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Thus all ve roots have the same magnitude, 2.07. To show the above results graphically, draw a circle with center at the origin of the complex plane, and radius of any convenient length to represent 2.07. The ve roots are distributed around the circumference of the circle, 728 apart, in accordance with the above angles. 107. We ordinarily say that 1 raised to a power is 1 ; thus, it might seem, o hand, that the ONLY possible answer is that 1 1=3 1. This, however, assumes that the answer has to be a real number, which is not true, because in mathematical applications one number is as valid as any other number on the total number plane of Fig. 108. Thus, as we ll now nd, the COMPLETE cube root of 1 is equal to the real number 1, plus two other roots, both complex numbers. To see why this is true, let us begin by noting that 1 1 j0; thus, 1 can be regarded as being a complex number, a jb, having a 1 and b 0. Then A 1 1=2 1 and  arctan 0 08 thus n 3 A1=n 1 1=6 1 by eq: 179 Then, since =n 0=n 0, eq. (176) becomes 1 1=3 1 cos 120k 8 j sin 120k 8 1= 120k 8 which, upon successively setting k 0; 1, and 2, gives the three cube roots of unity, thus for k 0: for k 1: for k 2: r1 cos 0 j sin 0 1:000; answer: r2 cos 1208 j sin 1208 0:500 j0:866; answer: r3 cos 2408 j sin 2408 0:500 j0:866; answer: by eq: 178 by eq: 177
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108. The purpose here is to emphasize that induced voltage depends not upon the amount of current, but only upon the rate of change of current. Here, in both (a) and (b), the current is changing at the same constant rate of 2 amperes per second, di=dt 2 amp/ sec. Hence, by eq. (181), in both cases, v 0:62 2 1:24 volts, answer. 109. By eq. (181), di=dt v=L 5:52=0:62 8:903 amp=sec; answer: 110. By eq. (181), L v= di=dt 0:048=76 6:316 10 4 henrys 631:6 microhenrys; answer: 111. Since 0.00065 meters 0:65 mm, and since E 3000 volts per mm, we have that v 3000 0:65 1950 volts; answer:
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112. 0:015 mF 1:5 10 2 10 6 F 1:5 10 8 F; thus, by eq. (184), q Cv 1:5 10 8 2:9 102 4:35 10 6 coulombs; answer: 113. (a) Using either eq. (188) or (189), you should nd that CT 0:0368 mF approx:; answer: (b) By eq. (191), CT 0:93 mF; answer: 114. We must use eq. (190), but to do this we rst need to know the value of CT . Upon using either eq. (188) or (189), you should nd that CT 0:03934 mF approx. Hence, for V 450 volts and CT 0:03934 mF, eq. (190) becomes Vx 17:703=Cx volts where, since CT is in mfd, Cx is also in mfd. Thus we have that V1 17:703=0:15 118:02 volts; V2 17:703=0:06 295:05 volts; V3 17:703=0:48 36:88 volts: Since none of the capacitor voltages will exceed 300 volts, it is theoretically proper to specify a capacitor voltage rating of 300 volts. 115. The three capacitors are, together, equivalent to a single capacitor of capacitance CT 0:03934 10 6 farads. Hence, by eq. (185), W 1 0:03934 10 6 450 2 3:983 10 3 joules; answer: 2 116. (a) " Here, V V 115 volts, R 28 ohms, !L 2 60 0:12 45:2389 ohms. Now substituting these values into eq. (197) we have " I hence 115 " jI j q 2:1615 amperes; answer: 2 28 45:2389 2 (b) By eq. (203),  arctan 45:2389 =28 58:2458, answer. (c) Voltmeters are calibrated to read rms volts; thus (see Fig. 129) VL !LI 45:2389 2:1615 97:784 volts rms; answer: 117. Here, " V V 95 volts, RT 30 ohms; !L 103 37 10 3 37 ohms 1 mH 10 3 H : 115 28 j45:2389
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