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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 .

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6. (a) Partial converses: interchange (2) and (3) or (1) and (3) Partial inverses: negate (1) and (3) or (2) and (3) (b) Partial converses: interchange (1) and (4) or (2) and (4) or (3) and (4) Partial inverses: negate (1) and (4) or (2) and (4) or (3) and (4) 7. (a) Necessary and sufficient; (b) necessary but not sufficient; (c) neither necessary nor sufficient; (d) sufficient but not necessary; (e) necessary and sufficient; ( f ) sufficient but not necessary; (g) necessary but not sufficient

Note that the high-frequency cuto point is independent of Cgs as long as the source internal impedance is negligible. (See Problem 8.40.)

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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

High-frequency models of transistors characteristically include a capacitor path from input to ouput, modeled as admittance YF in the two-port network of Fig. 8-12(a). This added conduction

1 1 1. (a) 6(72) or 294 yd2; (b) 2(8)(62) 2(8)(14) 2(62)(14) or 510 ft2; 1 (c) 4(3.14)302 or 11,304 m2; (d) (3.14)(10)(10 42) or 911 yd2

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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

Io + V2 _

2. (a) 363 or 46,656 in3; (b) 1003 or 1,000,000 cm3 3. (a) 27 in3; (b) 91 in3; (c) 422 in3; (d) 47 in3; (e) 2744 in3 4. (a) 3(82)(8) or 204 in3; (b) 2(9)(9) or 162 ft3; (c) 2(6)(6.4) or 13 ft3 5. (a) 904 m3; (b) 1130 ft3; (c) 18 ft3 6. (a) V 7. (a) 6e3

CHAP. 12]

1 2 3 pr h; 1 1

YF (1 _ KR)

Fig. 12-51

Fig. 12-52

1 2 3 s h;

Fig. 8-12

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

1 3 lwh;

path generally increases the di culty of analysis; we would like to replace it with an equivalent shunt element. Referring to Fig. 8-12(a) and using KCL, we have Yin But Substitution of (8.45) into (8.44) gives Yin I1 V1 V2 YF Y1 1 KF YF V1 V1 8:46 Ii I IF 1 V1 V1 8:44

Fig. 12-53

Fig. 12-54

(8.45)

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