176 Inductance, Capacitance, and Mutual Inductance Section 6.3 describes techniques used to simplify circuits with series or parallel combinations of capacitors or inductors. Energy can be stored in both magnetic and electric fields. Hence you should not be too surprised to learn that inductors and capacitors are capable of storing energy. For example, energy can be stored in an induc- tor and then released to fire a spark plug. Energy can be stored in a capac- itor and then released to fire a flashbulb. In ideal inductors and capacitors, only as much energy can be extracted as has been stored. Because induc- tors and capacitors cannot generate energy, they are classified as passive elements. In Sections 6.4 and 6.5 we consider the situation in which two circuits are linked by a magnetic field and thus are said to be magnetically cou- pled. 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 known as mutual inductance. The practical significance of magnetic coupling unfolds as we study the relationships between current, voltage, power, and several new parameters specific to mutual inductance. We introduce these relationships here and then describe their utility in a device called a trans- former in Chapters 9 and 10. 6.1 The Inductor Inductance is the circuit parameter used to describe an inductor. Inductance is symbolized by the letter L, is measured in henrys (H), and is represented graphically as a coiled wire—a reminder that inductance is a consequence of a conductor linking a magnetic field. Figure 6.1(a) shows an inductor. Assigning the reference direction of the current in the direction of the volt- age drop across the terminals of the inductor, as shown in Fig. 6.1(b), yields The inductor v - i equation • v = L—, (6.1) dt where v is measured in volts, L in henrys, i in amperes, and t in seconds. Equation 6.1 reflects the passive sign convention shown in Fig. 6.1(b); that is, the current reference is in the direction of the voltage drop across the inductor. If the current reference is in the direction of the voltage rise, Eq. 6.1 is written with a minus sign. Note from Eq. 6.1 that the voltage across the terminals of an inductor is proportional to the time rate of change of the current in the inductor. We can make two important observations here. First, if the current is con- stant, the voltage across the ideal inductor is zero. Thus the inductor behaves as a short circuit in the presence of a constant, or dc, current. Second, current cannot change instantaneously in an inductor; that is, the current cannot change by a finite amount in zero time. Equation 6.1 tells us that this change would require an infinite voltage, and infinite voltages are not possible. For example, when someone opens the switch on an inductive circuit in an actual system, the current initially continues to flow in the air across the switch, a phenomenon called arcing. The arc across the switch prevents the current from dropping to zero instantaneously. Switching inductive circuits is an important engineering problem, because arcing and voltage surges must be controlled to prevent equipment dam- age. The first step to understanding the nature of this problem is to master the introductory material presented in this and the following two chapters. Example 6.1 illustrates the application of Eq. 6.1 to a simple circuit. L (a) (b) Figure 6.1 • (a) The graphic symbol for an inductor with an inductance of L henrys. (b) Assigning reference voltage and current to the inductor, following the pas- sive sign convention. 