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* In regard to the use of degrees here, see last footnote in section 6.5.
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(Euler s formula), gives p " Z3 10 cos 18:43508 j sin 18:43508 3:000 j1:000 ohms; then, by eqs. (283) and (284), " " Z2 0:5 j13:5 ohms; and Z1 1:0 j3:0 ohms: Next, to nd the actual values of inductance and capacitance required, we make use of the reactance formulas XL !L and XC 1=!C, that is, L XL =! and C 1=!XC . Thus, using ! 105 and the reactance values found above, we have that " for Z3 ; " for Z2 ; " for Z1 ; XL 1:0 ohm; thus L 1=105 10 5 H henrys 10 mH; XL 13:5 ohms; thus L 13:5 10 5 H 135 mH; XC 3:0 ohms; thus C 1= 105 3 F farads 3:33 mF:
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Thus the complete equivalent T for Fig. 178 is as shown below, with resistance values in ohms.
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Answer to problem 156.
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157. It may be less confusing if we redraw the network in the form shown in the gure below.
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From the gure we see that Z1O * 11 ohms in parallel with 20 ohms 220=31 7:097 ohms; Z1S 48=14 60=17 6:958 ohms; Z2O 14 ohms in parallel with 17 ohms 14 17 =31 7:677 ohms:
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" * The overscore notation, Z, indicates that Z is, or may be, a complex number. Since, in this problem, the Zs can " only represent real numbers, we can, if we wish, write plain Z instead of Z.
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We now put the above values into eqs. (282), (283), and (284) to get the required answers; thus p Z3 7:677 7:097 6:958 1:033 ohms; Z2 7:677 1:033 6:644 ohms; Z1 7:097 1:033 6:064 ohms: The equivalent network is shown to the right. 158. Here we must use eqs. (285) through (289), and in using these equations the required divisions will generally be easier to do if the complex numbers are expressed in either the exponential or the polar form. Thus, if A=p and B=q are two complex numbers, then (eq. (165) in Chap. 6) A=p A =p q B=q B and for this reason we ve elected to write some of the following answers in both the polar and rectangular forms, with nal answers in rectangular form. First, by eq. (288), " Z 0 20 j12 8 j18 8 5 j3 4 j9 thus, " Z 0 8 47 j33 459:426=35:0748 ohms Next, by eq. (289), p p " Z 00 8 5 j3 5 j3 8 5 j3 16:492= 30:9648 ohms Then, " " Z2O Z 00 5:858 j3:515 6:832= 30:9658 ohms and " " Z1O Z 00 3:858 j20:485 20:845=79:3348 ohms Now substitute the above values, in polar form, into eqs. (285) through (287); then, by use of the polar formula C= C cos  j sin  , express the nal answers in rectangular form; thus by eq: 285 ; by eq: 286 ; by eq: 287 ; " ZA 67:246=66:0398 27:310 j61:451 ohms; answer: " ZB 27:858=66:0388 11:314 j25:457 ohms; answer: " ZC 22:040= 44:2608 15:785 j15:382 ohms; answer:
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159. In eq. (298), combine the two quantities on the left-hand side over the common " " " " " " denominator Z2 Z3 , then multiply both sides by Z2 Z3 , then replace Z2 Z3 with the right-hand side of eq. (299) to get " " " Z Z Z " " " " " " Z1 Z2 Z1 Z3 Z2 Z3 " A "B C " ZA ZB ZC A
or, if we wish, " " " " " " " Z1 Z2 Z1 Z3 Z2 Z3 ZA hence, by eq. (303), 
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 " " ZB ZC " " " ZA ZB ZC
" " " " " " " " Z1 Z2 Z1 Z3 Z2 Z3 ZA Z2 " which, upon solving for ZA , proves that eq. (304) is correct. Next, upon making use of eq. (A) above and eq. (301), we have that   " " ZA ZC " " " " " " " " "1 Z2 Z1 Z3 Z2 Z3 ZB Z ZB Z3 " " " ZA ZB ZC " which, upon solving for ZB , proves that eq. (305) is correct. Next, again making use of eq. (A) and also eq. (302), you can verify that eq. (306) is also correct. " 160. First, ! 2f 3:1416 106 rad/sec. Next, from Fig. 180 we have ZA 30 ohms, " "B j!L j62:832 ohms, ZC 20 ohms. Putting these values in equations (302), Z (303), and (301), we have " Z1 " Z2 " Z3 j1884:96 18:368 j14:617 ohms 50 j62:832 j1256:64 12:246 j9:745 ohms 50 j62:832 600 4:653 j5:847 ohms 50 j62:832
where the REAL PART of each answer represents resistance in ohms and the IMAGINARY PART represents reactance in ohms. Thus, since XL !L and XC 1=!C, we have that L XL =! and C 1=!XC , and therefore, using the " " " known value of ! and the above values of ZA ; ZB , and ZC , we nd that " Z1 consists of a resistance of 18.368 ohms in series with an inductor coil having 4:653 mH of inductance, answer. " Z2 consists of a resistance of 12.246 ohms in series with an inductor coil having 3:102 mH of inductance, answer. " Z3 consists of a resistance of 4.653 ohms in series with a capacitor having 5:444 10 8 F 0:05444 mF of capacitance, answer. " " " 161. In Fig. 181, Z1 j12; Z2 j9, and Z3 j6. Putting these values into eqs. (304), (305), and (306), you should nd that 18 " j2 ohms; answer; ZA j9 18 " j3 ohms; answer; ZB j6 18 " ZC j1:5 ohms; answer: j12
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