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143. The values in the following table were calculated using eq. (248), with freehand plots of the data shown to the right. As the curves show, the higher Q circuit is clearly superior to the lower Q circuit as far as MAXIMUM OUTPUT VOLTAGE and SELECTIVITY are concerned. Note, however, that the higher Q circuit has a relatively NARROW BANDWIDTH as compared with the low Q circuit, and this may or may not be an advantage, depending upon the rate of transmission of information through the circuit. (It is a fundamental fact of nature that, the more information that is to be transmitted through a system per unit time, the greater must be the bandwidth of the system.)
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A, for Q 10 2.71 3.45 4.76 6.10 7.35 8.80 9.90 10.00 9.71 8.36 6.82 5.55 4.22 2.92 2.19 A, for Q 20 2.76 3.56 5.12 7.00 9.22 13.08 18.74 20.00 18.40 12.54 8.68 6.47 4.61 3.05 2.25
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d 0.80 0.85 0.90 0.93 0.95 0.97 0.99 1.00 1.01 1.03 1.05 1.07 1.10 1.15 1.20
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144. Since the generator is operating at the resonant frequency de ned by eq. (252), it would see a pure resistance of R0 ohms, given by eq. (254). Thus, upon substituting the given values into eq. (254) (see conversion formulas following eq. (184) in section 7.6), we nd that R0 10 4 20,000 ohms; answer: 50 10 10
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Thus the LOW RESISTANCE load of 50 ohms appears to the generator as a HIGH RESISTANCE load of 20,000 ohms. This can be of great practical advantage if, for instance, the generator is a device having a relatively high internal resistance.
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145. For Q 20, eq. (269) becomes A q 1 400d 2 d 2 0:9975 2 d 2 400 d 2 1 2
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A 0.11 0.15 0.23 0.33 0.44 0.64 0.93 1.00
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A 0.93 0.65 0.46 0.35 0.24 0.18 0.14
The parallel circuit of Fig. 167 is widely used because it presents a very HIGH IMPEDANCE at, and near, its resonant frequency (as the gure illustrates). Thus, when used as a tuned load in an ampli er stage, the gain of the stage is great at, and in the immediate neighborhood of, the resonant frequency, but low at undesired frequencies away from the resonant frequency. 146. (a) Putting L 250 10 6 and C 4 10 9 into eq. (259), the approximate value of !0 is !0 106 radians=second; answer; or f0 !0 =2 159:155 kHz kilohertz : (b) Putting p values of L, C, and >R into eq. (252) gives the given !0 106 1 0:0064, which for almost all practical purposes can be taken to be !0 106 rad/sec, the same value as found in part (a). Thus the answer to the question is yes. The generator sees a pure resistance at resonance, given by eq. (254); thus R0 (b) 250 10 6 3125 ohms; answer: 20 4 10 9
147. (a)
By Ohms s law, at resonance " Ig Ig =08 90=3125 0:0288 amperes; answer: P V 2 =R0 8100=3125 2:592 watts, answer; or, if you wish,
2 P Ig R0 0:0288 2 3125 2:592 watts; same answer:
XC 1=!0 C 250 ohms. Hence, by Ohm s law, " IC 90 j0:36 amperes; leading V by 908; answer: j250 " IL 90 20 j250
XL !0 L 250 ohms. Hence, by Ohm s law,
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and thus, upon rationalizing, " IL 0:02862 j0:35771 amperes; lagging; answer: (f) By eq. (256), Q 250=20 12:5; answer: (g) " P jIL j2 R 0:12878 20 2:576 watts approx., answer. The answer here di ers, by a very small amount, from the answer found in part (c), because the true value of !0 is SLIGHTLY LESS than the value of !0 found by using eq. (259) (as was brought out in the solution to problem 146). " Hence the value of !0 L used in calculating IL in part (e) is SLIGHTLY MORE than the true value of !0 L. 148. (a)
Since the values of C; L, and R have not been changed, it follows that the values of both !0 and Q, de ned by eqs. (259) and (256), will have the same values as found in problems 146 and 147. Likewise, R0 will still have the same value, R0 L=RC 3125 ohms. Also note that, for this problem, by eq. (262), d 1:05. Thus, upon substituting R0 3125; Q 12:5, and d 1:05 into eq. (268), we have that ! 1 j1:429313 " Zp 3125 3125 0:364418 j0:520867 ohms 2:744102 hence, upon applying Ohm s law, then rationalizing, we have
V 90 " 0:025972 j0:037122 amperes; answer: Ig " Zp 3125 0:364418 j0:520867 (b) (c) 0:037122 55:0228; answer: 0:025972 The rst (and easiest) way is to multiply the generator voltage by the in phase component of the generator current (section 5.7); thus  arctan P 90 0:025972 2:338 watts approx:; answer: The second way is to make use of eq. (117) in Chap. 5; that is, P VI cos , where V and Ig are rms magnitudes of voltage and current. From part (a) you can verify that Ig 0:045305 amperes; thus P 90 0:045 305 cos 55:0228 2:338 watts; same answer: 149. We must rst determine whether to use eq. (270) or eq. (274). To do this, make use of eq. (271) or (275)); thus Rin L=RC 28:5 10 6 = 135 36 10 10 58:848 ohms This shows that Rin is less than R, and therefore we must use eq. (274), which gives the value !0 2:348 106 Hz 2:348 megahertz; answer:
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