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If Xj, j 1, 2, c, n, is normally distributed with mean (see Table 4-2, page 110) fj(v) The characteristic function of X1 f(v) X2 eimv and variance

f1(v)f2(v) c fn(v)

resistors in parallel. a b 60:0 2 30:0

A way to formalize this identi cation of subdomains is to build a matrix using the conditions that we can identify from the speci cation and then to systematically identify all combinations of these conditions as being true or false.

since the Xj are independent. Then, by Theorem 3-11, the characteristic function of # X is fX(v) X1 X2 n eimv

120:0 1 1 1 1 Req 20:0

But this is the characteristic function for a normal distribution with mean result follows from Theorem 3-13.

Now by Theorem 4-3, page 115, V is chi-square distributed with n degrees of freedom [as is seen on replacing # Xj by (Xj m)>s]. Also, by Problem 5.21, X is normally distributed with mean m and variance 2 >n. Therefore, from Theorem 4-3 with n # s 1 and X1 replaced by (X m)> 2s2 >n, we see that V2 is chisquare distributed with 1 degree of freedom. It follows from Theorem 4-5, page 115, that if V1 and V2 are independent, then Vl, is chi-square distributed with n 1 degrees of freedom. Since it can be shown that V1 and V2 are indeed independent, the required result follows.

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Note: For n identical resistors in parallel the equivalent resistance is given by R=n. Combinations of inductances in parallel have similar expressions to those of resistors in parallel: 1 1 1 Leq L1 L2 and, for two inductances, Leq L1 L2 L1 L2

5.23. (a) Use Theorem 5-6 to determine the expected number of samples in Problem 5.1 for which sample variances are greater than 7.2. (b) Check the result in (a) with the actual result.

Two inductances L1 3:0 mH and L2 6:0 mH are connected in parallel. 1 1 1 Leq 3:0 mH 6:0 mH and Leq 2:0 mH

Therefore, P(4r > Cr) P[(4 < C)r] 1 1 1 P(4 < C) [P(4) B 1 13 P(C) 1 4 1 R 52 P(4 > C)] 9 13

The conditions in the triangle problem might be (1) a b or a c or b c, (2) a b and b c, (3) a < b c and b < a c and c < a b, and (4) a > 0 and b > 0 and c > 0. These four conditions can be put on the rows of a matrix. The columns of the matrix will each be a subdomain. For each subdomain, a T will be placed in each row whose condition is true and an F when the condition is false. All

We could also get this by noting that the diagram favorable to this event is the complement of the event shown circled in Fig. 1-8. Since this complement has 52 16 36 sample points in it and each sample point is assigned probability 1 > 52, the required probability is 36 > 52 9 > 13.

With three capacitances in parallel, i C1 dv dv dv dv dv C2 C3 C1 C2 C3 Ceq dt dt dt dt dt

Fig. 1-8

For several parallel capacitors, Ceq C1 C2 , which is of the same form as resistors in series.

1.7. A ball is drawn at random from a box containing 6 red balls, 4 white balls, and 5 blue balls. Determine the probability that it is (a) red, (b) white, (c) blue, (d) not red, (e) red or white.

(a) Method 1 Let R, W, and B denote the events of drawing a red ball, white ball, and blue ball, respectively. Then P(R) ways of choosing a red ball total ways of choosing a ball 6 4 6 15 2 5

A set of series-connected resistors as shown in Fig. 3-5 is referred to as a voltage divider. The concept extends beyond the set of resistors illustrated here and applies equally to impedances in series, as will be shown in 9.

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