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Fig. 14-20 To place the dots on the circuit, consider only the coils and their winding sense. Drive a current into the top of the left coil and place a dot at this terminal. The corresponding ux is upward. By Lenz s law, the ux at the right coil must be upward-directed to oppose the rst ux. Then the natural current leaves this winding by the upper terminal, which is marked with a dot. See Fig. 14-21 for the complete dotted equivalent circuit, with currents I1 and I2 chosen for calculation of V.

\, it follows from Axiom 3 that P(S) P(\) or P(\) 0

Fig. 14-21 5 j5 5 j3 5 j3 10 j6 I1 I2 " 10 08 10 j10 #

1.5. Prove (a) Theorem 1-4, (b) Theorem 1-6. (a) We have A < Ar S. Then since A > Ar P(A < Ar) i.e., (b) We have from the Venn diagram of Fig. 1-7, (1) Then since the sets A and B A <B A < [B (A > B)] \, we have P(S) or P(Ar) 1 P(A) P(A) P(Ar) 1

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There are no superglue tokens, so the strong functional cohesion (SFC) is equal to zero. Out of 31 tokens, there are 12 glue tokens, so the weak functional cohesion is 12/31 or 0.387. There are four slices. Zero tokens have 100 percent adhesiveness. Four tokens are on three slices, so they have 75 percent adhesiveness. Eight tokens are on two slices, so they have 50 percent adhesiveness. The remaining tokens, 19, are on only one slice, so they have 25 percent adhesiveness. Adhesiveness is the average adhesiveness of all tokens, so 4 0:75 8 0:50 19 0:25 =31 11:25=31 0:363.

(A > B) are mutually exclusive, we have, using Axiom 3 and Theorem 1-1, P(A < B) P(A) P(A) P[B P(B) (A > B)] P(A > B)

Obtain the dotted equivalent for the circuit shown in Fig. 14-22 and use the equivalent to nd the equivalent inductive reactance.

Fig. 1-7

Fig. 14-22 Drive a current into the rst coil and place a dot where this current enters. The natural current in both of the other windings establishes an opposing ux to that set up by the driven current. Place dots where the natural current leaves the windings. (Some confusion is eliminated if the series connections are ignored while determining the locations of the dots.) The result is Fig. 14-23. Z j3 j5 j6 2 j2 2 j4 2 j3 j12 that is, an inductive reactance of 12 .

Calculation of probabilities 1.6. A card is drawn at random from an ordinary deck of 52 playing cards. Find the probability that it is (a) an ace, (b) a jack of hearts, (c) a three of clubs or a six of diamonds, (d) a heart, (e) any suit except hearts, (f) a ten or a spade, (g) neither a four nor a club.

Let us use for brevity H, S, D, C to indicate heart, spade, diamond, club, respectively, and 1, 2 , c, 13 for ace, two, c , king. Then 3 > H means three of hearts, while 3 < H means three or heart. Let us use the sample space of Problem 1.1(b), assigning equal probabilities of 1 > 52 to each sample point. For example, P(6 > C) 1 > 52. (a) P(1) P(1 > H or 1 > S or 1 > D or 1 > C ) P(1 > H) P(1 > S) P(1 > D) P(1 > C )

Fig. 14-23

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