) The coefficient of coupling is 0 when the two coils have no common flux; that is, when <£ 12 = ¢21 = 0. This condition implies that 8P 12 = 0, and Eq. 6.59 indicates that l/k 2 = 00 1 or k — 0. If there is no flux linkage between the coils, obviously M is 0. The coefficient of coupling is equal to 1 when c/> n and <f> 22 are O.This condition implies that all the flux that links coil 1 also links coil 2. In terms of Eq. 6.59, SPJI = ^22 = 0, which obviously represents an ideal state; in reality, winding two coils so that they share precisely the same flux is phys- ically impossible. Magnetic materials (such as alloys of iron, cobalt, and nickel) create a space with high permeance and are used to establish coef- ficients of coupling that approach unity. (We say more about this impor- tant quality of magnetic materials in Chapter 9.) NOTE: Assess your understanding of this material by trying Chapter Problems 6.48 and 6.49. Energy Calculations We conclude our first look at mutual inductance with a discussion of the total energy stored in magnetically coupled coils. In doing so, we will confirm two observations made earlier: For linear magnetic coupling, (1) M n = M 2l = M, and (2) M = kVL^L 2 , where 0 < k < 1. We use the circuit shown in Fig. 6.30 to derive the expression for the total energy stored in the magnetic fields associated with a pair of linearly coupled coils. We begin by assuming that the currents i { and / 2 are zero and that this zero-current state corresponds to zero energy stored in the coils. Then we let / } increase from zero to some arbitrary value I { and com- pute the energy stored when i l = /}. Because i 2 = 0, the total power input into the pair of coils is v^, and the energy stored is rW\ eh j dw = L x \ Jo ./0 i]di[. W t = -L.ll (6.62) Now we hold /] constant at I { and increase i 2 from zero to some arbitrary value / 2 . During this time interval, the voltage induced in coil 2 by ij is zero because I x is constant. The voltage induced in coil 1 by / 2 is M ]2 di 2 /dt. Therefore, the power input to the pair of coils is d'h p = I\M U — + i 2 v 2 . The total energy stored in the pair of coils when i 2 = f 2 is ,.\v /./, pf 2 / dw = / IiM ]2 di 2 + / L 2 i 2 di 2 , Jw, Jo Jo or W = W 1 + I X I 2 M {2 + -L 2 lh = -L,7? + -L 2 ll + I x I 2 M n . (6.63) If we reverse the procedure—that is, if we first increase i 2 from zero to / 2 and then increase i { from zero to 1 { — the total energy stored is W = \h x J\ + \L 2 1\ + /,/ 2 A#2i. (6.64) Equations 6.63 and 6.64 express the total energy stored in a pair of lin- early coupled coils as a function of the coil currents, the self-inductances, and the mutual inductance. Note that the only difference between these equations is the coefficient of the current product l\l 2 . We use Eq. 6.63 if / t is established first and Eq. 6.64 if / 2 is established first. When the coupling medium is linear, the total energy stored is the same regardless of the order used to establish /] and / 2 . The reason is that 6.5 A Closer Look at Mutual Inductance 199 in a linear coupling, the resultant magnetic flux depends only on the final values of/j and / 2 , not on how the currents reached their final values. If the resultant flux is the same, the stored energy is the same. Therefore, for lin- ear coupling, M\i = M 2 \. Also, because 1\ and J 2 are arbitrary values of /j and *" 2 , respectively, we represent the coil currents by their instantaneous values /*! and / 2 . Thus, at any instant of time, the total energy stored in the coupled coils is 1,1, w{t) = -L x i\ + -L 2 t2 + Miy 2 . (6.65) We derived Eq. 6.65 by assuming that both coil currents entered polarity-marked terminals. We leave it to you to verify that, if one current enters a polarity-marked terminal while the other leaves such a terminal, the algebraic sign of the term Mi{i 2 reverses. Thus, in general, 1,1, w(t) = —L\i\ + ~L 2 i 2 ± Mi { i 2 . (6.66) A Energy stored in magnetically-coupled coils We use Eq. 6.66 to show that M cannot exceed VL]L 2 . The magneti- cally coupled coils are passive elements, so the total energy stored can never be negative. If w(t) can never be negative, Eq. 6.66 indicates that the quantity 1,1, -L { q + -L 2 ii - Mi } i 2 must be greater than or equal to zero when /j and i 2 are either both posi- tive or both negative. The limiting value of M corresponds to setting the quantity equal to zero: 1,1, -Liq + ^2'2 - Milk = 0. (6.67) To find the limiting value of M we add and subtract the term / 1 / 2 VL 1 L 2 to the left-hand side of Eq. 6.67. Doing so generates a term that is a perfect square: + /V 2 VLJ7 2 - M = 0. (6.68) The squared term in Eq. 6.68 