6.1 The Inductor 177 Example 6.1 Determining the Voltage, Given the Current, at the Terminals of an Inductor The independent current source in the circuit shown in Fig. 6.2 generates zero current for t < 0 and a pulse 10/e~ 5 'A, for t > 0. /<0 / = 0, 100 mH i = 10/<T 5 'A, t > 0 Figure 6.2 • The circuit for Example 6.1. a) Sketch the current waveform. b) At what instant of time is the current maximum? c) Express the voltage across the terminals of the 100 mH inductor as a function of time. d) Sketch the voltage waveform. e) Are the voltage and the current at a maximum at the same time? f) At what instant of time does the voltage change polarity? g) Is there ever an instantaneous change in voltage across the inductor? If so, at what time? Solution a) Figure 6.3 shows the current waveform. b) di/dt = 10(-5te~ 5 ' + e~ 5c ) = 10e -5 ' (1- 5/) A/s; di/dt = 0 when t = 1 s. (See Fig. 6.3.) c) t; = Ldi/dt = (0.1)10e _5 '(l - 5/) = e~ St (1-5/) V,/ > 0;v = 0,/ < 0. d) Figure 6.4 shows the voltage waveform. e) No; the voltage is proportional to di/dt, not i. f) At 0.2 s, which corresponds to the moment when di/dt is passing through zero and changing sign. g) Yes, at t - 0. Note that the voltage can change instantaneously across the terminals of an inductor. Figure 6.3 A The current waveform for Example 6.1. Figure 6.4 A The voltage waveform for Example 6.1. Current in an Inductor in Terms of the Voltage Across the Inductor Equation 6.1 expresses the voltage across the terminals of an inductor as a function of the current in the inductor. Also desirable is the ability to express the current as a function of the voltage. To find i as a function of v, we start by multiplying both sides of Eq. 6.1 by a differential time dt: v dt = L dr. (6.2) Multiplying the rate at which i varies with /by a differential change in time generates a differential change in /, so we write Eq. 6.2 as v dt = L di. (6.3) 178 Inductance, Capacitance, and Mutual Inductance We next integrate both sides of Eq. 6.3. For convenience, we interchange the two sides of the equation and write L dx Jt(tn) v dr. (6.4) Note that we use x and r as the variables of integration, whereas i and / become limits on the integrals. Then, from Eq. 6.4, The inductor i - v equation • »to U, v dr 4- /(/ 0 ), (6.5) where /(/) is the current corresponding to /, and /(/ 0 ) is the value of the inductor current when we initiate the integration, namely, / 0 . In many practical applications, / 0 is zero and Eq. 6.5 becomes if wo /(/) = — / v dr + /(0). (6.6) Equations 6.1 and 6.5 both give the relationship between the voltage and current at the terminals of an inductor. Equation 6.1 expresses the voltage as a function of current, whereas Eq. 6.5 expresses the current as a function of voltage. In both equations the reference direction for the cur- rent is in the direction of the voltage drop across the terminals. Note that /(/()) carries its own algebraic sign. If the initial current is in the same direc- tion as the reference direction for /, it is a positive quantity. If the initial current is in the opposite direction, it is a negative quantity. Example 6.2 illustrates the application of Eq. 6.5. Example 6.2 Determining the Current, Given the Voltage, at the Terminals of an Inductor The voltage pulse applied to the 100 mH inductor shown in Fig. 6.5 is 0 for t < 0 and is given by the expression v{t) = 20/e" 10 ' V for / > 0. Also assume i = 0 for / < 0. a) Sketch the voltage as a function of time. b) Find the inductor current as a function of time. c) Sketch the current as a function of time. b) The current in the inductor is 0 at / = 0. Therefore, the current for / > 0 is hil : I 20T<T 10 7/T + 0 200 -10r 100 -(10T + 1) = 2(1 - \0te~ U)t - e~ mt ) A, / > 0. c) Figure 6.7 shows the current as a function of time. Solution a) The voltage as a function of time is shown in Fig. 6.6. y = 0, t<0 i i < 100 mH v = 20te- m V, r>0 Figure 6.5 A The circuit for Example 6.2. 6.1 The Inductor Figure 6.6 A The voltage waveform for Example 6.2. 