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Chapter 14

Inductive Transients

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14.0 Preview [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.

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14.1 Introduction [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 14-1 - 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.

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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. 14-2 [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

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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

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Example 14-1 [page 521]

• A coil and two resistors are connected to a 20-voltage VL 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/RT = 20 V/10 = 2 A and the voltage across R2 is

V2 = (2 A) (4 ) = 8 V. Since VL = V2, VL = 8 volts as well.

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14.2 Current Buildup Transients [page 523]

Current

VL + VR = E

Substituting VL = Ldi/dt and VR = Ri into Equation 14-1 yields

ERidtdiL

ALRteREi )/1(

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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 -e0) = 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.

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Example 14-2 [Cont’d] - solution

c. Values are plotted in Figure 14-8. 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.

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Circuit voltages [page 524]

• With i known, circuit voltages can be determined. Consider voltage V

R. Since VR = Ri, when you multiply R times Equation 14-3, you get

VR = E(1 - e-Rt/L) (V) (14-4)

Note that VR has exactly the same shape as the current. Now consider VL. Voltage VL can be found by subtracting VR from E as per Equation 14-1:

VL = E - VR = E - E(1 - e-Rt/L) = E - E + Ee-Rt/L

Thus,

VL = Ee-Rt/L (14-5)

An examination of Equation 14-5 shows that VL has an initial value of E and then decays exponentially to zero.

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Example 14-3 [page 524]

• Repeat Example 14-2 for voltage VL.

Solution

a. From equation 14-5,

VL = Ee-Rt/L = 50e-5t volts

b. At t = 0+ s, VL = 50e-5t = 50e0 = 50(1) = 50 V.

At t = 0.2 s, VL = 50e-5(0.2) = 50e-1 = 18.4 V.

At t = 0.4 s, VL = 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].

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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)

VL = Ee-t/ (V) (14-8)

VR = E(1 - e-t/ ) (V) (14-9)

Curves are plotted in Figure 14-10 [page 526] versus time constant. As expected, transitions take approximately 5 ; thus, for all practical purposes, inductive transients last five time constants.

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EXAMPLE 14-4

In a circuit where L = 2 mH, transients lasts 50s. 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 VL = 100e-5t V.

a. What are E and ?

b. What is R?

c. Determine L.

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EXAMPLE 14-5 (Cont’d)

Solution

a. From Equation 14-8, VL = 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. 14-11, page 526]

• larger L, longer transient (fixed R)

• larger R, shorter transient (fixed L)

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Learning Check: [page 527]

1.For the circuit of Figure 14-12, the switch is closed at t = 0 s .

a. Determine expressions for VL and i.

b. Compute VL and i at t = 0+, 10 s, 20 s, 30 s, 40 s, and 50 s.

c. Plot curves for VL and i.

Solution: [page 546]

14-3 Interrupt 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.

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The basic ideas

• Discharge Resistor R2 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

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14.4 De-Energizing Transients [page 529][Fig. 14-19, page 530]

KVL yields VL + VR1 + VR2 = 0

Substituting VL = Ldi/dt, VR1 = R1 ． i, and VR2 = R2 ． i yields

Ldi/dt +(R1 + R2)i = 0

i = IOe-t/’ (A)

τ’ = L / RT = L / (R1 + R2) (s)

initial current IO = E/R1

i = E / R1 ． e –t/τ’ (A)

Refer to Figure 14-19 - Circuit for studying decay transients.

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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, R1 = 30, R2 = 600, and L = 126 mH:

a. Determine IO.

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, IO = E/R1 = 4A.

b. Consider Figure 14-19(b).

’ = L/(R1 + R2) = 126 mH/630 = 0.2 ms.

c. i = Ioe-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 = 0.328 A.

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Refer to Figure 14-20 [page 531] for inductor voltage during decay phase.

VL = VOe-t/’ VO = -IO(R1 + R2) = -IORT

VL = - IORTe-t/’

if the current has reached steady state IO = E/R1,

VL = -E ( 1 + R2/R1 ) ‧ e –t/τ’

VR1 = R1IO e-t/’

VR2 = R2IO e-t/’

current has reached steady state before switching, these become

VR1 = Ee-t/’

VR2 = (R2/R1) ‧Ee-t/’

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14.5 More Complex Circuits [page 531]

Example 14-8 [page 532]

Determine iL 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/RTh = 5 H/200 = 25 ms. Now apply Equation 14-7. Thus,

iL = ETh / RTh ( 1 – e –t/τ) = 40/200 (1 – e –t/25ms ) = 0.2 (1 – e –40t) (A)

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Example 14-9 [page 532]

For the circuit of Example 14-7, at what time does current reach 0.12 amps?

Solution

iL = 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 0.4

-40t = - 0.916

t = 22.9 ms

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EXAMPLE 14-10 [page 533]

Refer to the circuit of Figure 14-23:

a. Close the switch and determine equations for iL and VL.

b. After the circuit has reached steady state, open the switch and determine equations

for iL and VL during the decay phase.

c. Sketch iL and VL.

Solution

a. Thévenize the circuit to the left of L. As indicated in Figure 14-24(a),

RTh = 60 30 + 80 = 100 . From (b), ETh = V2. Voltage V2 can be found by the voltage divider rule as

60 / (60 + 30 ) x 300V = 200V

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EXAMPLE 14-10 - cont‘d [page 534]

The Thevenin equivalent circuit is shown in Figure 14-25, = L/RTh = 50 ms. Thus,

iL = ETh / RTh ( 1 – e –t/τ) = 200/100 (1 – e –t/50ms ) = 2 (1 – e –20t) (A)

VL = ETh 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.

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EXAMPLE 14-10 - cont‘d [page 534]

Refer to Figure 14-25 [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 60 + 80 = 140 as shown in Figure 14-26.

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

iL = IOe-t/ ’ = 2e-t/35.7ms = 2e-28t A

Figure 14-26 [page 535] - The circuit of Figure 14-23 as it looks during the decay phase.

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EXAMPLE 14-10 - 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 VO = -280 V. Thus

VL = VOe-t/’ = - 280e-28t V

c. The waveforms are shown in Figure 14-27. The decay transient is shorter than the buildrup transient because the circuit resistance is larger and hence the decay time constant is smaller.

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Example 14-11 [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 iL at t = 0+ s.

c. Determine the expression for iL.

d. Determine VL at t = 0+ s.

e. Determine the expressio for VL.

f. How long does the transient last?

g. Sketch iL and VL.

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Example 14-11 - cont‘d [page 537]

Solution

a. When you close the switch, you short out E and R1, leaving the decay circuit of (b). Thus ’ = L/R2 = 100 mH/40 = 2.5 ms.

b. In steady state with the switch open, iL = IO = 100/50 = 2A. This is the current just before the switch is closed. Therefore, just after the switch is closed, iL will still be 2 A.

c. iL decays from 2 A to 0. From Equation 14-10,

iL = IOe-t/ ’ = 2e-t/2.5ms = 2e-400t A.

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Example 14-11 - cont‘d [page 537]

Solution

d. KVL yields VL = -VR2 = -R2IO = -(40) (2A) = -80 V.

Thus, VO = -80 V.

e. VL decays from -80V to 0. Thus, VL = VO e-t/ = -80e-400t V.

f. Transients last 5 =5(2.5ms) = 12.5 ms.

g. See Figure 14-29. [page 537]