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where r 0, 1, 2, . . . , and in this case there are formulas analogous to (23) and (24) in which m The relationship between these moments is given by mr mr r r mr 1 m 1 r m2 m3 m4 c 1, r ( 1) j mr j m j j r 2m3 6mr m2 2 c ( 1)rmr mr 0

Fig. 9-38

(26)

9.31 Two circuit elements in a series connection have current and total voltage i 13:42 sin 500t 53:48 Identify the two elements. 9.32 Ans: A v 150 sin 500t 108 V

The moment generating function of X is defined by MX (t) that is, assuming convergence, MX(t)

E(etX )

Two circuit elements in a series connection have current and total voltage i 4:0 cos 2000t 13:28 Identify the two elements. Ans: A v 200 sin 2000t 50:08 V

(28)

Risk estimation involves two tasks in rating a risk. The rst task is estimating the probability of the occurrence of a risk, called the risk probability, and the second task is estimating the cost of the risk event happening, often called the risk impact. Estimating the risk probability will be hard. Known risks are much easier to manage, and they become part of the software process. The new risks that are unique to the current project are those most important to manage. The cost of the risk may be easier to determine from previous experience with project failures.

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A series RC circuit, with R 27:5 and C 66:7 mF, has sinusoidal voltages and current, with angular frequency 1500 rad/s. Find the phase angle by which the current leads the voltage. Ans: 208 A series RLC circuit, with R 15 , L 80 mH, and C 30 mF, has a sinusoidal current at angular frequency 500 rad/s. Determine the phase angle and whether the current leads or lags the total voltage. Ans: 60:68, leads A capacitance C 35 mF is in parallel with a certain element. Identify the element, given that the voltage and total current are v 150 sin 3000t V Ans: R 30:1 . Determine the iT 16:5 sin 3000t 72:48 A

a etx f (x)

(discrete variable)

A two-element series circuit, with R 20 and L 20 mH, has an impedance 40:0 angle and the frequency. Ans: 608; 276 Hz

(29)

Determine the impedance of the series RL circuit, with R 25 and L 10 mH, at (a) 100 Hz, (b) 500 Hz, (c) 1000 Hz. Ans: a 25:8 14:18 ; b 40:1 51:58 ; c 67:6 68:38 Determine the circuit constants of a two-element series circuit if the applied voltage v 150 sin 5000t 458 results in a current i 3:0 sin 5000t 158 (A). Ans: V

MX (t)

(continuous variable)

A series circuit of R 10 and p 40 mF has an applied voltage v 500 cos 2500t 208 (V). Find the C resulting current i. Ans: 25 2 cos 2500t 258 (A) p Three impedances are in series: Z1 3:0 458 , Z2 10 2 458 , Z3 5:0 908 . Find the applied voltage V, if the voltage across Z1 is 27:0 108 V. Ans: 126:5 24:68 V For the three-element series circuit in Fig. 9-39, (a) nd the current I; (b) nd the voltage across each impedance and construct the voltage phasor diagram which shows that V1 V2 V3 100 08 V. Ans: a 6:28 9:178 A; (b see Fig. 9-40.

Risk exposure is the expected value of the risk event. This is calculated by multiplying the risk probability by the cost of the risk event.

(30)

Fig. 9-39

We can show that the Taylor series expansion is [Problem 3.15(a)] MX (t) 1 mt mr 2 t2 2! c mr r tr r! c (31)

Since the coefficients in this expansion enable us to find the moments, the reason for the name moment generating function is apparent. From the expansion we can show that [Problem 3.15(b)] mr r i.e., mr is the rth derivative of MX (t) evaluated at t r stead of MX (t). dr M (t) 2 dtr X t (32)

CHAP. 9]

0. Where no confusion can result, we often write M(t) in-

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