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j 1 " R Z R !C j!C
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as in eq: 217
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CHAPTER 8 Reactance and Impedance
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" In the above, either expression for Z can be used, but we ll more often use the form " R j=!C R jXC , where XC 1=!C. Z " If, in a given case, ONLY RESISTANCE is present, then Z R, a real number; or, if " j=!C 1=j!C, an imaginary number. ONLY CAPACITANCE is present, then Z The simplest case consists of a series connection of n such impedances (see Fig. 131). In such a case the current is given by eq. (204) in section 8.2, in which each impedance now " has the general form Z R j=!C (or R 1=j!C). Hence, for the case of capacitance (instead of inductance), eq. (205) in section 8.2 would become " " V V "   222 IT " j 1 1 1 ZT R1 R2 Rn ! C1 C2 Cn " " where, to summarize the notation, V and I are vector volts and amperes (being represented as complex numbers), R and C are in ohms and farads, and ! 2f is frequency in radians per second ( f being frequency in cycles per second). Problem 122 A certain series circuit consists of two resistances, both of 18 ohms, and three capacitors, each of 0.12 F. If the applied sinusoidal reference voltage is 75 volts rms and ! 500,000 rad/sec, nd (a) (b) (c) magnitude of rms current, phase angle of current, magnitude of voltage drop across each capacitor.
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Now consider the case of a purely PARALLEL connection of n such impedances (see " Fig. 132). In this case the total impedance ZT seen by the generator is given by eq. (207) or " (208), each individual impedance now having the general form Z R j=!C R 1=j!C. (For the special case of two impedances in parallel, eqs. (207) and (208) reduce " to the convenient form of eq. (209).) After ZT is found, the value of the total (generator) " is then, by Ohm s law, equal to current IT " V " IT " ZT Problem 123 In Fig. 140, the reference generator voltage is 60 volts rms, as shown. If the frequency is 100,000 rad/sec, nd magnitude of generator current, " (b) phase angle of current IT . The symbol  (capital omega ) denotes ohms. (a)
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Fig. 140
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In practical work we often encounter networks containing not only resistance but also both inductive and capacitive reactances. To nd the sinusoidal-steady-state response of such a network, we simply replace the inductors and capacitors with the imaginary quantities, jXL j!L and jXC j=!C 1=j!C, where ! 2f radians per second. If the network is complicated we ll generally use the method of loop currents, applying the Kirchho voltage and current laws and the algebra of complex numbers, as discussed in connection with Fig. 135. Problem 124 In the series circuit of Fig. 141, the R, L, and C values are in ohms, microhenrys, and microfarads. Given that f 28 kHz (kilohertz), nd the vector voltage at point a with respect to ground.* " (Answer (in polar coord.), Va 22:43=143:958 volts)
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Fig. 141
Problem 125 In Fig. 142, the R, L, and C values are in ohms, microhenrys, and microfarads. As always, the generator voltage is in rms vector volts. Given that ! 106 radians/ second, nd the vector voltage at point y with respect to ground. " (Answer (in polar coord.), Vy 6:19221= 26:5668 volts)
Fig. 142
Problem 126 Making use of the work already done in problem 125, nd, in Fig. 142, the vector voltage at point x with respect to ground.
* See footnote with problem 22, Chap. 2.
CHAPTER 8 Reactance and Impedance
Problem 127 Here we wish to apply the fundamental principle of Thevenin s theorem (section 4.6) to the circuit of Fig. 143, in which we ll assume the internal resistance of the gen" erator is either negligibly small or is included in the resistance R. Find the voltage V 0 " 0 of the equivalent Thevenin generator, as indicated in Fig. and internal impedance Z 144. (Final answers in terms of R, C, and !.)
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