226 Response of First-Order RL and RC Circuits constant after the switch has been closed, the current will have reached approximately 63% of its final value, or i(T) v t V R R s -i 0.6321 R (7.37) If the current were to continue to increase at its initial rate, it would reach its final value at / = T; that is, because di dt VJ-l R L = -lV'A (7.38) the initial rate at which i{t) increases is >"£ (7.39) If the current were to continue to increase at this rate, the expression for i would be L ' (7.40) from which, at t = T, L R R (7.41) K v s R V 0.632 -^ 0 0 /l / -, , V, _ JY '"> = it 1 ' II T 2r 3T V s 1 4T 1 5T Figure 7.17 A The step response of the RL circuit shown in Fig. 7.16 when / () = 0. 0.368 V, 0 r 2r 3T 4r 5T Figure 7.18 A Inductor voltage versus time. Equations 7.36 and 7.40 are plotted in Fig. 7.17. The values given by Eqs. 7.37 and 7.41 are also shown in this figure. The voltage across an inductor is Ldi/dt, so from Eq. 7.35, for t >. 0 + , v = L -R L /o R fl-W - (V s - I Q R)e-W L K (7.42) The voltage across the inductor is zero before the switch is closed. Equation 7.42 indicates that the inductor voltage jumps to V s - l () R at the instant the switch is closed and then decays exponentially to zero. Does the value of v at t = 0 + make sense? Because the initial current is /() and the inductor prevents an instantaneous change in current, the current is / 0 in the instant after the switch has been closed. The voltage drop across the resistor is I Q R, and the voltage impressed across the induc- tor is the source voltage minus the voltage drop, that is, V s — I { )R. When the initial inductor current is zero, Eq. 7.42 simplifies to v = V s e- (R/L)t . (7.43) If the initial current is zero, the voltage across the inductor jumps to V s . We also expect the inductor voltage to approach zero as t increases, because the current in the circuit is approaching the constant value of VJR. Figure 7.18 shows the plot of Eq. 7.43 and the relationship between the time constant and the initial rate at which the inductor voltage is decreasing. 7.3 The Step Response of RL and RC Circuits 227 If there is an initial current in the inductor, Eq. 7.35 gives the solution for it. The algebraic sign of / 0 is positive if the initial current is in the same direction as /'; otherwise, / 0 carries a negative sign. Example 7.5 illustrates the application of Eq. 7.35 to a specific circuit. | Determining the Step Response of an RL Circuit The switch in the circuit shown in Fig. 7.19 has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a make-before-break type; that is, the connection at position b is established before the connection at position a is broken, so there is no interruption of current through the inductor. a) Find the expression for /(/•) for t ^ 0. b) What is the initial voltage across the inductor just after the switch has been moved to position b? c) How many milliseconds after the switch has been moved does the inductor voltage equal 24 V? d) Does this initial voltage make sense in terms of circuit behavior? e) Plot both i(t) and v(t) versus t. t = 0 Figure 7.19 A The circuit for Example 7.5. Solution a) The switch has been in position a for a long time, so the 200 mH inductor is a short circuit across the 8 A current source. Therefore, the inductor carries an initial current of 8 A. This current is oriented opposite to the reference direction for i; thus / 0 is -8 A. When the switch is in position b, the final value of i will be 24/2, or 12 A. The time constant of the circuit is 200/2, or 100 ms. Substituting these values into Eq. 7.35 gives i= 12 + (-8 - 12)e~' /01 = 12 - 20e~ la A, t > 0. b) The voltage across the inductor is r di v = L — dt = 0.2(200<T 1( ") = 40<T ll) 'V, t > 0 + . The initial inductor voltage is v{0 + ) = 40 V. c) Yes; in the instant after the switch has been moved to position b, the inductor sustains a cur- rent of 8 A counterclockwise around the newly formed closed path. This current causes a 16 V drop across the 2 fl resistor. This voltage drop adds to the drop across