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Fig. 11-29 144:3 08 144:3 08 42:0 9:98 A 5 08 10 308 p p P 3VL eff IL eff cos 3 176:8 29:7 cos 9:98 8959 IL

Method 1 If X is binomially distributed, f (x) Then the moment generating function is given by M(t) E(etx) a etxf (x)

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The angle on IL , 9:98, is the negative of the angle on the equivalent impedance of the parallel combination of 5 08 and 10 308 . Therefore, 9:98 in the formulas of Section 11.13. W1 VL eff IL eff cos 308 176:8 29:7 cos 39:98 4028 W W2 VL eff IL eff cos 308 176:8 29:7 cos 20:18 4931 W As a check, W1 W2 8959 W, in agreement with Problem 11.8.

(q Method 2 For a sequence of n Bernoulli trials, define Xj e 0 1 if failure in jth trial if success in jth trial X1 et0 q X2 et1 p c q

Fig. 8-8.

11.10 A three-phase supply, with an e ective line voltage 240 V, has an unbalanced -connected load shown in Fig. 11-30. Obtain the line currents and the total power.

1, 2, . . . , n)

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Then the Xj are independent and X Mj (t) Then by Theorem 3-9, page 80,

Fig. 11-30 The power calculations can be performed without knowledge of the sequence of the system. e ective values of the phase currents are IAB eff 240 9:6 A 25 IBC eff 240 16 A 15 ICA eff 240 12 A 20 The

Xn. For the moment generating function of Xj , we have pet (j 1, 2, . . . , n)

Hence, the complex powers in the three phases are SAB 9:6 2 25 908 2304 908 0 j2304 SBC 16 2 15 308 3840 308 3325 j1920 SCA 12 2 20 08 2880 08 2880 j0

M(t)

M1(t)M2(t) cMn(t)

and the total complex power is their sum, ST 6205 j4224 that is, PT 6205 W and QT 4224 var (inductive). To obtain the currents a sequence must be assumed; let it be ABC. IAB IBC ICA 339:4 1208 13:6 308 A 25 908 339:4 08 22:6 308 A 15 308 339:4 2408 17:0 2408 A 20 08

4.8. Prove that the mean and variance of a binomially distributed random variable are, respectively, 2 npq.

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2. Draw an object model for the dental o ce problem (see Problem 4.2). See Fig. 8-9.

Then, using Fig. 11-7(a),

Proceeding as in Method 2 of Problem 4.7, we have for j E(Xj) Var (Xj) E[(Xj p2q Then s2 m E(X ) 0q p)2] q2p 1, 2, . . . , n, 1p (0 pq( p E( X2) Var ( X2) p p)2q q) c c (1 pq E(Xn) np npq p)2p

E( X1) Var ( X1)

The line currents are obtained by applying KCL at the junctions. IA IAB IAC 13:6 308 17:0 2408 29:6 46:78 A

Var (X )

IB IBC IBA 22:6 308 13:6 308 19:7 66:78 A IC ICA ICB 17:0 2408 22:6 308 28:3 173:18 A

Var ( Xn)

where we have used Theorem 3-7 for 2. The above results can also be obtained (but with more difficulty) by differentiating the moment generating function (see Problem 3.38) or directly from the probability function.

Fig. 8-9.

11.11 Obtain the readings of wattmeters placed in lines A and B of the circuit of Problem 11.10 (Line C is the potential reference for both meters.)

4.9. If the probability of a defective bolt is 0.1, find (a) the mean, (b) the standard deviation, for the number of defective bolts in a total of 400 bolts.

29:6 WA Re VAC eff I eff Re 240 608 p 46:78 A 2 Re 5023 13:38 4888 W 19:7 WB Re VBC eff I eff Re 240 08 p 66:78 B 2 Re 3343 66:78 1322 W Note that WA WB 6210 W, which agrees with PT as found in Problem 11.10.

40, i.e., we can expect 40 bolts to be defective. 36. Hence, the standard deviation s !36 6.

(400)(0.1)(0.9)

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