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005 vout (V)
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Bridge output voltage
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Figure 440 Displacement input and bridge output voltage for capacititve displacement transducer
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Admittance In 3, it was suggested that the solution of certain circuit analysis problems was handled more easily in terms of conductances than resistances In AC circuit analysis, an analogous quantity may be de ned, the reciprocal of complex impedance Just as the conductance, G, of a resistive element was de ned as the inverse of the resistance, the admittance of a branch is de ned as follows: 1 S (466) Z Note immediately that whenever Z is purely real that is, when Z = R + j 0 the admittance Y is identical to the conductance G In general, however, Y is the complex number Y = Y = G + jB (467)
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where G is called the AC conductance and B is called the susceptance; the latter plays a role analogous to that of reactance in the de nition of impedance Clearly, G and B are related to R and X However, this relationship is not as simple as an inverse Let Z = R + j X be an arbitrary impedance Then, the corresponding admittance is 1 1 Y = = (468) Z R + jX In order to express Y in the form Y = G + j B, we multiply numerator and denominator by R j X: Y = 1 R jX R jX = 2 R + jX R jX R + X2
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R X = 2 j 2 2 R +X R + X2 and conclude that R G= 2 R + X2 X B= 2 R + X2
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(469)
(470)
Notice in particular that G is not the reciprocal of R in the general case! The following example illustrates the determination of Y for some common circuits
EXAMPLE 414 Admittance
Problem
Find the equivalent admittance of the two circuits shown in Figure 441
Solution
C = 3 F
Known Quantities: = 2 103 rad/s; R1 = 150
; L = 16 mH; R2 = 100
a R1 Yab L b (a) a
4
AC Network Analysis
Find: The equivalent admittance of the two circuits Analysis: Circuit (a): First, determine the equivalent impedance of the circuit:
Zab = R1 + j L Then compute the inverse of Zab to obtain the admittance: Yab = R1 j L 1 = 2 R1 + j L R1 + ( L)2 1 1 = 3968 10 3 j 7976 10 3 S = 50 + j 2 103 50 + j 1005 R2 1 = j C 1 + j R2 C
Substituting numerical values gives Yab =
Circuit (b): First, determine the equivalent impedance of the circuit: Zab = R2
b (b)
Then compute the inverse of Zab to obtain the admittance: Yab = 1 + j R2 C 1 = + j C = 001 + j 0019 S R2 R2
Comments: Note that the units of admittance are siemens, that is, the same as the units
of conductance
Focus on Computer-Aided Tools: You will nd the solution to the same example
computed by MathCad in the electronic les that accompany this book
Check Your Understanding
414 Add the sinusoidal voltages v1 (t) = A cos( t + ) and v2 (t) = B cos( t + ) using phasor notation, and then convert back to time-domain form, for: a A = 15 V, = 10 ; B = 32 V, = 25 b A = 50 V, = 60 ; B = 24, = 15 415 Add the sinusoidal currents i1 (t) = A cos( t + ) and i2 (t) = B cos( t + ) for:
a A = 009 A, = 72 ; B = 012 A, = 20 b A = 082 A, = 30 ; B = 05 A, = 36
416 Compute the equivalent impedance of the circuit of Example 413 for = 1,000 and 100,000 rad/s 417 Compute the equivalent admittance of the circuit of Example 413 418 Calculate the equivalent series capacitance of the parallel R2 -C circuit of Example
413 at the frequency = 10 rad/s
AC CIRCUIT ANALYSIS METHODS
This section will illustrate how the use of phasors and impedance facilitates the solution of AC circuits by making it possible to use the same solution methods
Part I
Circuits
developed in 3 for DC circuits The AC circuit analysis problem of interest in this section consists of determining the unknown voltage (or currents) in a circuit containing linear passive circuit elements (R, L, C) and excited by a sinusoidal source Figure 442 depicts one such circuit, represented in both conventional time-domain form and phasor-impedance form
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