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(a) Referring to Problem 3.2, we have E(X) E(X2) Then, by Theorem 3-4, Var (X) E(Y2) 1 12 6 E(Y) 1 22 6 91 6 1 > 2. Moreover, c 1 62 6 35 12

91 6

The instantaneous power entering a resistor varies sinusoidally between zero and 2RI 2 , with twice the frequency of excitation, and with an average value of P RI 2 . v t and pR t are plotted in Fig. 10-5(a). EXAMPLE 10.7 Find the ac power entering an inductor L. p p pL t viL V 2 cos !t I 2 cos !t 908 2VI cos !t sin !t VI sin 2!t L!I 2 sin 2!t Thus, V2 sin 2!t L! P 0 Q VI V2 L!I 2 L!

Var (Y)

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The instantaneous power entering an inductor varies sinusoidally between Q and Q, with twice the frequency of the source, and an average value of zero. See Fig. 10-5(b). EXAMPLE 10.8 Find the ac power delivered in a capacitor C. p p pC t viC V 2 cos !t I 2 cos !t 908 2VI cos !t sin !t VI sin 2!t C!V 2 sin 2!t Thus, I2 sin 2!t C! P 0 Q VI I2 C!V 2 C!

7 2

35 6

Date 1/1/01 1/3/01 1/4/01

Like an inductor, the instantaneous power entering a capacitor varies sinusoidally between Q and Q, with twice the frequency of the source, and an average value of zero. See Fig. 10-5(c).

1 2.

If an inductor and a capacitor are fed in parallel by the same ac voltage source or in series by the same current source, the power entering the capacitor is 1808 out of phase with the power entering the inductor. This is explicitly re ected in the opposite signs of reactive power Q for the inductor and the capacitor. In such cases, the inductor and the capacitor will exchange some energy with each other, bypassing the ac source. This reduces the reactive power delivered by the source to the LC combination and consequently improves the power factor. See Sections 10.8 and 10.9.

m)2]

Another method By Theorem 3-4, Var (X)

EXAMPLE 10.9 Find the total instantaneous power p t , the average power P, and the reactive power Q, delivered p from v V 2 cos !t to a parallel RLC combination. The total instantaneous power is pT vi v iR iL iC pR pL pC

1 (2e 2 m)2]

CHAP. 10]

E(X2)

Start 09:00 09:00 09:00 12:00

2Var (X)

Fig. 10-5

1 2

Fig. 10-5 (cont.) Substituting the values of pR , pL , and pC found in Examples 10.6, 10.7, and 10.8, respectively, we get V2 1 1 cos 2!t V 2 pT C! sin 2!t R L! The average power is PT PR V 2 =R The reactive power is QT QL QC V 2 For 1=L! C! 0, the total reactive power is zero. factor. 1 C! L! 12

E [(X a)2] E [5(X E [(X E [(X E [(X since E(X (m a)2 m) E(X) m m. E m) m)2 m)2] m)2] (m 2(X 2(m (m a)6 2] m)(m a)E(X a)2 a)2] occurs when a) m) (m (m a)2] a)2

3.12. If X*

The two components P and Q of power play di erent roles and may not be added together. However, they may conveniently be brought together in the a vector quantity called complex power S p form of and de ned by S P jQ. The magnitude jSj P2 Q2 Veff Ieff is called the apparent power S and is expressed in units of volt-amperes (VA). The three scalar quantities S, P, and Q may be represented geometrically as the hypotenuse, horizontal and vertical legs, respectively, of a right triangle

m) > s is a standardized random variable, prove that (a) E(X*)

Stop 15:30 14:00 11:30 14:00

CHAP. 10]

E(X*) X s m 1 s [E(X m)] 1 s [E(X) m]

0, (b) Var(X*)

(called a power triangle) as shown in Fig. 10-6(a). The power triangle is simply the triangle of the 2 impedance Z scaled by the factor Ieff as shown in Fig. 10-6(b). Power triangles for an inductive load and a capacitive load are shown in Figs. 10-6(c) and (d), respectively.

since E(X) (b)

Fig. 10-6

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