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Note that the current in L1 is equal to the current in L2 . Thus, from our work in sections 10.2 and 10.3, the equations are V j !L1 !L2 !M !M Iaid j! L1 L2 2M Iaid and V j !L1 !L2 !M !M Iopp j! L1 L2 2M Iopp hence L1 L2 2M inductance measured in AIDING case Laid and L1 L2 2M inductance measured in OPPOSING case Lopp Thus, subtracting the second equation from the rst, then solving for M, we have M 1=4 Laid Lopp hence, by eq. (371), k Laid Lopp p ; answer: 4 L1 L2
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The above result is of considerable practical importance because it allows us, working in the laboratory, to physically adjust the spacing between two coils until a required value of k is obtained. 199. (a) Here T 4; hence, by eq. (418) " " Zin T 2 ZL 16 3 j5 48 j80 ohms; answer: (b) By eq. (419), V2 V1 N2 =N1 240=4 60 volts; hence, by Ohm s law, the secondary current is " I2 (c) 60 60 3 j5 5:294 j8:824 amperes; answer: 3 j5 34
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One way is to use the formula P RI 2 , where I is the magnitude of the current. Thus we can take the square of the magnitude of the secondary current times the resistance in the secondary circuit, P 3 10:290 2 317:65 watts; answer: A second way is to multiply the secondary voltage by the in-phase component of the secondary current (section 8.5). Thus p A third way is to use the power factor, cos  R=Z 3= 34 0:5145, thus P V2 I2 cos  60 10:290 0:5145 317:65 watts; answer: P 60 5:294 317:65 watts; answer:
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" 200. Zin is equal to the denominator of eq. (386), where M 2 k2 L1 L2 , and where, in this " " case, Z1 j!L1 and Z2 j!L2 . Thus !2 k2 L 1 L 2 " j!L1 1 k2 ; answer: Zin j!L1 j!L2 201. Setting ! 2f 376:99; L1 4, and k 0:995 in the answer to problem 200, we " nd that Zin j15:04 ohms. Thus by Ohm s law, " I1 120=j15:04 j7:98 amperes; answer: (Note that in the ideal case, k 1, the shorted secondary would entirely neutralize the primary inductance, causing theoretically in nite primary line current to ow.) 202. Because the INSTANTANEOUS SUM of the three generator voltages around the loop is always equal to zero. This can be shown as follows. Let us, in Fig. 247, select one of the three generators to be the reference generator, and let the equation of this voltage wave be given by V sin !t, where V is the peak voltage (V and ! having the same values in all three generators). Now, for convenience, let V 1 volt and ! 1 rad/sec. Then, letting v be the sum of the three voltages at any time t, we would have that, at any instant, v sin t sin t 1208 sin t 2408 * Now let us make use of the following trigonometric identity sin x y sin x cos y cos x sin y note 7 in Appendix Now, in the above identity set x t; y 1208 in one case, and x t, y 2408 in the second case. Doing this, and remembering that sin h sin h and cos h cos h, you should nd that the preceding equation for v becomes v sin t sin t cos 1208 cos t sin 1208 sin t cos 2408 cos t sin 2408 thus v sin t 1 0:5 0:5 cos t 0:8660 0:8660 0 showing that the instantaneous sum of the voltages around the closed loop of Fig. 247 is always equal to zero. Thus NO CURRENT can ow under such conditions. Note that the same conclusion is reached if we are thinking in terms of sinusoidal steady-state vector representation. In that case we would be dealing with the vector sum of the three voltages around the loop, which (as is evident from inspection of Fig. 252) will add up to zero. 203. (a) (b) By eq. (435), Vp 3300=1:732 1905:3 volts; answer: First, the phase voltage on the Y-connected secondary side would, by eq. (435), be equal to Vp 66;000=1:732 38,106.2 volts. This voltage would then, going from right to left in Fig. 254, be stepped down by a factor of 12; thus the line voltage on the delta-connected primary side would be 38,106.2/12 3175.5 volts, making the generator phase voltage equal to 3175.5/1.732 1833.4 volts, answer.
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* Here !t t radians, so that 1208 and 2408 should really be expressed in radians instead of degrees. For the special demonstration here, however, no di culty arises.
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