can never be negative, but it can be zero. Therefore w{t) 2t 0 only if VL { L 2 > M, (6.69) which is another way of saying that M = kVLxL 2 (O^H 1). We derived Eq. 6.69 by assuming that i\ and i 2 are either both positive or both negative. However, we get the same result if i\ and i 2 are of opposite 200 Inductance, Capacitance, and Mutual Inductance sign, because in this case we obtain the limiting value of M by selecting the plus sign in Eq. 6.66. NOTE: Assess your understanding of this material by trying Chapter Problems 6.44 and 6.45. Practical Perspective Proximity Switches At the beginning of this chapter we introduced the capacitive proximity switch. There are two forms of this switch. The one used in table lamps is based on a single-electrode switch. It is left to your investigation in Problem 6.52. In the example here, we consider the two-electrode switch used in elevator call buttons. EXAMPLE The elevator call button is a small cup into which the finger is inserted, as shown in Fig. 6.31. The cup is made of a metal ring electrode and a circular plate electrode that are insulated from each other. Sometimes two concentric rings embedded in insulating plastic are used instead. The electrodes are covered with an insulating Layer to prevent direct contact with the metal. The resulting device can be modeled as a capacitor, as shown in Fig. 6.32. (a) (b) Figure 6.31 • An elevator call button, (a) Front view, (b) Side view. Figure 6.32 • A capacitor model of the two-electrode proximity switch used in elevator call buttons. c 2 Q Figure 6.33 A A circuit model of a capacitive proximity switch activated by finger touch. Unlike most capacitors, the capacitive proximity switch permits you to insert an object, such as a finger, between the electrodes. Because your fin- ger is much more conductive than the insulating covering surrounding the eLectrodes, the circuit responds as though another electrode, connected to ground, has been added. The result is a three-terminal circuit containing three capacitors, as shown in Fig. 6.33. The actual values of the capacitors in Figs. 6.32 and 6.33 are in the range of 10 to 50 pF, depending on the exact geometry of the switch, how Practical Perspective the finger is inserted, whether the person is wearing gloves, and so forth. For the following problems, assume that all capacitors have the same value of 25 pF. Also assume the elevator call button is placed in the capacitive equivalent of a voltage-divider circuit, as shown in Fig. 6.34. a) Calculate the output voltage with no finger present. b) Calculate the output voltage when a finger touches the button. Solution a) Begin by redrawing the circuit in Fig. 6.34 with the call button replaced by its capacitive model from Fig. 6.32. The resulting circuit is shown in Fig. 6.35. Write the current equation at the single node: _ d(v - v s ) _ dv Ci . ' + C 2 — = 0. dt dt (6.70) *M(Z) Button ( A ) Fixed capacitor' ;25pF v(t) Figure 6.34 • An elevator call button circuit. Vs(t)(l) Button ] Fixed capacitor :c, + • C 2 v(t) r - Figure 6.35 A A model of the elevator call button Rearrange this equation to produce a differential equation for the output Q Tm \i W1 - t h no finger present, voltage v(t): dv dt Cj dv s C, + C 7 dt' (6.71) Finally, integrate Eq. 6.71 to find the output voltage: v(t) Ci c, +c -v s (t) + v(0). (6.72) The result in Eq. 6.72 shows that the series capacitor circuit in Fig. 6.35 forms a voltage divider just as the series resistor circuit did in Chapter 3. In both voltage-divider circuits, the output voltage does not depend on the component values but only on their ratio. Here, C { = C 2 = 25 pF, so the capacitor ratio is Ci/C 2 = 1. Thus the output voltage is v(t) = 0.5^(0 + v(0). (6.73) The constant term in Eq. 6.73 is due to the initial charge on the capacitor. We can assume that v(0) - 0 V, because the circuit that senses the out- put voltage eliminates the effect of the initial capacitor charge. Therefore, the sensed output voltage is v(t) = OMt). (6.74) 202 Inductance, Capacitance, and Mutual Inductance b) Now we replace the call button of Fig. 6.34 with the model of the acti- vated switch in Fig. 6.33. The result is shown in Fig. 6.36. Again, we cal- culate the currents leaving the output node: _ d(v - v s ) dv „ dv Q-^-—- + C 2 — + C 3 — = 0. dt dt • dt Rearranging to write a differential equation for v(t) results in dv _ C\ dv s ~dt ~ Q + C 2 + C 3 dt ' Finally, solving the differential equation in Eq. 6.76, we see (6.75) (6.76) v(t) C, v s (t) + v(0). C x + C 2 + C 3 If Q = C 2 = C 3 = 25 pF, v(t) = 0333v s (t) + v(0). (6.77) (6.78) As before, the sensing circuit eliminates v(Q), so