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EXAMPLE 15.6 Write data statements for the current-controlled sources in Fig. 15-6. Introduce Vdmy (Vdmy) with current i entering it at node 1. Vdmy The data statements for the controlled sources are Source CCVS CCCS hnamei H1 F1 hnodesi 4 3 5 6 hcontroli Vdmy Vdmy hgaini k3 k4 1 7 DC 0

(or Z 0.9). The probability of this is the area under the normal curve to the right of Z 0.9, which is 0.5000 0.3159 0.1841. Therefore, the odds against A winning are (1 0.1841) : 0.1841 0.8159 : 0.1841, or 4.43 to 1.

5.16. Two distances are measured as 27.3 inches and 15.6 inches, with standard deviations (standard errors) of 0.16 inches and 0.08 inches, respectively. Determine the mean and standard deviation of (a) the sum, (b) the difference of the distances.

CHAP. 15]

15.6 2(0.16)2 15.6 2(0.16)2

Fig. 15-6

5.17. A certain type of electric light bulb has a mean lifetime of 1500 hours and a standard deviation of 150 hours. Three bulbs are connected so that when one burns out, another will go on. Assuming the lifetimes are normally distributed, what is the probability that lighting will take place for (a) at least 5000 hours, (b) at most 4200 hours

mL2 s2 2 L (5000

(5,5,5 equilateral)

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Fig. 15-7 Number the nodes and name the elements as in Fig. 15-7(b). DC analysis with dependent source, Fig. 15-7 Vs 1 0 DC 12 R1 1 2 1k R2 0 3 2k R3 0 4 500 Vdmy 2 3 0 F1 4 3 Vdmy 100 .END The results in the output le are NODE (1) VOLTAGE 12.0000 NODE (2) VOLTAGE 11.9410 NODE (3) VOLTAGE 11.9410 NODE (4) VOLTAGE 2:9557 Then, the netlist is

23(150)2 1.92.

VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 5:911E 05 Vdmy 5:911E 05 TOTAL POWER DISSIPATION 7:09E 04 WATTS

1.92)

(4200

CONTROL AND OUTPUT STATEMENTS IN DC ANALYSIS Certain statements control actions or the output format. Examples are: The syntax is hstep sizei

1.15)

:OP prints the dc operating point of all independent sources. :DC sweeps the value of an independent dc source. :DC hnamei hinitial valuei hfinal valuesi

Sampling distribution of variances 5.18. With reference to Problem 5.1, find (a) the mean of the sampling distribution of variances, (b) the standard deviation of the sampling distribution of variances, i.e., the standard error of variances.

(6,4,2 not a triangle) (4,6,2 not a triangle) (1,2,3 not a triangle)

(a) The sample variances corresponding to each of the 25 samples in Problem 5.1 are 0 0.25 4.00 9.00 20.25 0.25 0 2.25 6.25 16.00 4.00 2.25 0 1.00 6.25 9.00 6.25 1.00 0 2.25 20.25 16.00 6.25 2.25 0

v iR1 iR2 iR3 i R1 R2 R3 iReq where a single equivalent resistance Req replaces the three series resistors. between i and v will pertain. For any number of resistors in series, we have Req R1 R2 . If the three passive elements are inductances, v L1 di di di L2 L3 dt dt dt di L1 L2 L3 dt di Leq dt The same relationship

The mean of sampling distribution of variances is mS 2 sum of all variances in the table above 25 1)(s2)>n, since for n 135 25 2 and s2 5.40 10.8 [see Problem 5.1(b)], the

Extending this to any number of inductances in series, we have Leq L1 L2 . If the three circuit elements are capacitances, assuming zero initial charges so that the constants of integration are zero, 1 1 1 v i dt i dt i dt C1 C2 C3 1 1 1 i dt C1 C2 C3 1 i dt Ceq The equivalent capacitance of several capacitances in series is 1=Ceq 1=C1 1=C2 .

This illustrates the fact that mS2 (n right-hand side is 1 (10.8) 5.4. 2

The result indicates why a corrected variance for samples is often defined as S 2 then follows that mS 2 s2 (see also remarks on page 158).

. What must be the ohmic resistance of the third resistor Req R1 R2 R3 750:0 40:0 410:0 R3

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