the source, producing a 40 V drop across the inductor. d) We find the time at which the inductor voltage equals 24 V by solving the expression 24 = A0e~ m for/: 1 , 40 t= w ]n T4 = 51.08 X 10" 3 = 51.08 ms. e) Figure 7.20 shows the graphs of i(t) and v{t) versus t. Note that the instant of time when the current equals zero corresponds to the instant of time when the inductor voltage equals the source volt- age of 24 V, as predicted by Kirchhoff s voltage law. v(y)i(A) 40 32 16 12 8 4 -4 — l\ I X f 100 -8 1 200 1—(- 300 _ t 400 i \ 500 t (ms) Figure 7.20 • The current and voltage waveforms for Example 7.5. 228 Response of First-Order RL and RC Circuits I/ASSESSMENT PROBLE Objective 2—Be able to determine the step response of both RL and RC circuits 7.5 Assume that the switch in the circuit shown in Fig. 7.19 has been in position b for a long time, and at t = 0 it moves to position a. Find (a) /-(0 + ); (b) v(0 + ); (c) T ,t > 0; (d) i(t), t > 0; and (e) v(t), t > 0 + . NOTE: Also try Chapter Problems 7.35-737. Answer: (a) 12 A; (b) -200 V; (c) 20 ms; (d) -8 + 20<T 50 ' A, t > 0; (e) -200<T 50f V, t > 0 + . We can also describe the voltage v(t) across the inductor in Fig. 7.16 directly, not just in terms of the circuit current. We begin by noting that the voltage across the resistor is the difference between the source voltage and the inductor voltage. We write Yrt V * V{t) l{t) = R-lf> (7.44) where V s is a constant. Differentiating both sides with respect to time yields (7.45) di dt ]_dv R dt' Then, if we multiply each side of Eq. 7.45 by the inductance L, we get an expression for the voltage across the inductor on the left-hand side, or v — - L dv R dt' (7.46) Putting Eq. 7.46 into standard form yields dv R — + — v ~ 0. dt L (7.47) You should verify (in Problem 7.38) that the solution to Eq. 7.47 is identi- cal to that given in Eq. 7.42. At this point, a general observation about the step response of an RL circuit is pertinent. (This observation will prove helpful later.) When we derived the differential equation for the inductor current, we obtained Eq. 7.29. We now rewrite Eq. 7.29 as di R V s — + —/=—. dt L L (7.48) Observe that Eqs. 7.47 and 7.48 have the same form. Specifically, each equates the sum of the first derivative of the variable and a constant times the variable to a constant value. In Eq. 7.47, the constant on the right-hand side happens to be zero; hence this equation takes on the same form as the natural response equations in Section 7.1. In both Eq. 7.47 and Eq. 7.48, the constant multiplying the dependent variable is the reciprocal of the time constant, that is, R/L = 1/r. We encounter a similar situation in the derivations for the step response of an RC circuit. In Section 7.4, we will use these observations to develop a general approach to finding the natu- ral and step responses of RL and RC circuits. The Step Response of an RC Circuit 7.3 The Step Response of RL and RC Circuits We can find the step response of a first-order RC circuit by analyzing the circuit shown in Fig. 7.21. For mathematical convenience, we choose the Norton equivalent of the network connected to the equivalent capacitor. Summing the currents away from the top node in Fig. 7.21 generates the differential equation dv c v c C—- + — = I dt R s (7.49) Division of Eq. 7.49 by C gives dv c v c _ I s ~dt ~RC~~C' (7.50) Figure 7.21 A A circuit used to illustrate the step response of a first-order RC circuit. Comparing Eq. 7.50 with Eq. 7.48 reveals that the form of the solution for Vc is the same as that for the current in the inductive circuit, namely, Eq. 7.35. Therefore, by simply substituting the appropriate variables and coefficients, we can write the solution for i? c directly. The translation requires that l s replace Vg, C replace L, 1/JR replace R, and V Q replace / () . We get v c = I.R + (V 0 - I s R)e-'/ RC , t > 0. (7.51) -^ Step response of an RC circuit A similar derivation for the current in the capacitor yields the differential equation di 1 (7.52) Equation 7.52 has the same form as