the sensed output voltage is v(t) = 0.333^,(0- (6.79) Comparing Eqs. 6.74 and 6.79, we see that when the button is pushed, the output is one third of the input voltage. When the button is not pushed, the output voltage is one half of the input voltage. Any drop in output voltage is detected by the elevator's control computer and ulti- mately results in the elevator arriving at the appropriate floor. NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 6.51 and 6.53. Button Fixed capacitor Figure 6.36 A A model of the elevator call button circuit when activated by finger touch. Summary Summary 203 Inductance is a linear circuit parameter that relates the voltage induced by a time-varying magnetic field to the current producing the field. (See page 176.) Capacitance is a linear circuit parameter that relates the current induced by a time-varying electric field to the voltage producing the field. (See page 182.) Inductors and capacitors are passive elements; they can store and release energy, but they cannot generate or dissipate energy. (See page 176.) The instantaneous power at the terminals of an inductor or capacitor can be positive or negative, depending on whether energy is being delivered to or extracted from the element. An inductor: • does not permit an instantaneous change in its termi- nal current • does permit an instantaneous change in its teminal voltage • behaves as a short circuit in the presence of a constant terminal current (See page 188.) A capacitor: • does not permit an instantaneous change in its termi- nal voltage • does permit an instantaneous change in its terminal current • behaves as an open circuit in the presence of a con- stant terminal voltage (See page 183.) Equations for voltage, current, power, and energy in ideal inductors and capacitors are given in Table 6.1. Inductors in series or in parallel can be replaced by an equivalent inductor. Capacitors in series or in parallel can be replaced by an equivalent capacitor. The equa- tions are summarized in Table 6.2. See Section 6.3 for a discussion on how to handle the initial conditions for series and parallel equivalent circuits involving induc- tors and capacitors. TABLE 6.1 Terminal Equations for Ideal Inductors and Capacitors Inductors i) = J - 0 u dt i = if v dr + J fa p = vi = Li% w = I Li 2 Capacitors v = £ 1 idr + Jfa 1 *- dt p = vi = Cv'j§ w = \Cv 2 Kh) v(U)) (V) (A) (W) (J) (V) (A) (W) (J) TABLE 6.2 Equations for Series- and Parallel-Connected Inductors and Capacitors Series-Connected ^ecj = L\ + Li + • • • + L n J- =-L + _L c„ Parallel-Connected J_ = i_ + X + + G, Mutual inductance, M, is the circuit parameter relating the voltage induced in one circuit to a time-varying cur- rent in another circuit. Specifically, dU di 2 Vl = Ll - + M l2 - di] dU » 2 = M 2 ,- + L 2 -, 204 Inductance, Capacitance, and Mutual Inductance where V\ and i { are the voltage and current in circuit 1, and v 2 and i 2 are the voltage and current in circuit 2. For coils wound on nonmagnetic cores, M 12 = M 2 \ = M (See page 190.) The dot convention establishes the polarity of mutually induced voltages: When the reference direction for a current enters the dotted terminal of a coil, the reference polar- ity of the voltage that it induces in the other coil is positive at its dotted terminal. Or, alternatively, When the reference direction for a current leaves the dotted terminal of a coil, the reference polar- ity of the voltage that it induces in the other coil is negative at its dotted terminal. (See page 190.) The relationship between the self-inductance of each winding and the mutual inductance between windings is M = kVLJ7 2 . The coefficient of coupling, k, is a measure of the degree of magnetic coupling. By definition, 0 < k < 1. (See page 197.) The energy stored in magnetically coupled coils in a lin- ear medium is related to the coil currents and induc- tances by the relationship 1 , 1 9 w = -L x i\ + -L 2 ij ± Mi { i 2 . (See page 199.) Problems Section 6.1 6.1 The current in the 2.5 mH inductor in Fig. P6.1 is known to be 1 A for t < 0. The inductor voltage for t ^ 0 is given by the expression v L (t) = 3e~ 4 ' mV, 0 + < t < 2 s v L {t) = -3e" 4(/_2) mV, 2 s < t < oo Sketch v L {t) and i L {t) for 0 < t < oo. Figure P6.1 '7.(0 «/.(01 |2.5mH PSPICE MULTISIM 6.2 The current in a 50 /xH inductor is known to be i L = 18te~ 10t A for t > 0. a) Find the voltage across the inductor for t > 0. (Assume the passive sign convention.) b) Find the power (in microwatts) at the terminals of the inductor when t = 200 ms. c) Is the inductor absorbing or delivering power at 200 ms? d) Find the energy (in microjoules) stored in the inductor at 200 ms. e) Find the maximum energy (in microjoules) stored in the inductor and the time (in milli- seconds) when it occurs. 