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11.12 A three-phase, four-wire, ABC system, with line voltage VBC 294:2 08 V, has a Y-connected load of ZA 10 08 , ZB 15 308 , and ZC 10 308 (Fig. 11-31). Obtain the line and neutral currents.

The law of large numbers for Bernoulli trials 4.10. Prove Theorem 4-1, the (weak) law of large numbers for Bernoulli trials.

Fig. 11-31

By Chebyshev s inequality, page 83, if X is any random variable with finite mean (1) P( u X mu ks) 1 k2 np, s 1 k2 1 k2 !npq and (1) becomes and variance

3. Draw a DFD for the B&B system (See Problem 4.3). See Fig. 8-10.

In particular, if X is binomially or Bernoulli distributed, then m (2) or (3) pq , (3) becomes An X Pa 2 n p2 k pq b An P( u X np u k!npq )

CHAP. 11]

and taking the limit as n S ` we have, as required,

169:9 908 16:99 908 A 10 08 169:9 308 11:33 608 A IB 15 308 169:9 1508 16:99 1208 A IC 10 308 IN IA IB IC 8:04 69:58 A IA

4.11. Give an interpretation of the (weak) law of large numbers for the appearances of a 3 in successive tosses of a fair die.

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The law of large numbers states in this case that the probability of the proportion of 3s in n tosses differing from 1> 6 by more than any value P 0 approaches zero as n S ` .

11.13 The Y-connected load impedances ZA 10 08 , Z 15 308 , and ZC 10 308 , in Fig. 11-32, are supplied by a three-phase, three-wire, ABC system in which VBC 208 08 V. Obtain the voltages across the impedances and the displacement neutral voltage VON .

The normal distribution 4.12. Find the area under the standard normal curve shown in Fig. 4-3 (a) between z 0 and z (b) between z 0.68 and z 0, (c) between z 0.46 and z 2.21, (d) between z 0.81 and z (e) to the right of z 1.28.

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Fig. 11-32

1.2, 1.94,

(a) Using the table in Appendix C, proceed down the column marked z until entry 1.2 is reached. Then proceed right to column marked 0. The result, 0.3849, is the required area and represents the probability that Z is between 0 and 1.2 (Fig. 4-3). Therefore, P(0 Z 1.2) 1 22p

The method of Example 11.7 could be applied here and one node-voltage equation solved. However, the mesh currents I1 and I2 suggested in Fig. 11-32 provide another approach. I1 10 08 15 308 208 1208 15 308 15 308 10 308 I2 15 308 208 08 Solving, I1 14:16 86:098 A and I2 10:21 52:418 A. IA I1 14:16 86:098 A The line currents are then IC I2 10:21 127:598 A

30 e

Now the phasor voltages at the load may be computed. VAO IA ZA 141:6 86:098 V VBO IB ZB 120:2 18:938 V VCO IC ZC 102:1 157:598 V VON VOA VAN 141:6 93:918 120:1 908 23:3 114:538 V The phasor diagram is given in Fig. 11-33.

Fig. 4-3

(b) Required area area between z 0 and z 0.68 (by symmetry). Therefore, proceed downward under column marked z until entry 0.6 is reached. Then proceed right to column marked 8.

11.14 Obtain the total average power for the unbalanced, Y-connected load in Problem 11.13, and compare with the readings of wattmeters in lines B and C.

Fig. 8-10.

The result, 0.2517, is the required area and represents the probability that Z is between (Fig. 4-4). Therefore, P( 0.68 Z 0)

Fig. 11-33 14:16 p 10 1002:5 W 2 8:01 2 PB IB eff RB p 15 cos 308 417:0 W 2 10:21 2 2 PC IC eff RC p 10 cos 308 451:4 W 2

Fig. 4-4

Fig. 4-5

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