Eq. 7.47, hence the solution for i is obtained by using the same translations used for the solution of Eq. 7.50. Thus i = I, - -i IRC t > 0 + , (7.53) where V () is the initial value of v c , the voltage across the capacitor. We obtained Eqs. 7.51 and 7.53 by using a mathematical analogy to the solution for the step response of the inductive circuit. Let's see whether these solutions for the RC circuit make sense in terms of known circuit behavior. From Eq. 7.51, note that the initial voltage across the capacitor is V {h the final voltage across the capacitor is I S R, and the time constant of the circuit is RC. Also note that the solution for Vc is valid for t 2: 0. These observations are consistent with the behavior of a capacitor in parallel with a resistor when driven by a con- stant current source. Equation 7.53 predicts that the current in the capacitor at t = 0 + is h ~ K)/^- This prediction makes sense because the capacitor voltage can- not change instantaneously, and therefore the initial current in the resistor is VQ/R. The capacitor branch current changes instantaneously from zero at t = 0~ to I s - V()/R at t = 0 + . The capacitor current is zero at t = oo. Also note that the final value of v = I S R. Example 7.6 illustrates how to use Eqs. 7.51 and 7.53 to find the step response of a first-order RC circuit. 230 Response of First-Order RL and RC Circuits Example 7.6 Determining the Step Response of an RC Circuit The switch in the circuit shown in Fig. 7.22 has been in position 1 for a long time. At t = 0, the switch moves to position 2. Find a) v 0 (t) for t > 0 and b) i a {t) for t > 0 + . The value of the Norton current source is the ratio of the open-circuit voltage to the Thevenin resistance, or -60/(40 X 10 3 ) = -1.5 mA. The resulting Norton equivalent circuit is shown in Fig. 7.23. From Fig. 7.23, 1 K R = -60 V and RC = \0ms. We have already noted that v o (0) = 30 V, so the solution for v 0 is 20 kH K /oSkfi 40 kO -t Uv t = 0 /4^-VvV—•- 40 V %60kfi 0.25 jaFPp: »„ X 160kO ( ^ )75 V Figure 7.22 • The circuit for Example 7.6. v 0 = -60 + [30 - (-60)]e~ mu = -60 + 90e~ 1(,( " V, t > 0. b) We write the solution for i„ directly from Eq. 7.53 by noting that 7 V = -1.5 mA and VJR = (30/40) X 10" 3 , or 0.75 mA: i a = -2.25e" ,uu ' mA, t > 0 + . Solution a) The switch has been in position 1 for a long time, so the initial value of v 0 is 40(60/80), or 30 V. To take advantage of Eqs. 7.51 and 7.53, we find the Norton equivalent with respect to the terminals of the capacitor for t > 0. To do this, we begin by computing the open-circuit voltage, which is given by the -75 V source divided across the 40 kfl and 160 ktt resistors: Vu 160 x 10- 1 (40 + 160) X 10 j (-75) = -60 V. We check the consistency of the solutions for v () and i a by noting that i a = C 1 ^- = (0.25 X 10- 6 )(-9000fr i00 0 at = -225e- mt mA. Because dv o (0 )/dt = 0, the expression for i Q clearly is valid only for t > 0 + . Next, we calculate the Thevenin resistance, as seen to the right of the capacitor, by shorting the -75 V source and making series and parallel combinations of the resistors: ttn, = 8000 + 40,000 || 160,000 = 40 101 Figure 7.23 • The equivalent circuit for t > 0 for the circuit shown in Fig. 7.22. 7.4 A General Solution for Step and Natural Responses 231 ^ASSESSMENT PROBLEM Objective 2—Be able to determine the step response of both RL and RC circuits 7.6 a) Find the expression for the voltage across the 160 kfi resistor in the circuit shown in Fig. 7.22. Let this voltage be denoted v A , and assume that the reference polarity for the voltage is positive at the upper terminal of the 160 kH resistor. NOTE: Also try Chapter Problems 7.51 and 7.53. b) Specify the interval of time for which the expression obtained in (a) is valid. Answer: (a) -60 + 12e~ m)t V; (b) t > 0*. 