qr code generator vb.net * See eqs. (161) and (163) in section 6.6. in .NET

Creator Code128 in .NET * See eqs. (161) and (163) in section 6.6.

* See eqs. (161) and (163) in section 6.6.
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In section 8.1 we introduced the basic series RL circuit of Fig. 127, to which is applied a sinusoidal voltage of V volts rms. We found that, by writing the combined e ects of R and
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CHAPTER 8 Reactance and Impedance
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" L in the form of a complex number, Z R j!L (called the impedance ), then applying the algebra of complex numbers, we were able to nd the magnitude and phase angle of the rms current owing in the circuit. Actually, the procedure of section 8.1 is not limited to the simple series circuit of Fig. 127 but applies, as well, to ANY type of series, parallel, or series-parallel connection of R and L components. The procedure is as follows. First, in drawing circuit diagrams, let us, for convenience, agree to represent the series " " , which we ll label Z , as shown impedance, Z R j!L, by the single symbol below,
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" where Z R j!L. For the case where L 0, the symbol then represents the real number R (a pure resistance) or, if R 0, the symbol then represents the imaginary number j!L (a pure inductive reactance). The simplest case consists of a series connection of n such impedances, as shown in Fig. 131.
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Fig. 131
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" In such a case the generator sees a total impedance, ZT , equal to the sum of all the impedances; thus, by Ohm s law (eq. (199)) we have, for Fig. 131, that " " V " V I " " 204 "2 Zn " ZT Z1 Z " in which each impedance has the general form Z R j!L. Or, by the law of addition of complex numbers (section 6.2), the above equation can also be written in the form " " V " V 205 I " ZT R1 R2 Rn j! L1 L2 Ln Problem 117 A certain series circuit consists of three resistances of 8, 10, and 12 ohms, and two inductor coils of 12 and 25 millihenrys (mH), where 1 mH=0.001 henry. If the applied sinusoidal reference voltage is 95 volts rms, and ! 2f 1000 radians/ second, nd (a) magnitude of rms current, (b) phase angle of current, (c) voltage drop across the 25 mH coil.
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CHAPTER 8 Reactance and Impedance
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Fig. 132
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Next consider the case of a purely PARALLEL network, as shown in Fig. 132. " " " " " Now let IT generator current I1 I2 In , and ZT total impedance seen by the generator. By Ohm s law: " V " " " " IT " I1 I2 In ZT 206
" Or, since the same voltage V is applied to all the branches, eq. (206) becomes, again by Ohm s law, " " " " V V V V " " " " ZT Z1 Z2 Zn that is, " V thus, 1 1 1 1 "T Z1 Z2 Zn " " " Z 207  1 "T Z   1 1 1 "1 Z2 Zn " " Z 
" V
Thus, in words, eq. (207) shows that in a purely PARALLEL network, the RECIPROCAL of the total impedance seen by the generator is equal to the SUM OF THE RECIPROCALS of the impedances of the individual branches. Or, inverting both sides of eq. (207), we have the equivalent result that " ZT 1 " " " 1=Z1 1=Z2 1=Zn 208
In words, eq. 208 says that, in a PARALLEL circuit, the total impedance seen by the generator is equal to the RECIPROCAL of the SUM OF THE RECIPROCALS of the individual impedances. If (as often happens in practical work) the network consists of just two parallel impedances, then eq. (207) becomes 1 1 1 " " " ZT Z1 Z2 " " which (after multiplying both sides by Z1 Z2 and then inverting) gives the special formula " " Z Z " ZT " 1 2 " Z1 Z2 for the impedance looking into a parallel connection of two impedances. 209
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