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FREQUENCY RESPONSE, FILTERS, AND RESONANCE
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Fig. 12-45
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Fig. 12-46
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The semicircular admittance locus of branch 2 has the radius r 1=2R 0:125 S. The total admittance is the sum of the xed admittance Y1 and the variable admittance Y2 . In Fig. 12-47, the semicircular locus is added to the xed complex number Y1 . The circuit resonance occurs at points a and b, where YT is real. YT 0:417 j0:088 1 4 jXC
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Fig. 12-47 which is real if
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2 XC 11:36XC 16 0
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or XC1 9:71 , XC2 1:65 .
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With ! 5000 rad/s, C1 20:6 mF C2 121 mF
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12.21 Show by locus diagrams that the magnitude of the voltage between points A and B in Fig. 12-48 is always one-half the magnitude of the applied voltage V as L is varied.
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Fig. 12-48
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FREQUENCY RESPONSE, FILTERS, AND RESONANCE
[CHAP. 12
Branch-1 current I1 passes through two equal resistors R. Thus A is the midpoint on the phasor V, as shown in Fig. 12-49.
Fig. 12-49 Branch 2 has a semicircular Y-locus [see Fig. 12-28(c)]. Then the current locus is also semicircular, as shown in Fig. 12.50(a). The voltage phasor diagram, Fig. 12-50(b), consists of the voltage across the inductance, VBN , and the voltage across R1 , VMB . The two voltages add vectorially, V VMN VBN VMB
Fig. 12-50 Because I2 lags VBN by 908, VBN and VMB are perpendicular for all values of L in Fig. 12-50(b). As L varies from 0 to 1, point B moves from N toward M along the semicircle. Figures 12-49 and 12-50(b) are superimposed in Fig. 12-50(c). It is clear that VAB is a radius of the semicircle and therefore, jVAB j 1 jVj 2 Further, the angle  by which VAB lags V is equal to 2, where  tan 1 !L=R1 .
Supplementary Problems
12.22 A high-pass RL circuit has R1 50 k and L2 0:2 mH. (a) Find ! if the magnitude of the voltage transfer function is jHv1 j 0:90. (b) With a load R 1 M across L2 , nd jHv j at ! 7:5 108 rad/s. Ans: a 5:16 108 rad/s; (b) 0.908 Obtain Hv1 for a high-pass RL circuit at ! 2:5!x , R 2 k, L 0:05 H. Ans: 0:928 21:808
12.23 12.24
A low-pass RC circuit under no-load has R1 5 k. (a) Find C2 if jHv j 0:5 at 10 kHz. (b) Obtain Hv at 5 kHz. (c) What value of C2 results in jHv j 0:90 at 8 kHz (d) With C2 as in (a), nd a new value for R1 to result in jHv j 0:90 at 8 kHz. Ans: a 5:51 mF; (b) 0:756 40:898; (c) 1:93 mF; d 1749  A simple voltage divider would consist of R1 and R2 . If stray capacitance Cs is present, then the divider would generally be frequency-dependent. Show, however, that V2 =V1 is independent of frequency for the circuit of Fig. 12-51 if the compensating capacitance C1 has a certain value. Ans: C1 R2 =R1 Cs
CHAP. 12]
FREQUENCY RESPONSE, FILTERS, AND RESONANCE
Fig. 12-51
Fig. 12-52
Assume that a sinusoidal voltage source with a variable frequency and Vmax 50 V is applied to the circuit shown in Fig. 12-52. (a) At what frequency f is jIj a minimum (b) Calculate this minimum current. (c) What is jIC j at this frequency Ans: a 2:05 kHz; (b) 2.78 mA; (c) 10.8 mA A 20-mF capacitor is in parallel with a practical inductor represented by L 1 mHz in series with R 7 . Find the resonant frequency, in rad/s and in Hz, of the parallel circuit. Ans: 1000 rad/s, 159.2 Hz What must be the relationship between the values of RL and RC if the network shown in Fig. 12-53 is to be resonant at all frequencies Ans: RL RC 5 
Fig. 12-53
Fig. 12-54
For the parallel network shown in Fig. 12-54, (a) nd the value of R for resonance; (b) convert the RC branch to a parallel equivalent. Ans: a 6:0 ; b Rp 6:67 ; XCp 20  For the network of Fig. 12-55(a), nd R for resonance. Obtain the values of R 0 , XL , and XC in the parallel equivalent of Fig. 12-55(b). Ans: R 12:25 ; R 0 7:75 ; XL 25 ; XC 25 
Fig. 12-55 12.31 Branch 1 of a two-branch parallel circuit has an impedance Z1 8 j6  at ! 5000 rad/s. Branch 2 contains R 8:34  in series with a variable capacitance C. (a) Find C for resonance. (b) Sketch the admittance locus diagram. Ans: a 24 mF b See Fig. 12-56
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