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t < 1 ms 1 t 5 ms 5 t 9 ms 9 t 13 ms t > 13 ms
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The inductor current is plotted in Figure 410
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Assumptions: iL (t = 0) 0
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01 + vL (t) 008 iL (t) (mA) iL (t) L 006 004 002 0 0 5 10 Time (ms) 15
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Analysis: Using the de ning differential relationship for the inductor, we may obtain the
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voltage by differentiating the current: diL (t) dt Piecewise differentiating the expression for the inductor current, we obtain: t < 1 ms 0V 025 V 1 < t 5 ms 0V 5 < t 9 ms vL (t) = 025 V 9 < t 13 ms 0V t > 13 ms vL (t) = L The inductor voltage is plotted in Figure 411
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03 02 01 0 01 02 03 04
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vL (t) (V)
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10 Time (ms)
Comments: Note how the inductor voltage has the ability to change instantaneously! Focus on Computer-Aided Tools: The MatlabTM m- les used to generate the plots of
Figures 410 and 411 may be found in the CD-ROM that accompanies this book
EXAMPLE 45 Calculating Inductor Current from Voltage
Problem
Calculate the current through the inductor from knowledge of the terminal voltage and of the initial current
Solution
Known Quantities: Inductor voltage; initial condition (current at t = 0); inductance
value
Find: Inductor current Schematics, Diagrams, Circuits, and Given Data:
t <0s 0V 10 mV 0<t 1s v(t) = 0V t >1s L = 10 mH; iL (t = 0) = I0 = 0 A
The terminal voltage is plotted in Figure 412(a)
Part I
Circuits
0 1 2 3 4 5 6 7 8 9 10 05
05 Time (s) (a)
0 01 02 03 04 05 06 07 08 09 1 05
v (t) (mV)
iL (t) (A)
05 Time (s) (b)
Figure 412 Assumptions: iL (t = 0) = I0 = 0 Analysis: Using the de ning integral relationship for the inductor, we may obtain the
voltage by integrating the current: iL (t) = 1 L
v(t) dt +iL (t0 )
t t0
t 10 2 1 ( 10 10 3 ) dt +I0 = t + 0 = t A L 0 10 2 iL (t) = 1 A The inductor current is plotted in Figure 412b
0 t 1s t >1s
Comments: Note how the inductor voltage has the ability to change instantaneously! Focus on Computer-Aided Tools: The MatlabTM m- les used to generate the plots of
Figures 412(a) and (b) may be found in the CD-ROM that accompanies this book
Energy Storage in Inductors The magnetic energy stored in an ideal inductor may be found from a power calculation by following the same procedure employed for the ideal capacitor The instantaneous power in the inductor is given by PL (t) = iL (t)vL (t) = iL (t)L diL (t) d = dt dt 1 2 Li (t) 2 L
(415)
Integrating the power, we obtain the total energy stored in the inductor, as shown in the following equation: WL (t) = PL (t ) dt = d dt 1 2 Li (t ) dt 2 L
(416)
4
AC Network Analysis
WL (t) =
1 2 Li (t) 2 L
Energy stored in an inductor (J)
Note, once again, the duality with the expression for the energy stored in a capacitor, in equation 48
EXAMPLE 46 Energy Storage in an Ignition Coil
Problem
Determine the energy stored in an automotive ignition coil
Solution
Known Quantities: Inductor current initial condition (current at t = 0); inductance value Find: Energy stored in inductor Schematics, Diagrams, Circuits, and Given Data: L = 10 mH; iL = I0 = 8 A Analysis:
WL =
1 2 1 Li = 10 2 64 = 32 10 2 = 320 mJ 2 L 2
Comments: A more detailed analysis of an automotive ignition coil is presented in
5 to accompany the discussion of transient voltages and currents
FOCUS ON MEASUREMENTS
Analogy between Electrical and Hydraulic Circuits
A useful analogy can be made between the ow of electrical current through electrical components and the ow of incompressible uids (eg, water, oil) through hydraulic components The analogy starts with the observation that the volume ow rate of a uid in a pipe is analogous to current ow in a conductor Similarly, the pressure drop across the pipe is analogous to the voltage drop across a resistor Figure 413 depicts this relationship graphically The uid resistance presented by the pipe to the uid ow is analogous to an electrical resistance: The pressure difference between the two ends of the pipe, (P1 P2 ), causes uid ow, qf , much like a potential difference across a resistor forces a current ow: 1 qf = (p1 p2 ) Rf i= 1 (v1 v2 ) R
Part I
Circuits
v1 i
v2 Rf
v1 i + p2 C v _ qf v2 p2 gas P1 Cf qf
p2 + p _
Figure 413 Analogy between electrical and uid resistance
Figure 414 Analogy between uid capacitance and electrical capacitance
The analogy between electrical and hydraulic circuits can also be extended to include energy storage effects corresponding to capacitance and inductance If the uid enters a vessel that has some elasticity (compressibility), energy can be stored in the expansion and contraction of the vessel walls (if this reminds you of a mechanical spring, you are absolutely right!) This phenomenon gives rise to a uid capacitance effect very similar to the electrical capacitance phenomenon we have just introduced Energy is stored in the compression and expansion of the gas; this form of energy storage is of the potential energy type Figure 414 depicts a so-called gas bag accumulator, which consists of a two-chamber arrangement that permits uid to displace a membrane separating the incompressible uid from a compressible uid (eg, air) If, for a moment, we imagine that the reference pressure, p2 , is zero (think of this as a ground or reference pressure), and that the voltage is the reference or ground voltage, we can create an analogy between an electrical capacitor and a uid capacitor (the gas-bag accumulator) as shown in Figure 414 qf = Cf i=C d p dp1 = Cf dt dt
d v dv1 =C dt dt The nal element in the analogy is the so-called uid inertance parameter, which is analogous to inductance in the electrical circuit Fluid inertance, as the name suggests, is caused by the inertial properties, ie, the mass, of the uid in motion As you know from physics, a particle in motion has kinetic energy associated with it; uid in motion consists of a collection of particles, and it also therefore must have kinetic energy storage properties If you wish to experience the kinetic energy contained in a uid in motion, all you have to do is hold a re hose and experience the reaction force caused by the uid in motion on your body! Figure 415 depicts the analogy between electrical inductance and uid inertance These analogies and the energy equations are summarized in Table 42 p = p1 p2 = If v = v1 v2 = L dqf dt
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