0 0.1 0.2 0.3 Figure 6.7 • The current waveform for Example 6.2. Note in Example 6.2 that i approaches a constant value of 2 A as t increases. We say more about this result after discussing the energy stored in an inductor. Power and Energy in the Inductor The power and energy relationships for an inductor can be derived directly from the current and voltage relationships. If the current refer- ence is in the direction of the voltage drop across the terminals of the inductor, the power is VI. (6.7) Remember that power is in watts, voltage is in volts, and current is in amperes. If we express the inductor voltage as a function of the inductor current, Eq. 6.7 becomes r di (6.8) A Power in an inductor We can also express the current in terms of the voltage: 1 /; = v l^k v (IT + /(/•()) (6.9) Equation 6.8 is useful in expressing the energy stored in the inductor. Power is the time rate of expending energy, so dw di p = —^ = Li —. dt dt (6.10) Multiplying both sides of Eq. 6.10 by a differential time gives the differen- tial relationship dw — Li di. (6.11) Both sides of Eq. 6.11 are integrated with the understanding that the ref- erence for zero energy corresponds to zero current in the inductor. Thus dx = L I y dy, Jo w = -Lr. 2 (6.12) «4 Energy in an inductor 180 Inductance, Capacitance, and Mutual Inductance As before, we use different symbols of integration to avoid confusion with the limits placed on the integrals. In Eq. 6.12, the energy is in joules, inductance is in henrys, and current is in amperes. To illustrate the appli- cation of Eqs. 6.7 and 6.12, we return to Examples 6.1 and 6.2 by means of Example 6.3. Example 6.3 Determining the Current, Voltage, Power, and Energy for an Inductor a) For Example 6.1, plot i, v.p, and w versus time. Line up the plots vertically to allow easy assess- ment of each variable's behavior. b) In what time interval is energy being stored in the inductor? c) In what time interval is energy being extracted from the inductor? d) What is the maximum energy stored in the inductor? e) Evaluate the integrals 0.2 p dt and p dt, 0.2 and comment on their significance. f) Repeat (a)-(c) for Example 6.2. g) In Example 6.2, why is there a sustained current in the inductor as the voltage approaches zero? Solution a) The plots of /, v,p, and w follow directly from the expressions for i and v obtained in Example 6.1 and are shown in Fig. 6.8. In particular, p - vi, and w = (f)Ii 2 . b) An increasing energy curve indicates that energy is being stored. Thus energy is being stored in the time interval 0 to 0.2 s. Note that this corre- sponds to the interval when p > 0. c) A decreasing energy curve indicates that energy is being extracted. Thus energy is being extracted in the time interval 0.2 s to oo. Note that this cor- responds to the interval when p < 0. d) From Eq. 6.12 we see that energy is at a maximum when current is at a maximum; glancing at the graphs confirms this. From Example 6.1, maximum current = 0.736 A. Therefore, w max = 27.07 mJ. e) From Example 6.1, / = 10fe" 5 'A and v = e _5t (l - 50 V. Therefore, p = vi = lOte' 101 - 50t 2 e~ m W. 0.2 0.4 0.6 0.8 1.0 Figure 6.8 A The variables /', v, p, and w versus / for Example 6.1. 6.1 The Inductor 181 Thus 0.2 p dt = 10 .,-1 Of ion (-10? - 1) t 2 e~ m 2 0.2 0 -K)r 100 (-10/ - 1) g) The application of the voltage pulse stores energy in the inductor. Because the inductor is ideal, this energy cannot dissipate after the volt- age subsides to zero. Therefore, a sustained cur- rent circulates in the circuit. A lossless inductor obviously is an ideal circuit element Practical inductors require a resistor in the circuit model. (More about this later.) -2 _ Q2e~ z = 27.07 mJ, p dt = 10 0.2 -l()f 100 (-10* - 1) , t 2 e~ m 2 0,2 -10/ 100 (-10/ - 1) 0.2 = -0.2e -2 = -27.07 mJ. Based on the definition of p, the area under the plot of p versus t represents the energy expended over the interval of integration. Hence the integration of the power between 0 and 0.2 s represents the energy stored in the inductor during this time interval. The integral of p over the interval 0.2 s - oo is the energy extracted. Note that in this time interval, all the energy originally stored is removed; that is, after the current peak has passed, no energy is stored in the inductor. f) The plots of v, i, p, and w follow directly from the expressions for v and i given in Example 6.2 and are shown in Fig. 6.9. Note that in this case the power is always positive, and hence energy is always being stored during the volt- age pulse. 0 0.1 0.2 0.3 0.4 0.5 0.