7.4 A General Solution for Step and Natural Responses The general approach to finding either the natural response or the step response of the first-order RL and RC circuits shown in Fig. 7.24 is based on their differential equations having the same form (compare Eq. 7.48 and Eq. 7.50). To generalize the solution of these four possible circuits, we let x(t) represent the unknown quantity, giving x(t) four possible values. It can represent the current or voltage at the terminals of an inductor or the current or voltage at the terminals of a capacitor. From Eqs. 7.47, 7.48, 7.50, and 7.52, we know that the differential equation describing any one of the four circuits in Fig. 7.24 takes the form £ + £ = K - dt T (7.54) where the value of the constant K can be zero. Because the sources in the circuit are constant voltages and/or currents, the final value of x will be constant; that is, the final value must satisfy Eq. 7.54, and, when x reaches its final value, the derivative dxjdt must be zero. Hence x f = KT, (7.55) where xy represents the final value of the variable. We solve Eq. 7.54 by separating the variables, beginning by solving for the first derivative: dt T (x - Kr) "(x - X f ) (7.56) L<n Figure 7.24 • Four possible first-order circuits. (a) An inductor connected to a Thevenin equivalent. (b) An inductor connected to a Norton equivalent. (c) A capacitor connected to a Thevenin equivalent. (d) A capacitor connected to a Norton equivalent. 232 Response of First-Order RL and RC Circuits General solution for natural and step responses of RL and RC circuits • In writing Eq. 7.56, we used Eq. 7.55 to substitute Xf for Kr. We now mul- tiply both sides of Eq. 7.56 by dt and divide by x - Xt to obtain dx -1 = —dt. X — Xf T (7.57) Next, we integrate Eq. 7.57. To obtain as general a solution as possible, we use time t Q as the lower limit and t as the upper limit. Time t {) corresponds to the time of the switching or other change. Previously we assumed that t 0 = 0, but this change allows the switching to take place at any time. Using u and v as symbols of integration, we get m du = — / dv. x(t a ) » - Xf Carrying out the integration called for in Eq. 7.58 gives x(t) = x f + [x(t 0 ) - x f ]e- {t -^\ (7.58) (7.59) The importance of Eq. 7.59 becomes apparent if we write it out in words: the unknown variable as a function of time the final value of the variable + the initial the final value of the — value of the variable variable \y „ -[(-(time of switching)! e (time constant) (7.60) Calculating the natural or step response of RL or RC circuits • In many cases, the time of switching—that is, r 0 —is zero. When computing the step and natural responses of circuits, it may help to follow these steps: 1. Identify the variable of interest for the circuit. For RC circuits, it is most convenient to choose the capacitive voltage; for RL circuits, it is best to choose the inductive current. 2. Determine the initial value of the variable, which is its value at t Q . Note that if you choose capacitive voltage or inductive current as your variable of interest, it is not necessary to distinguish between t = ?o and t = tQ. 2 This is because they both are continuous vari- ables. If you choose another variable, you need to remember that its initial value is defined at t — $. 3. Calculate the final value of the variable, which is its value as t —> oo. 4. Calculate the time constant for the circuit. With these quantities, you can use Eq. 7.60 to produce an equation describing the variable of interest as a function of time. You can then find equations for other circuit variables using the circuit analysis techniques introduced in Chapters 3 and 4 or by repeating the preceding steps for the other variables. Examples 7.7-7.9 illustrate how to use Eq. 7.60 to find the step response of an RC or RL circuit. 2 The expressions / () and $ are analogous to 0 and 0'. Thus x(t {) ) is the limit of x(t) as t from the left, and X(IQ) is The limit of x(t) as t —»z () from the right. 