qr code vb.net source Solutions to Problems in VS .NET

Printing Code-128 in VS .NET Solutions to Problems

150. Here R 16 ohms and Rin 75 ohms, so that we must use the L-section of Fig. 170. Hence, substituting the given values into eqs. (272) and (273), we nd that p L 16 10 6 3:6875 30:725 10 6 H 30:725 mH; answer: C 10 6 p 3:6875 0:0256 10 6 F 0:0256 mF; answer: 75

151. Here R 125 ohms and Rin 85 ohms, so that we must use the reverse L section of Fig. 171. Also note that, here, !0 2f0 2:26195 106 rad/sec, and upon substituting all these values into eqs. (276) and (277) you should nd that L 25:778 10 6 H 25:778 mH; answer: C 2:426 10 9 F 0:002426 mF; answer; or 2426 pF picofarads : " 152. Let Zin denote the impedance looking to the right, into terminals (1, 1), in Fig. 171. Since the inductor L is in series with the parallel combination of C and R, we have, jRXC " Zin jXL R jXC which, after rationalizing the fraction, then separating real and imaginary parts, becomes ! 2 2 "in RXC j XL R XC Z 2 2 R2 XC R2 XC Looking to the right, into terminals (1, 1), we are to see, at the resonant frequency !0 , a pure resistance; this means that, at the resonant frequency, the IMAGINARY PART of the above equation must have the value ZERO. Thus, equating the above imaginary part equal to zero gives the equation

which is true at the resonant frequency !0 , at which XL !0 L, and XC 1=!0 C. Now, making these substitutions into the last equation, then multiplying by !0 and solving for !0 , should give eq. (274), answer. " 153. Set XL !0 L and XC 1=!0 C in the equation for Zin found in the solution to problem (152). Doing this, the imaginary part of the equation vanishes, leaving " Zin Rin ; thus Rin R=!2 C2 R 0 R2 1=!2 C 2 1 !2 R2 C 2 0 0

Now substitute into the last equation the value of !0 given by eq. (274); doing this gives the value L=RC, which is eq. (275), answer. 154. By eq. (275), C L=Rin R, and substituting this value of C into eq. (274) gives, after a bit of algebra, eq. (276). Then, also by eq. (275), L Rin RC, and substituting this value of L into eq. (274) gives eq. (277), answers.

155. By inspection, " Z1O 3 9 9 =18 7:500 ohms; " Z1S 3 9 4 =13 5:769 ohms; " Z2O 5 13 =18 3:611 ohms; thus by eq: 282 ; by eq: 283 ; by eq: 284 ; " Z3 p 6:251 2:500 ohms; answer; " Z2 3:611 2:500 1:111 ohms; answer; " Z1 7:500 2:500 5:000 ohms; answer:

In regard to the question asked, the answer is yes, because the value of a pure resistance is theoretically the same at all frequencies. 156. The reactances are j=!C j5 ohms and, j!L j15 ohms. Then, by inspection of Fig. 178, we have 10 5 j5 10 1 j 10 1 j 3 j " 4 j2 ohms Z1O 15 j5 3 j 3 j 3 j " Next, to nd the value of Z1S , rst note that 5 ohms in parallel with j15 ohms is 5 j15 j15 4:5 j1:5 ohms; 5 j15 1 j3 and with this in mind, inspection of Fig. 178 then shows that 10 4:5 j3:5 775:0 j350:0 " 3:483 j1:573 ohms Z1S 14:5 j3:5 222:50 Next, looking into terminals (2, 2) with (1, 1) open-circuited, we see that 5 10 j5 " 3:5 j14:5 ohms Z2O j15 15 j5 " " " The nal step is to substitute the values of Z1O ; Z1S ; Z2O , just found, into eqs. (282) through (284). Let us begin with eq. (282); thus p p " Z3 3:5 j14:5 0:517 j0:427 8:00 j6:00 To nd the indicated square root, let us write the complex number 8:00 j6:00 in the exponential form A j , as follows. First, 8:00 j6:00 10 j . Then, since 8 j6 lies in the rst quadrant of the complex plane, we have  arctan 6=8 36:869 908, and hence 8:00 j6:00 10 j36:869 908 * " Now, substituting this value into the above value of Z3 , and remembering that   p p p   A j A j=2 A cos j sin 2 2