6.3 The voltage at the terminals of the 200 /xH inductor PSPICE i n pig. P6.3(a) is shown in Fig. P6.3(b). The inductor 1M current i is known to be zero for t < 0. a) Derive the expressions for i for f > 0. b) Sketch i versus t for 0 < t < oo. Figure P6.3 v s (mV) 200 /xH- t (ms) (a) (b) 6.4 The triangular current pulse shown in Fig. P6.4 is ™± l . applied to a 20 mH inductor. MULTISIM a) Write the expressions that describe /(f) in the four intervals t < 0, 0^/^5 ms, 5 ms ^ t ^ 10 ms, and t > 10 ms. b) Derive the expressions for the inductor volt- age, power, and energy. Use the passive sign convention. Figure P6.4 /(mA) 250 t (ms) Problems 205 6.5 The current in and the voltage across a 5 H inductor are known to be zero for t < 0. The voltage across the inductor is given by the graph in Fig. P6.5 for t > 0. a) Derive the expression for the current as a function of time in the intervals 0 < t < 1 s, Is < t < 3 s, 3 s < t < 5 s, 5 s < t < 6 s, and 6 s < t < oo. b) For t > 0, what is the current in the inductor when the voltage is zero? c) Sketch i versus t for 0 < t < oo. Figure P6.5 v (V) 100 - / v 6.9 a) Find the inductor current in the circuit in PSP.CE Fig. P6.9 if v = -50 sin 250 t V, L = 20 mH, MULTISIM and/(0) = 10 A. b) Sketch v, i, p s and w versus t. In making these sketches, use the format used in Fig. 6.8. Plot over one complete cycle of the voltage waveform. c) Describe the subintervals in the time interval between 0 and Sn ms when power is being absorbed by the inductor. Repeat for the subintervals when power is being delivered by the inductor. Figure P6.9 6.10 The current in a 4 H inductor is i = 10 A, t < 0; i = {B\ cos At + B 2 sin 4t)e~ tf2 A, t > 0. The voltage across the inductor (passive sign con- vention) is 60 V at t = 0. Calculate the power at the terminals of the inductor at t = 1 s. State whether the inductor is absorbing or delivering power. 6.11 Evaluate the integral for Example 6.2. Comment on the significance of the result. 6.12 The expressions for voltage, power, and energy derived in Example 6.5 involved both integration and manipulation of algebraic expressions. As an engineer, you cannot accept such results on faith alone. That is, you should develop the habit of ask- ing yourself, "Do these results make sense in terms of the known behavior of the circuit they purport to describe?" With these thoughts in mind, test the expressions of Example 6.5 by performing the fol- lowing checks: a) Check the expressions to see whether the volt- age is continuous in passing from one time inter- val to the next. b) Check the power expression in each interval by selecting a time within the interval and see- ing whether it gives the same result as the cor- responding product of v and i. For example, test at 10 and 30 (JLS. c) Check the energy expression within each interval by selecting a time within the interval and seeing whether the energy equation gives the same result as \Cv 2 . Use 10 and 30 /xs as test points. 6.13 Initially there was no energy stored in the 5 H inductor in the circuit in Fig. P6.13 when it was placed across the terminals of the voltmeter. At 6.6 The current in a 20 mH inductor is known to be i = 40 mA, t < 0; i = A,e~ mm[ + yW- 4()(X,0 'A, 0. The voltage across the inductor (passive sign con- vention) is 28 V at t = 0. a) Find the expression for the voltage across the inductor for t > 0. b) Find the time, greater than zero, when the power at the terminals of the inductor is zero. 6.7 Assume in Problem 6.6 that the value of the voltage across the inductor at t = 0 is -68 V instead of 28 V. a) Find the numerical expressions for i and v for t >0. b) Specify the time intervals when the inductor is storing energy and the time intervals when the inductor is delivering energy. c) Show that the total energy extracted from the inductor is equal to the total energy stored. 6.8 The current in a 25 mH inductor is known to be PSPICE -io A for / < 0 and -(10 cos 400r - 5 sin 400/)<T 2,)() ' A for t Sr 0. Assume the passive sign convention. a) At what instant of time is the voltage across the inductor maximum? b) What is the maximum voltage? . voltage divider just as the series resistor circuit did in Chapter 3. In both voltage-divider circuits, the output voltage does not depend on the component values but only on their ratio. Here,. Capacitance is a linear circuit parameter that relates the current induced by a time-varying electric field to the voltage producing the field. (See page 182.) Inductors and capacitors are