( /(A) 2.0 (s) 0 0.1 0.2 0.3 0.4 0.5 0.6 p(mW) 600 (s) 0 0.1 0.2 0.3 0.4 0.5 0.6 •w (mJ) f(s) 200 100 0 - - 0.1 I 0.2 1 0.3 1 0.4 1 0.5 1 0.6 r(s) Figure 6.9 • The variables v, i, p, and w versus t for Example 6.2. 182 Inductance, Capacitance, and Mutual Inductance I/ASSESSMENT PROBLEM Objective 1—Know and be able to use the equations for voltage, current, power, and energy in an inductor 6.1 The current source in the circuit shown gener- ates the current pulse WO = °< f < 0, i g (t) = 8e-™' - 8e~ im * A, t > 0. Find (a) v(0); (b) the instant of time, greater than zero, when the voltage v passes through zero; (c) the expression for the power delivered to the inductor; (d) the instant when the power delivered to the inductor is maximum; (e) the maximum power; (f) the instant of time when the stored energy is maximum; and (g) the max- imum energy stored in the inductor. NOTE: Also try Chapter Problems 6.1 and 6.4. Answer: M4m (a) 28.8 V; (b) 1.54 ms; (c) -76.8<T 600 ' + 384e- 1500 ' -307.2<T 2400f W, t > 0; (d) 411.05 jus; (e) 32.72 W; (f) 1.54 ms; (g) 28.57 mJ. 6.2 The Capacitor (a) C + v / (b) Figure 6.10 • (a) The circuit symbol for a capacitor, (b) Assigning reference voltage and current to the capacitor, following the passive sign convention. The circuit parameter of capacitance is represented by the letter C is measured in farads (F), and is symbolized graphically by two short paral- lel conductive plates, as shown in Fig. 6.10(a). Because the farad is an extremely large quantity of capacitance, practical capacitor values usually lie in the picofarad (pF) to microfarad (/xF) range. The graphic symbol for a capacitor is a reminder that capacitance occurs whenever electrical conductors are separated by a dielectric, or insulating, material. This condition implies that electric charge is not transported through the capacitor. Although applying a voltage to the terminals of the capacitor cannot move a charge through the dielectric, it can displace a charge within the dielectric. As the voltage varies with time, the displacement of charge also varies with time, causing what is known as the displacement current. At the terminals, the displacement current is indistinguishable from a conduction current. The current is proportional to the rate at which the voltage across the capacitor varies with time, or, mathematically. Capacitor I - v equation • . n dv (6.13) where /' is measured in amperes, C in farads, v in volts, and t in seconds. Equation 6.13 reflects the passive sign convention shown in Fig. 6.10(b); that is, the current reference is in the direction of the voltage drop across the capacitor. If the current reference is in the direction of the voltage rise, Eq. 6.13 is written with a minus sign. 6.2 The Capacitor 183 Two important observations follow from Eq. 6.13. First, voltage cannot change instantaneously across the terminals of a capacitor. Equation 6.13 indicates that such a change would produce infinite current, a physical impossibility. Second, if the voltage across the terminals is constant, the capacitor current is zero. The reason is that a conduction current cannot be established in the dielectric material of the capacitor. Only a time-varying voltage can produce a displacement current.Thus a capacitor behaves as an open circuit in the presence of a constant voltage. Equation 6.13 gives the capacitor current as a function of the capaci- tor voltage. Expressing the voltage as a function of the current is also use- ful. To do so, we multiply both sides of Eq. 6.13 by a differential time dt and then integrate the resulting differentials: •'•(') i ft i dt = C dv or I dx = — I i dr. h(t ( >) c Jh Carrying out the integration of the left-hand side of the second equa- tion gives v(t) =-/ idr + v(t 0 ). (6.14) < Capacitor v - i equation In many practical applications of Eq. 6.14, the initial time is zero; that is, t {) = 0. Thus Eq. 6.14 becomes i*0 =^fidr + v(0). (6.15) We can easily derive the power and energy relationships for the capacitor. From the definition of power, p = vi = O dv It* (6.16) -4 