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barcode reader code in asp.net Figure 434 Impedances of R, L, and C in the complex plane in Software
Figure 434 Impedances of R, L, and C in the complex plane Recognize QRCode In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Paint QR Code 2d Barcode In None Using Barcode drawer for Software Control to generate, create QR Code 2d barcode image in Software applications. where R is called the AC resistance and X is called the reactance The frequency dependence of R and X has been indicated explicitly, since it is possible for a circuit to have a frequencydependent resistance Note that the reactances of equations 460 and 464 have units of ohms, and that inductive reactance is always positive, while capacitive reactance is always negative The following examples illustrate how a complex impedance containing both real and imaginary parts arises in a circuit Reading QR Code 2d Barcode In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Encoding Quick Response Code In C# Using Barcode maker for .NET Control to generate, create QR Code ISO/IEC18004 image in .NET framework applications. EXAMPLE 411 Impedance of a Practical Capacitor
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Making Code39 In None Using Barcode generator for Software Control to generate, create Code 39 Extended image in Software applications. Printing EAN / UCC  13 In None Using Barcode encoder for Software Control to generate, create EAN13 image in Software applications. A practical capacitor can be modeled by an ideal capacitor in parallel with a resistor (Figure 435) The parallel resistance represents leakage losses in the capacitor and is usually quite large Find the impedance of a practical capacitor at the radian frequency = 377 rad/s How will the impedance change if the capacitor is used at a much higher frequency, say 800 MHz Generate UPCA Supplement 2 In None Using Barcode generator for Software Control to generate, create UPCA Supplement 2 image in Software applications. Barcode Generator In None Using Barcode creator for Software Control to generate, create barcode image in Software applications. Solution
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Part I
Circuits
Substituting numerical values, we nd Z1 ( = 377) = 106 106 = 1 + j 377 106 01 10 6 1 + j 377
= 26516 104 15443 The impedance of the capacitor alone at this frequency would be: ZC1 ( = 377) = 1 = 2653 103 /2 j 377 01 10 6 If the frequency is increased to 800 MHz, or 1600 106 rad/s a radio frequency in the AM range we can recompute the impedance to be: Z1 ( = 1600 106 ) = = 106 1 + j 1600 106 01 10 6 106 106 = 0002 15708 1 + j 160 106 1 = 0002 /2 j 1600 106 01 10 6 The impedance of the capacitor alone at this frequency would be: ZC1 ( = 1600 106 ) = Comments: Note that the effect of the parallel resistance at the lower frequency
(corresponding to the wellknown 60Hz AC power frequency) is signi cant: The effective impedance of the practical capacitor is substantially different from that of the ideal capacitor On the other hand, at much higher frequency, the parallel resistance has an impedance so much larger than that of the capacitor that it effectively acts as an open circuit, and there is no difference between the ideal and practical capacitor impedances This example suggests that the behavior of a circuit element depends very much in the frequency of the voltages and currents in the circuit We should also note that the inductance of the wires may become signi cant at high frequencies EXAMPLE 412 Impedance of a Practical Inductor
Problem
A practical inductor can be modeled by an ideal inductor in series with a resistor Figure 436 shows a toroidal (doughnutshaped) inductor The series resistance represents the resistance of the coil wire and is usually small Find the range of frequencies over which the impedance of this practical inductor is largely inductive (ie, due to the inductance in the circuit) We shall consider the impedance to be inductive if the impedance of the inductor in the circuit of Figure 437 is at least 10 times as large as that of the resistor Toroid
Leads a n turns
025 cm
Solution
Known Quantities: L = 0098 H; lead length = lc = 2 10 cm; n = 250 turns; wire is 30 gauge Resistance of 30 gauge wire = 0344 /m Find: The range of frequencies over which the practical inductor acts nearly like an ideal 05 cm Cross section
Figure 436 A practical inductor
inductor
4
AC Network Analysis
Analysis: We rst determine the equivalent resistance of the wire used in the practical inductor using the cross section as an indication of the wire length, lw , used in the coil: lw = 250 (2 025 + 2 05) = 375 cm l = Total length = lw + lc = 375 + 20 = 395 cm
The total resistance is therefore
R = 0344 /m 0395 m = 0136
Thus, we wish to determine the range of radian frequencies, , over which the magnitude of j L is greater than 10 0136 : L > 136, or > 136/L = 136/0098 = 139 rad/s Alternatively, the range is f = /2 > 022 Hz

