1 Chapter 14 Inductive Transients

2 14.0Preview [page 519] Capacitive circuits capacitor voltage cannot change instantanously. Inductive circuits inductor current cannot change instantanously. Inductive transients result when circuits containing inductance are disturbed Potential destructive and dangerous - extremely large and damaging voltage may result when breaking current in inductive circuit.

3 14.1Introduction [page 520] transient occur because current in inductance cannot change instantanously [Fig. 14-1, page 520] when switch is closed, counter emf appears across the inductance attempting to stop the current from changing Figure Transient is due to inductance. For fixed resistance, the larger the inductance, the longer the transient lasts. (a) No transient occurs in a purely resistive circuit (b) Adding inductance causes a transient to appear. R is held constant here.

4 Continuity of current [page 520] step changed current cannot be occurred Inductor voltage [page 521] inductor voltage jumps from 0 to E just after switch is closed and then decays to 0V In Fig [page 521, we can see (a) Inductive circuit with switch open, current i = 0. (b) Inductive circuit just after the Switch has been closed. Current is still equal to zero. (c) Voltage across L

5 Open-circuit equivalent of an inductance [Fig. 14-3, page 521] an inductor with zero initial current looks an open circuit at the instant of switching Figure 14-3 Inductor swith zero initial current looks like an open circuit at the instant the switch is closed. Initial condition circuits [page 521] by replacing inductors with open circuits

6 Example 14-1 [page 521] A coil and two resistors are connected to a 20-voltage V L at the instant the switch is closed. Refer to Figure 14-4 [page 522]: (a) Original circuit (b) Initial condition network Solution Replace the inductance with an open circuit. This yields the network shown in Figure 14-4 (b). Thus i = E/R T = 20 V/10  = 2 A and the voltage across R 2 is V 2 = (2 A) (4  ) = 8 V. Since V L = V 2, V L = 8 volts as well.

7 14.2Current Buildup Transients [page 523] Current V L + V R = E Substituting V L = Ldi/dt and V R = Ri into Equation 14-1 yields

8 Example 14-2 [page 523] For the circuit of Figure 14-7, suppose E = 50 V, R = 10 , and L = 2 H : a.Determine the expression for i b.Compute and tabulate values of i at t = 0 +, 0.2, 0.4, 0.6, 0.8, and 1.0 s. c.Using these values, plot the current. Solution a.Substituting the values into Equation 14-3 yields i = E / R( 1 - e -Rt/L ) = 50V/10  ( 1 – e -10t/2 ) = 5 ( 1 – e -5t ) A At t = 0 + s, i = 5(1-e -5t ) = 5(1 -e 0 ) = 5(1-1) = 0 A. At t = 0.2 s, i = 5(1-e -5(0.2) ) = 5(1 - e -1 ) = 3.16 A At t = 0.4 s, i = 5(1-e -5(0.4) ) = 5(1 - e -2 ) = 4.32 A. Continuing in this manner, you get Table 14-1.

9 Example 14-2 [Cont’d] - solution c.Values are plotted in Figure Note that this curve looks exactly like the curves we determined intuitively in Figure 14-1(b). Example 14-2 [page 524] Refer to Table 14-1 and Figure 14-8 for current buildup transient.

10 Circuit voltages [page 524] With i known, circuit voltages can be determined. Consider voltage V R. Since V R = Ri, when you multiply R times Equation 14-3, you get V R = E(1 - e -Rt/L ) (V)(14-4) Note that V R has exactly the same shape as the current. Now consider V L. Voltage V L can be found by subtracting V R from E as per Equation 14-1: V L = E - V R = E - E(1 - e -Rt/L ) = E - E + Ee -Rt/L Thus, V L = Ee -Rt/L (14-5) An examination of Equation 14-5 shows that V L has an initial value of E and then decays exponentially to zero.

11 Example 14-3 [page 524] Repeat Example 14-2 for voltage V L. Solution a.From equation 14-5, V L = Ee -Rt/L = 50e -5t volts b.At t = 0 + s, V L = 50e -5t = 50e 0 = 50(1) = 50 V. At t = 0.2 s, V L = 50e -5(0.2) = 50e -1 = 18.4 V. At t = 0.4 s, V L = 50e -5(0.4) = 50e -2 = 6.77 V. Continuing in this manner, you get Table 14-2 [page 525].. c.The waveform is shown in Figure 14-9 [page 525].

