# barcode in vb.net 2005 The impedance vector is therefore R in Software Encoding QR Code in Software The impedance vector is therefore R

The impedance vector is therefore R
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When you re confronted with a parallel RLC circuit, and you want to know the complex impedance R jX, take these steps: 1. Find the conductance G = 1/R for the resistor. (It will be positive or zero.) 2. Find the susceptance BL of the inductor using the appropriate formula. (It will be negative or zero.) 3. Find the susceptance BC of the capacitor using the appropriate formula. (It will be positive or zero.) 4. Find the net susceptance B = BL + BC. (It might be positive, negative, or zero.) 5. Compute R and X in terms of G and B using the appropriate formulas. 6. Assemble the vector R + jX.
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A resistor of 10.0 , a capacitor of 820 pF, and a coil of 10.0 H are in parallel. The frequency is 1.00 MHz. What is the impedance R jX
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Reducing complicated RLC circuits 295 Proceed by the steps as numbered above. 1. G 1/R 1/10.0 0.100. 1/(6.28fL) 1/(6.28 1.00 10.0) = 0.0159. 2. BL 3. BC 6.28fC 6.28 1.00 0.000820 0.00515. (Remember to convert the capacitance to microfarads, to go with megahertz.) 0.0159 0.00515 0.0108. 4. B BL BC 5. First find G 2 B2 0.1002 (-0.0108)2 0.010117. (Go to a couple of extra places to be on the safe side.) Then R G/0.010117 = 0.100/0.010117 = 9.88, and X B/0.010117 0.0108/0.010117 1.07. 6. The vector R jX is therefore 9.88 j1.07. This is the complex impedance of this parallel RLC circuit.
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A resistor of 47.0 , a capacitor of 500 pF, and a coil of 10.0 H are in parallel. What is their complex impedance at a frequency of 2.252 MHz Proceed by the steps as numbered above. 1. 2. 3. 4. 5. G 1/R 1/47.0 0.0213. 1/(6.28fL) 1/(6.28 2.252 10.0) 0.00707. BL BC 6.28fC 6.28 2.252 0.000500 0.00707. B BL BC 0.00707 0.00707 0. Find G2 B 2 0.02132 0.0002 0.00045369. (Again, go to a couple of extra places.) Then R G/0.00045369 0.0213/0.00045369 46.9, and X B/0.00045369 0. 6. The vector R jX is therefore 46.9 j0. This is a pure resistance, almost exactly the value of the resistor in the circuit.
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Reducing complicated RLC circuits
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Sometimes you ll see circuits in which there are several resistors, capacitors, and/or coils in series and parallel combinations. It is not the intent here to analyze all kinds of bizarre circuit situations. That would fill up hundreds of pages with formulas, diagrams, and calculations, and no one would ever read it (assuming any author could stand to write it). A general rule applies to complicated RLC circuits: Such a circuit can usually be reduced to an equivalent circuit that contains one resistor, one capacitor, and one inductor.
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Series combinations
Resistances in series simply add. Inductances in series also add. Capacitances in series combine in a somewhat more complicated way. If you don t remember the formula, it is 1/C 1/C1 1/C2 1/Cn
where C1, C2, , Cn are the individual capacitances and C is the total capacitance. Once you ve found 1/C, take its reciprocal to obtain C.
296 RLC circuit analysis An example of a complicated series RLC circuit is shown in Fig. 16-8A. The equivalent circuit, with just one resistor, one capacitor, and one coil, is shown in Fig. 16-813.
16-8 At A, a complicated series RLC circuit; at B, the same circuit simplified.
Parallel combinations
In parallel, resistances and inductances combine the way capacitances do in series. Capacitances just add up. An example of a complicated parallel RLC circuit is shown in Fig. 16-9A. The equivalent circuit, with just one resistor, one capacitor, and one coil, is shown in Fig. 16-9B.
Complicated, messy nightmares
Some RLC circuits don t fall neatly into either of the above categories. An example of such a circuit is shown in Fig. 16-10. Complicated really isn t the word to use here! How would you find the complex impedance at some frequency, such as 8.54 MHz You needn t waste much time worrying about circuits like this. But be assured, given a frequency, a complex impedance does exist. In real life, an engineer would use a computer to solve this problem. If a program didn t already exist, the engineer would either write one, or else hire it done by a professional programmer.