barcode in vb.net 2005 Adding impedance vectors in Software

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Adding impedance vectors
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Often, there is resistance, as well as reactance, in an ac series circuit containing a coil and capacitor. This occurs when the coil wire has significant resistance (it s never a perfect conductor). It might also be the case because a resistor is deliberately connected into the circuit. Whenever the resistance in a series circuit is significant, the impedance vectors no longer point straight up and straight down. Instead, they run off towards the northeast (for the inductive part of the circuit) and southeast (for the capacitive part). This is illustrated in Fig. 16-2. When vectors don t lie along a single line, you need to use vector addition to be sure that you get the correct resultant. Fortunately, this isn t hard. In Fig. 16-3, the geometry of vector addition is shown. Construct a parallelogram, using the two vectors Z1 R1 jX1 and Z2 R2 jX2 as two of the sides. The diagonal is the resultant. In a parallelogram, opposite angles have equal measure. These equalities are indicated by single and double arcs in the figure.
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Complex impedances in series 287
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16-2 When resistance is present along with reactance, impedance vectors point northeast or southeast.
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16-3 Parallelogram method of vector addition.
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Formula for complex impedances in series
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Given two impedances, Z1 R1 jX1 and Z2 in series is their vector sum, given by Z (R1 R2) R2 jX2, the net impedance Z of these X2)
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The reactances X1 and X2 might both be inductive; they might both be capacitive; or one might be inductive and the other capacitive.
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288 RLC circuit analysis Calculating a vector sum using the formula is easier than doing it geometrically with a parallelogram. The arithmetic method is also more nearly exact. The resistance and reactance components add separately. That s all there is to it.
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Series RLC circuits
When a coil, capacitor, and resistor are connected in series (Fig. 16-4), the resistance R can be thought of as all belonging to the coil, when you use the above formulas. (Thinking of it all as belonging to the capacitor will also work.) Then you have two vectors to add, when finding the impedance of a series RLC circuit: Z (R R jXL) (0 jXC) j(XL XC)
16-4 A series RLC circuit.
Problem 16-5
A resistor, coil, and capacitor are connected in series with R 50 , XL 22 , and XC 33 . What is the net impedance, Z Consider the resistor to be part of the coil, obtaining two complex vectors, 50 j22 and 0 j33. Adding these gives the resistance component of 50 0 50, and the reactive component of j22 j33 j11. Therefore, Z 50 j11.
Problem 16-6
A resistor, coil, and capacitor are connected in series with R 600 , XL 444 , and 444 . What is the net impedance, Z XC Again, consider the resistor to be part of the inductor. Then the vectors are 600 j444 and 0 j444. Adding these, the resistance component is 600 0 600, and the reactive component is j444 j444 j0. Thus, Z 600 j0. This is a purely resistive impedance, and you can rightly call it 600 .
Problem 16-7
A resistor, coil, and capacitor are connected in series. The resistor has a value of 330 , the capacitance is 220 pF, and the inductance is 100 H. The frequency is 7.15 MHz. What is the net complex impedance First, you need to calculate the inductive and capacitive reactances. Remembering the formula XL 6.28fL, multiply to obtain jXL j(6.28 7.15 100) j4490
Complex admittances in parallel 289 Megahertz and microhenrys go together in the formula. As for XC, recall the formula 1/(6.28fC). Convert 220 pF to microfarads to go with megahertz in the formula XC C 0.000220 F. Then jXC j(l/(6.28 7.15 0.000220)) j101
Now, you can consider the resistance and the inductive reactance to go together, so one of the impedance vectors is 330 j4490. The other is 0 j101. Adding these gives 330 j4389; this rounds off to Z 330 j4390.
Problem 16-8
A resistor, coil, and capacitor are in series. The resistance is 50.0 , the inductance is 10.0 H, and the capacitance is 1000 pF. The frequency is 1592 kHz. What is the complex impedance of this series RLC circuit at this frequency 6.28fL. Convert the frequency to megahertz; 1592 kHz First, calculate XL 1.592 MHz. Then jXL j(6.28 1.592 10.0) j100
Then calculate XC 1/(6.28fC). Convert picofarads to microfarads, and use megahertz for the frequency. Therefore, jXC j(l/(6.28 1.592 0.001000)) j100
Let the resistance and inductive reactance go together as one vector, 50.0 j100. Let the capacitance alone be the other vector, 0 j100. The sum is 50.0 j100 j100 50.0 j0. This is a pure resistance of 50.0 . You can correctly say that the impedance is 50.0 in this case. This concludes the analysis of series RLC circuit impedances. What about parallel circuits To deal with these, you must calculate using conductance, susceptance, and admittance, converting to impedance only at the very end.
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