12 Time constant [page 525] τ = L / R(s) (14-6)  has units of seconds. (This is left as an exercise for the student.) In terms of , equations 14-3, 14-4, and 14-5 may be written as i = E / R( 1 - e -t/τ ) (A) (14-7) V L = Ee -t/  (V)(14-8) V R = E(1 - e -t/  ) (V)(14-9) Curves are plotted in Figure [page 526] versus time constant. As expected, transitions take approximately 5  ; thus, for all practical purposes, inductive transients last five time constants.

13 EXAMPLE 14-4 In a circuit where L = 2 mH, transients lasts 50  s. What is R? Solution Transients last five time constants. Thus,  = 50  s/5 = 10  s. Now  = L/R. Therefore, R = L/  = 2 mH/10  s = 200 . EXAMPLE 14-5 For an RL circuit, i = 40(1 - e -5t ) A and V L = 100e -5t V. a. What are E and  ? b. What is R? c.Determine L.

14 EXAMPLE 14-5 (Cont’d) Solution a.From Equation 14-8, V L = Ee -t/  = 100e -5t. Therefore, E = 100 V and  = = 0.2s. b.From Equation 14-7, i = E / R( 1 – e -t/τ ) = 40 ( 1 – e -5t ) A Therefore, E/R = 40 A and R = E/40 A = 100 V/40 A = 2.5 . c.  = L/R. Therefore, L = R  = (2.5)(0.2) = 0.5 H. Effect of RL on transient [Fig , page 526] larger L, longer transient (fixed R) larger R, shorter transient (fixed L)

15 Learning Check: [page 527] 1.For the circuit of Figure 14-12, the switch is closed at t = 0 s. a. Determine expressions for V L and i. b. Compute V L and i at t = 0 +, 10  s, 20  s, 30  s, 40  s, and 50  s. c. Plot curves for V L and i. Solution: [page 546] 14-3Interrupt current in an inductive circuit [page 527] when inductive current is interrupted, a great deal of energy is released in a very short time. This will create a huge voltage referred to as inductive kickback. This voltage may damage equipment and can create a shock hazard. Flashovers are generally undesirable; however, they can be controlled through proper engineering design. On the other hand, the large voltages created by breaking inductive currents have their uses, one is the ignition system of automobiles.

16 The basic ideas Discharge Resistor R 2 helps limit the size of the induced voltage when the switch is opened. The inductor at switching [page 529] Because inductor current is the same just after switching, an inductor with an initial current looks like a current source at the instant of switching

De-Energizing Transients [page 529] [Fig , page 530] KVL yields V L + V R1 + V R2 = 0 Substituting V L = Ldi/dt, V R1 = R 1 ． i, and V R2 = R 2 ． i yields Ldi/dt +(R 1 + R 2 )i = 0 i = I O e -t/  ’ (A) τ’ = L / R T = L / (R 1 + R 2 ) (s) initial current I O = E/R 1 i = E / R 1 ． e –t/τ’ (A) Refer to Figure Circuit for studying decay transients.

18 Example 14-6 [page 530] For Figure 14-19, assume the current has reached steady state with the switch closed. Suppose that E = 120 V, R 1 = 30 , R 2 = 600 , and L = 126 mH: a. Determine I O. b. Determine the decay time constant. c. Determine the equation for the current decay. d. Compute the current i at t = 0 + s and t = 0.5 ms. Solution a. Consider Figure 14-19(a). Since the inductor looks like a short circuit to dc, I O = E/R 1 = 4A. b. Consider Figure 14-19(b).  ’ = L/(R 1 + R 2 ) = 126 mH/630  = 0.2 ms. c. i = I o e -t/  ’ = 4e -t/0.2 ms A. d. At t = 0 + s, i = 4e -0 = 4 A. At t = 0.5 ms, i = 4e -0.5 ms/0.2 ms = 4e -2.5 = A.

