# barcode in vb.net 2010 A parallel RLC in Software Making Code 3/9 in Software A parallel RLC

17-12 A parallel RLC
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278 Power and Resonance in Alternating-Current Circuits
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Considering = 3.14 to three significant figures, this formula can be simplified to: fo = 0.159/(LC )1/2 The 1 2 power of a quantity represents the positive square root of that quantity. The preceding formulas are valid for series-resonant and parallel-resonant RLC circuits. The formula will also work if you want to find fo in megahertz (MHz) when L is given in microhenrys ( H) and C is in microfarads ( F). These values are far more common than hertz, henrys, and farads in electronic circuits. Just remember that millions of hertz go with millionths of henrys, and with millionths of farads.
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The Effects of R and G Interestingly, the value of R or G does not affect the resonant frequency in either type of circuit. But these quantities are significant, nevertheless! The presence of nonzero resistance in a series-resonant circuit, or nonzero conductance in a parallel-resonant circuit, makes the resonant frequency less welldefined. Engineers say that the resonant frequency response becomes more broad or less sharp. In a series circuit, the resonant frequency response becomes more broad as the resistance increases. In a parallel circuit, the resonant frequency response becomes more broad as the conductance increases. The sharpest possible responses occur when R = 0 in a series circuit, and when G = 0 (that is, R = ) in a parallel circuit. Problem 17-10 Find the resonant frequency of a series circuit with an inductance of 100 H and a capacitance of 100 pF. First, convert the capacitance to microfarads: 100 pF = 0.000100 F. Then find the product LC = 100 0.000100 = 0.0100. Take the square root of this, getting 0.100. Finally, divide 0.159 by 0.100, getting fo = 1.59 MHz. Problem 17-11 Find the resonant frequency of a parallel circuit consisting of a 33- H coil and a 47-pF capacitor. Again, convert the capacitance to microfarads: 47 pF = 0.000047 F. Then find the product LC = 33 0.000047 = 0.00155. Take the square root of this, getting 0.0394. Finally, divide 0.159 by 0.0394, getting fo = 4.04 MHz. Problem 17-12 Suppose you want to design a circuit so that it has fo = 9.00 MHz. You have a 33-pF fixed capacitor available. What size coil will be needed to get the desired resonant frequency Use the formula for the resonant frequency, and plug in the values. This will allow you to use simple arithmetic to solve for L. Convert the capacitance to microfarads: 33 pF = 0.000033 F. Then calculate as follows:
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fo = 0.159/(LC )1/2 9.00 = 0.159/(L 0.000033)1/2 9.002 = 0.1592/(0.000033 L) 81.0 = 0.0253/(0.000033 L)
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Resonance 279
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81.0 0.000033 L = 0.0253 0.00267 L = 0.0253 L = 0.0253/0.00267 = 9.48 H
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Problem 17-13 Suppose a circuit must be designed to have fo = 455 kHz. A coil of 100 H is available. What size capacitor is needed Convert the frequency to megahertz: 455 kHz = 0.455 MHz. Then the calculation proceeds in the same way as with the preceding problem:
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fo = 0.159/(LC )1/2 0.455 = 0.159/(100 C )1/2 0.4552 = 0.1592/(100 C ) 0.207 = 0.0253/(100 C ) 0.207 100 C = 0.0253 20.7 C = 0.0253 C = 0.0253/20.7 = 0.00122 F = 1220 pF In practical circuits, variable inductors and/or variable capacitors are often placed in tuned circuits, so that small errors in the frequency can be compensated for. The most common approach is to design the circuit for a frequency slightly higher than fo, and to use a padder capacitor in parallel with the main capacitor (Fig. 17-13).
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