Capacitor power equation or P = l L C,, (1 i dr + v(t () ) (6.17) 184 Inductance, Capacitance, and Mutual Inductance Combining the definition of energy with Eq. 6.16 yields dw = Cv dv, from which f>W A» / dx = C I y dy, Jo J[) or Capacitor energy equation • w = —Or. 2 (6.18) In the derivation of Eq. 6.18, the reference for zero energy corresponds to zero voltage. Examples 6.4 and 6.5 illustrate the application of the current, voltage, power, and energy relationships for a capacitor. Example 6.4 Determining Current, Voltage, Power, and Energy for a Capacitor The voltage pulse described by the following equa- tions is impressed across the terminals of a 0.5 /xF capacitor: Solution a) From Eq. 6.13, 0, / < 0 s; { At V, 0 s < t < 1 s; 4^^ V, />ls. (0.5 X Kr 6 )(0) = 0, t < 0s; (0.5 X 10" 6 )(4) = 2 yuA, 0 s < / < 1 s; (0.5 x 10 _6 )(-4e"<'~ 1) ) = -2<r (,_1 VA, t > 1 s. a) Derive the expressions for the capacitor current, power, and energy. b) Sketch the voltage, current, power, and energy as functions of time. Line up the plots vertically. c) Specify the interval of time when energy is being stored in the capacitor. d) Specify the interval of time when energy is being delivered by the capacitor. e) Evaluate the integrals p dt and / p dt and comment on their significance. P = w The expression for the power is derived from Eq.6.16: 0, (4/)(2) = 8/ /iW, ( 4 <,-(/-i)) ( _ 2£ >-('-i)) = t < 0 s; 0 s < t < 1 s; -8e _2( ' _1 Vw, t > 1 s. The energy expression follows directly from Eq.6.18: 0 t < 0 s; 1(0.5)16/ 2 = 4f 2 /xJ, 0 s < t < 1 s; ±(0.5) 16e- 2( ' _1) = Ae-^-VfiJ, t > 1 s. 6.2 The Capacitor 185 b) Figure 6.11 shows the voltage, current, power, and energy as functions of time. c) Energy is being stored in the capacitor whenever the power is positive. Hence energy is being stored in the interval 0-1 s. d) Energy is being delivered by the capacitor when- ever the power is negative. Thus energy is being delivered for all / greater than 1 s. e) The integral of p dt is the energy associated with the time interval corresponding to the limits on the integral. Thus the first integral represents the energy stored in the capacitor between 0 and 1 s, whereas the second integral represents the energy returned, or delivered, by the capacitor in the interval 1 s to oo: l ,i p dt = / 8t dt = At 2 4/AJ, f 00 IT 2 *'" 1 ) pdt= I (-8^-^ = (-8) — J\ _ 2 -4/AJ. The voltage applied to the capacitor returns to zero as time increases without limit, so the energy returned by this ideal capacitor must equal the energy stored. P 0*w) 4 2 0 - 1 1 2 - 1 3 1 4 l 5 1 6 Us) Figure 6.11 A The variables v, i, p, and w versus t for Example 6.4. Example 6.5 Finding v, p, and w Induced by a Triangular Current Pulse for a Capacitor An uncharged 0.2 /AF capacitor is driven by a trian- gular current pulse. The current pulse is described by m = < 0, fsO; 5000* A, 0 < t < 20 /AS; 0.2 - 5000* A, 20 < t < 40 jus; {0, t > 40 /AS. a) Derive the expressions for the capacitor voltage, power, and energy for each of the four time inter- vals needed to describe the current. b) Plot i, v, p, and w versus t. Align the plots as spec- ified in the previous examples. c) Why does a voltage remain on the capacitor after the current returns to zero? Solution a) For t < 0, v, p, and w all are zero. ForO < t •& 20/is, v 5 : : 10 h I (5000r) dr + 0 = 12.5 X 10V V, /o vi = 62.5 X lO'VW, w = -Cv 2 = 15.625 X lO'Vj. 2 For 20 /AS < r < 40 /AS, v = 5 X 10 6 / (0.2 - 5000T)^T + 5. . reminder that capacitance occurs whenever electrical conductors are separated by a dielectric, or insulating, material. This condition implies that electric charge is not transported through. voltage to the terminals of the capacitor cannot move a charge through the dielectric, it can displace a charge within the dielectric. As the voltage varies with time, the displacement of charge. or parallel combinations of capacitors or inductors. Energy can be stored in both magnetic and electric fields. Hence you should not be too surprised to learn that inductors and capacitors are