19 Refer to Figure [page 531] for inductor voltage during decay phase. V L = V O e -t/  ’ V O = -I O (R 1 + R 2 ) = -I O R T V L = - I O R T e -t/  ’ if the current has reached steady state I O = E/R 1, V L = -E ( 1 + R 2 /R 1 ) ‧ e –t/τ’ V R1 = R 1 I O e -t/  ’ V R2 = R 2 I O e -t/  ’ current has reached steady state before switching, these become V R1 = Ee -t/  ’ V R2 = (R 2 /R 1 ) ‧ Ee -t/  ’

More Complex Circuits [page 531] Example 14-8 [page 532] Determine i L for the circuit of Figure 14-21(a) if L = 5 H. Solution The circuit can be reduced to its Thévenin equivalent (b) as you saw in Chapter 11 (Section 11.5). For the circuit,  = L/R Th = 5 H/200  = 25 ms. Now apply Equation Thus, i L = E Th / R Th ( 1 – e –t/τ ) = 40/200 (1 – e –t/25ms ) = 0.2 (1 – e –40t ) (A)

21 Example 14-9 [page 532] For the circuit of Example 14-7, at what time does current reach 0.12 amps? Solution i L = 0.2(1 - e -40t ) (A) Thus, 0.12 = 0.2 (1 - e -40t ) (Figure 14-22) 0.6 = 1 - e -40t e -40t = 0.4 Taking the natural log of both sides, In e -40t = In t = t = 22.9 ms

22 EXAMPLE [page 533] Refer to the circuit of Figure 14-23: a. Close the switch and determine equations for i L and V L. b. After the circuit has reached steady state, open the switch and determine equations for i L and V L during the decay phase. c. Sketch i L and V L. Solution a.Thévenize the circuit to the left of L. As indicated in Figure 14-24(a), R Th = 60  = 100 . From (b), E Th = V 2. Voltage V 2 can be found by the voltage divider rule as 60  / (60  + 30  ) x 300V = 200V

23 EXAMPLE cont‘d [page 534] The Thevenin equivalent circuit is shown in Figure 14-25,  = L/R Th = 50 ms. Thus, i L = E Th / R Th ( 1 – e –t/τ ) = 200/100 (1 – e –t/50ms ) = 2 (1 – e –20t ) (A) V L = E Th e –t/τ = 200 e –20t (A) In both equations, t is measured from the instant the switch is closed. Current rises to a steady state value of 2 A.

24 EXAMPLE cont‘d [page 534] Refer to Figure [page 535] for circuit and current for the buildup phase. b.The current has an initial value of 2 A when the switch is opened, as shown in Figure 14-25(b). It then decays to zero through a resistance of = 140  as shown in Figure Thus,  ’ = 5H/140  = 35.7 ms. If t = 0 s is redefined as the instant the switch is opened, the equation for the decay current is i L = I O e -t/  ’ = 2e -t/35.7ms = 2e -28t A Figure [page 535] - The circuit of Figure as it looks during the decay phase.

25 EXAMPLE cont‘d [page 534] Now consider voltage. As indicated in Figure 14-26(b), the voltage across L just after the switch is open is V O = -280 V. Thus V L = V O e -t/  ’ = - 280e -28t V c. The waveforms are shown in Figure The decay transient is shorter than the buildrup transient because the circuit resistance is larger and hence the decay time constant is smaller.

26 Example [page 536] The circuit of Figure 14-28(a) is in steady state with the switch open. At t = 0 s, the switch is closed. a. Sketch the circuit as it looks after the switch is closed and determine  ’. b. Determine current i L at t = 0 + s. c. Determine the expression for i L. d. Determine V L at t = 0 + s. e. Determine the expressio for V L. f. How long does the transient last? g. Sketch i L and V L.

27 Example cont‘d [page 537] Solution a.When you close the switch, you short out E and R 1, leaving the decay circuit of (b). Thus  ’ = L/R 2 = 100 mH/40  = 2.5 ms. b.In steady state with the switch open, i L = I O = 100/50 = 2A. This is the current just before the switch is closed. Therefore, just after the switch is closed, i L will still be 2 A. c. i L decays from 2 A to 0. From Equation 14-10, i L = I O e -t/  ’ = 2e -t/2.5ms = 2e -400t A.

28 Example cont‘d [page 537] Solution d.KVL yields V L = -VR 2 = -R 2 I O = -(40  ) (2A) = -80 V. Thus, V O = -80 V. e.V L decays from -80V to 0. Thus, V L = V O e -t/  = -80e -400t V. f.Transients last 5  =5(2.5ms) = 12.5 ms. g.See Figure [page 537]