7.4 A General Solution for Step and Natural Responses 233 Using the General Solution Method to Find an RC Circuit's Step Response The switch in the circuit shown in Fig. 7.25 has been in position a for a long time. At t = 0 the switch is moved to position b. a) What is the initial value of v c l b) What is the final value of v c l c) What is the time constant of the circuit when the switch is in position b? d) What is the expression for v c (t) when t a 0? e) What is the expression for i(t) when t > 0 + ? f) How long after the switch is in position b does the capacitor voltage equal zero? g) Plot v c (t) and i(t) versus t. Solution a) The switch has been in position a for a long time, so the capacitor looks like an open circuit. Therefore the voltage across the capacitor is the voltage across the 60 H resistor. From the voltage- divider rule, the voltage across the 60 fl resistor is 40 X [60/(60 + 20)], or 30 V. As the refer- ence for VQ is positive at the upper terminal of the capacitor, we have v c (0) — —30 V. b) After the switch has been in position b for a long time, the capacitor will look like an open circuit in terms of the 90 V source. Thus the final value of the capacitor voltage is + 90 V. c) The time constant is r = RC = (400 X 10 3 )(0.5 X 10 -6 ) = 0.2 s. d) Substituting the appropriate values for vp v(0), and t into Eq. 7.60 yields v c {t) = 90 + (-30 - 90)e -5 ' = 90 - 120e _5 'V, t > 0. e) Here the value for r doesn't change. Thus we need to find only the initial and final values for the current in the capacitor. When obtaining the initial value, we must get the value of l(0 + ), because the current in the capacitor can change instantaneously. This current is equal to the cur- rent in the resistor, which from Ohm's law is [90 - (-30)]/(400 X 10 3 ) = 300 fxA. Note that when applying Ohm's law we recognized that the 400 kO h 20 n Figure 7.25 • The circuit for Example 7.7. capacitor voltage cannot change instantaneously. The final value of i(t) = 0, so '(0 0 + (300 - 0)e -5 ' 300e~ 5 VA, t > 0 + . We could have obtained this solution by dif- ferentiating the solution in (d) and multiplying by the capacitance. You may want to do so for your- self. Note that this alternative approach to finding i{t) also predicts the discontinuity at t = 0. f) To find how long the switch must be in position b before the capacitor voltage becomes zero, we solve the equation derived in (d) for the time when v c {t) = 0: 120e" 90 or e 5l = 120 90' so ' = Mf = 57.54 ms. Note that when v c = 0, i = 225 JXA and the voltage drop across the 400 kft resistor is 90 V. g) Figure 7.26 shows the graphs of v c (t) and i(t) versus t. i(/*A) v c (V) 300 120 250 100 200 80 150 60 100 40 50 20 0 -20 _ry' ~i \y -1 /\ -1 / y ' / 200 r -30 i 400 v C —H 1 600 800 t (ms) Figure 7.26 A The current and voltage waveforms for Example 7.7, 234 Response of First-Order RL and RC Circuits Example 7.8 Using the General Solution Method with Zero Initial Conditions The switch in the circuit shown in Fig. 7.27 has been open for a long time. The initial charge on the capacitor is zero. At t = 0, the switch is closed. Find the expression for a) i(t) for t > 0 + and b) v{t) when f > 0 + . 0.1 /xF -A^Ht 7.5 mA I »(0S20kQ '(0 :3()kn Figure 7.27 A The circuit for Example 7.8. Solution a) Because the initial voltage on the capacitor is zero, at the instant when the switch is closed the current in the 30 kO branch will be /(0 + ) = (7.5)(20) 50 = 3 mA. The final value of the capacitor current will be zero because the capacitor eventually will appear as an open circuit in terms of dc current. Thus if = 0. The time constant of the circuit will equal the product of the Thevenin resistance (as seen from the capacitor) and the capacitance. Therefore T = (20 + 30)1(^(0.1) X 10 -6 = 5 ms. Substituting these values into Eq. 7.60 generates the expression /(f) = 0 -f (3 - 0)e' f/5xl(r3 = 3e- 2(,0f mA, t > 0 + . b) To find v(t), we note from the circuit that it equals the sum of the voltage across the capaci- tor and the voltage across the 30 kf! resistor. To find the capacitor voltage (which is a drop in the direction of the current), we note that its initial value is zero and its final value is (7.5)(20), or 150 V. The time constant is the same as before, or 5 ms. Therefore we use Eq. 7.60 to write v c (t) = 150 + (0- I50)e" mi = (150 - 150<r 2()( ") V, t > 0. Hence the expression for the voltage v(t) is v(t) = 150 - 150^ 200 ' + (30)(3K 2,)( " = (150 - 60<T 20()f )V, f>0 + . As one check on this expression, note that it pre- dicts the initial value of the voltage across the 20 LI resistor as 150 - 60, or 90 V. The instant the switch is closed, the current in the 20 kO resistor is (7.5)(30/50), or 4.5 mA. This current produces a 90 V drop across the 20 kil resistor, confirming the value predicted by the solution. Example 7.9 Using the General Solution Method to Find an RL Circuit's Step Response The switch in the circuit shown in Fig. 7.28 has been open for a long time. At t = 0 the switch is closed. Find the expression for a) v{t) when t > 0 + and b) i(i) when t > 0. 20 V Figure 7.28 • The circuit for Example 7.9. v(t) j80mH Solution a) The switch has been open for a long time, so the initial current in the inductor is 5 A, oriented from top to bottom. Immediately after the switch closes, the current still is 5 A, and therefore the initial voltage across the inductor becomes 20 - 5(1), or 15 V. The final value of the induc- tor voltage is 0 V. With the switch closed, the time constant is 80/1, or 80 ms. We use Eq. 7.60 to write the expression for v(t): v(t) = 0 + (15 - 0)e-' /80x10 " s = 15<T 12 - 5 ' V, t > 0 + . b) We have already noted that the initial value of the inductor current is 5 A. After the switch has 7.4 A General Solution for Step and Natural Responses 235 been closed for a long time, the inductor current reaches 20/1, or 20 A. The circuit time constant is 80 ms, so the expression for /(/) is /(/) = 20 + (5- 20)<r 12J5f = (20 - 15e _l2 - 5 ')A, t > 0. We determine that the solutions for v(t) and i(t) agree by noting that V{t) = L Jt = 80 X 10- 3 [15(12.5K 12 - 5/ ] = 15fT 12 - 5 'V, / > 0 + . NOTE: Assess your understanding of the general solution method by trying Chapter Problems 7.55 and 7.56. Example 7.10 shows that Eq. 7.60 can even be used to find the step response of some circuits containing magnetically coupled coils. Example 7.10 Determining Step Response of a Circuit with Magnetically Coupled Coils There is no energy stored in the circuit in Fig. 7.29 at the time the switch is closed. a) Find the solutions for i m v t) , i u and / 2 . b) Show that the solutions obtained in (a) make sense in terms of known circuit behavior. Solution a) For the circuit in Fig. 7.29, the magnetically cou- pled coils can be replaced by a single inductor having an inductance of L X L 2 - M 2 45 - 36 18 - 12 1.5 H. (See Problem 6.41.) It follows that the circuit in Fig. 7.29 can be simplified as shown in Fig. 7.30. By hypothesis the initial value of / ( , is zero. From Fig. 7.30 we see that the final value of i a will be 120/7.5 or 16 A. The time constant of the circuit is 1.5/7.5 or 0.2 s. It follows directly from Eq. 7.60 that i 0 = 16 - 16e~ 5 'A, / > 0. The voltage v 0 follows from Kirchhoff s voltage law. Thus, v a - 120 — 7.5/ ( , = 120«""* V, t > 0 + . To find i[ and / 2 we first note from Fig. 7.29 that du dh di\ di 7 3-^ + 6-^ = 6-r + 15- 1 dt dt dt dt or dt dt ' 7.5 fl 1 = 0 120 V •I ^6H^« "p3Hi* 15 H^ Figure 7.29 A The circuit for Example 7.10. 120 V 7.5 ft Hr >'„ ^1.5 H Figure 7.30 • The circuit in Fig. 7.29 with the magnetically coupled coils replaced by an equivalent coil. It also follows from Fig. 7.29 that because to = 'I + «2» di 0 di\ di2 dt dt dt ' Therefore a** = -2^. dt Because / 2 (0) is zero we have Jo = -8 + 8e~ 5t A, / > 0. Using Kirchhoffs current law we get it = 24 - 24e" 5/ A, / > 0. b) First we observe that i o (0), /*i(0), and / 2 (0) are all zero, which is consistent with the statement that . responses of circuits, it may help to follow these steps: 1. Identify the variable of interest for the circuit. For RC circuits, it is most convenient to choose the capacitive voltage; for RL circuits,. the same form (compare Eq. 7.48 and Eq. 7.50). To generalize the solution of these four possible circuits, we let x(t) represent the unknown quantity, giving x(t) four possible values. It can