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Figure 46 Response of a capacitive displacement transducer
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b Circuit model
Bridge configuration
Figure 47 Capacitive pressure transducer, and related bridge circuit
with a conducting material) and of a de ecting plate (typically made of steel) sandwiched between the glass disks Pressure inlet ori ces are provided, so that the de ecting plate can come into contact with the uid whose pressure it is measuring When the pressure on both sides of the de ecting plate is the same, the capacitance between terminals b and d, Cbd , will be equal to that between terminals b and c, Cbc If any pressure differential exists, the two capacitances will change, with an increase on the side where the de ecting plate has come closer to the xed surface and a corresponding decrease on the other side This behavior is ideally suited for the application of a bridge circuit, similar to the Wheatstone bridge circuit illustrated in Example 212, and also shown in Figure 47 In the bridge circuit, the output voltage, vout , is precisely balanced when the differential pressure across the transducer is zero, but it will deviate from zero whenever the two capacitances are not identical because of a pressure differential across the transducer We shall analyze the bridge circuit later
The Ideal Inductor The ideal inductor is an element that has the ability to store energy in a magnetic eld Inductors are typically made by winding a coil of wire around a core, which can be an insulator or a ferromagnetic material, as shown in Figure 48 When a
4
AC Network Analysis
Magnetic flux lines
Iron core inductor
current ows through the coil, a magnetic eld is established, as you may recall from early physics experiments with electromagnets2 In an ideal inductor, the resistance of the wire is zero, so that a constant current through the inductor will ow freely without causing a voltage drop In other words, the ideal inductor acts as a short circuit in the presence of DC currents If a time-varying voltage is established across the inductor, a corresponding current will result, according to the following relationship: diL dt
vL (t) = L
(49)
where L is called the inductance of the coil and is measured in henrys (H), where
i (t) L + vL (t) = Ldi dt _
1 H = 1 V-s/A
(410)
Circuit symbol
Figure 48 Iron-core inductor
Henrys are reasonable units for practical inductors; millihenrys (mH) and microhenrys ( H) are also used It is instructive to compare equation 49, which de nes the behavior of an ideal inductor, with the expression relating capacitor current and voltage: dvC (411) dt We note that the roles of voltage and current are reversed in the two elements, but that both are described by a differential equation of the same form This duality between inductors and capacitors can be exploited to derive the same basic results for the inductor that we already have for the capacitor simply by replacing the capacitance parameter, C, with the inductance, L, and voltage with current (and vice versa) in the equations we derived for the capacitor Thus, the inductor current is found by integrating the voltage across the inductor: iC (t) = C iL (t) = 1 L
t
vL dt
(412)
If the current owing through the inductor at time t = t0 is known to be I0 , with I0 = iL (t = t0 ) = 1 L
t0
vL dt
(413)
then the inductor current can be found according to the equation iL (t) = 1 L
t t0
vL dt + I0
t t0
(414)
Series and parallel combinations of inductors behave like resistors, as illustrated in Figure 49, and stated as follows:
Inductors in series add Inductors in parallel combine according to the same rules used for resistors connected in parallel
2 See
also 15
Part I
Circuits
L1 LEQ =
1 1 + 1 + 1 L L L 2 3 1
LEQ = L1 + L2 + L3
L3 Inductances in series add
Inductances in parallel combine like resistors in parallel
Figure 49 Combining inductors in a circuit
EXAMPLE 44 Calculating Inductor Voltage from Current
Problem
Calculate the voltage across the inductor from knowledge of its current
Solution
Known Quantities: Inductor current; inductance value Find: Inductor voltage Schematics, Diagrams, Circuits, and Given Data:
0 01 01 + t 4 4 01 iL (t) = 13 01 01 t 4 4 0 L = 10 H
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