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4.116. For a Student s distribution with 15 degrees of freedom, find the value of t1 such that (a) the area to the right of t1 is 0.01, (b) the area to the left of t1 is 0.95, (c) the area to the right of t1 is 0.10, (d) the combined area to the right of t1 and to the left of t1 is 0.01, (e) the area between t1 and t1 is 0.95. 4.117. Find the values of t for which the area of the right-hand tail of the t distribution is 0.01, if the number of degrees of freedom v is equal to (a) 4, (b) 12, (c) 25, (d) 60, (e) 150. 4.118. Find the values of t1 for Student s distribution that satisfy each of the following conditions: (a) the area between t1 and t1 is 0.90 and v 25, (b) the area to the left of t1 is 0.025 and v 20, (c) the combined area to the right of t1 and left of t1 is 0.01 and v 5, (d) the area to the right of t1 is 0.55 and v 16. 4.119. If a variable U has a Student s distribution with v 10, find the constant c such that (a) P(U (b) P( c U c) 0.98, (c) P(U c) 0.20, (d) P(U c) 0.90. c) 0.05,

14b

4.120. Evaluate each of the following: (a) F0.95,15,12; (b) F0.99,120,60; (c) F0.99,60,24; (d) F0.01,30,12; (e) F0.05,9,20; (f) F0.01,8,8.

Fig. 2-10.

4.61. (a) 1> 64 (b) 3 > 32 (c) 15 > 64 (d) 5 > 16 (e) 15 > 64 (f) 3 > 32 (g) 1> 64 4.62. (a) 57 > 64 (b) 21> 32 4.64. (a) 250 (b) 25 (c) 500 4.67. 193 > 512 4.63. (a) 1> 4 (b) 5 > 16 (c) 11> 16 (d) 5 > 8 4.65. (a) 17> 162 (b) 1> 324 4.66. 64 > 243

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An ideal transformer is a hypothetical transformer in which there are no losses and the core has in nite permeability, resulting in perfect coupling with no leakage ux. In large power transformers the losses are so small relative to the power transferred that the relationships obtained from the ideal transformer can be very useful in engineering applications. Referring to Fig. 14-13, the lossless condition is expressed by

4.68. (a) 32 > 243 (b) 192 > 243 (c) 40 > 243 (d) 242 > 243 0.1127 (d) 2.927 6pq) 3n2p2q2 4.72. (a) 1.5, 1.6 (b) 72, 90

(see Section 10.7).

p) (b) npq(1

4.73. (a) 75.4 (b) 9

and so, a being real, V1 I2 a V2 I1 The input impedance is readily obtained from relations (15): Zin V1 aV2 V a2 2 a2 Z L I1 I2 =a I2 16 15

4.74. (a) 0.8767 (b) 0.0786 (c) 0.2991

Fig. 14-13

The functions in the ovals are methods of the classes in the object model. The patron object can borrow and return copies. The librarian actor is not an object on the object model. The librarian in the use case shows that some functions for instance, catalog and shelve books are not functions available to the patron.

4.75. (a) 0.0375 (b) 0.7123 (c) 0.9265 (d) 0.0154 (e) 0.7251 (f) 0.0395 4.76. (a) 0.9495 (b) 0.9500 (c) 0.6826 4.77. (a) 0.75 (b) 4.78. 0.995 1.86 (c) 2.08 (d) 1.625 or 0.849 (e) 1.645

CHAP. 14]

4.79. 0.0668

4.80. (a) 20 (b) 36 (c) 227 (d) 40 4.81. (a) 93% (b) 8.1% (c) 0.47% (d) 15% 4.82. 84

EXAMPLE 14.6 The ideal transformer may be considered as the limiting case of the linear transformer of Section 14.7. Thus, in (14a) set R1 R2 X11 X22 0 (no losses) and then let XM ! 1 (in nite core permeability), to obtain jXM =a ZL Zin lim a2 a2 ZL XM !1 jXM =a ZL in agreement with (16)

4.83. (a) 61.7% (b) 54.7% 4.85. (a) 1.15 (b) 0.77 4.87. (a) 0.2511 (b) 0.1342 4.89. 0.0089

Ampere-Turn Dot Rule Since a N1 =N2 in (15), N1 I1 N2 I2 that is, the ampere turns of the primary equal the ampere turns of the secondary. A rule can be formulated which extends this result to transformers having more than two windings. A positive sign is applied to an ampere-turn product if the current enters the winding by the dotted terminal; a negative sign is applied if the current leaves by the dotted terminal. The ampere-turn dot rule then states that the algebraic sum of the ampere-turns for a transformer is zero.

4.84. (a) 95.4% (b) 23.0% (c) 93.3% 4.86. (a) 0.9962 (b) 0.0687 (c) 0.0286 (d) 0.0558 4.88. (a) 0.0567 (b) 0.9198 (c) 0.6404 (d) 0.0079

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