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. Draw the simpli ed circuit and nd v for b 50; 100; 200. Compare with corresponding values obtained in Problem 3.27 and note that in the present case v is less sensitive to variations in b. Ans. v 0:90; 1; 1:04 V A nonlinear element is modeled by the following terminal characteristic. 10v when v ! 0 i 0:1v when v 0 Find the element s current if it is connected to a voltage source with (a) v 1 sin t and (b) v 1 sin t. See Fig. 3-26(a). Ans. (a) i 10 1 sin t ; (b) i 0:1 1 sin t

Then mZ has the 95% confidence interval 0.5366 to 1.5606. If If mZ mZ 1.1513 log a 1.1513 log a 1 1 1 1 r b r r b r 0.5366, 1.5606, r r 0.4904. 0.9155.

Fig. 3-26

8.51. Two correlation coefficients obtained from samples of size n1 28 and n2 35 were computed to be r1 0.50 and r2 0.30, respectively. Is there a significant difference between the two coefficients at a 0.05 level

Place a 1-

mZ ) and (H1: mZ 2 mZ ).

12.2.5 METRIC 5: RESPONSE FOR A CLASS (RFC)

mZ )

linear resistor between the nonlinear element of Problem 3.29 and the voltage source. See Fig. 3-26(b). Find the element s current if the voltage source is (a) v 1 sin t and (b) v 1 sin t. Ans. (a) i 0:91 1 sin t ; (b) i 0:091 1 sin t

0.3095 0.2669

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4.1 THE BRANCH CURRENT METHOD In the branch current method a current is assigned to each branch in an active network. Then Kirchho s current law is applied at the principal nodes and the voltages between the nodes employed to relate the currents. This produces a set of simultaneous equations which can be solved to obtain the currents.

1.96. Using a two-tailed test of the normal distribution, we would reject H0 only if z 1.96 or z Therefore, we cannot reject H0, and we conclude that the results are not significantly different at a 0.05 level.

For the least-squares line we have, from Problems 8.20 and 8.21, s2 y.x But by definition, a( y n and, by (6) on page 267, a (x a (x s2 xy s2 x s2 c 1 y x)( y # y) # sxy s2 x s2 (1 y r2) y)2 # s2 y a (x x)( y # n y) # sxy a( y n y)2 # b a (x x)( y # n y) #

EXAMPLE 4.1 Obtain the current in each branch of the network shown in Fig. 4-1 using the branch current method.

An analogous formula holds for the population (see Problem 8.54).

The response set of a class, {RS}, is the set of methods that can potentially be executed in response to a message received by an object of that class. It is the union of all methods in the class and all methods called by methods in the class. It is only counted on one level of call. RFC jRSj This metric is calculated for each class.

Fig. 4-1 Currents I1 ; I2 , and I3 are assigned to the branches as shown. Applying KCL at node a, I1 I2 I3 1

Yest)2]

# Y)2] for the case of (a) a least-squares line,

The voltage Vab can be written in terms of the elements in each of the branches; Vab 20 I1 5 , Vab I3 10 and Vab I2 2 8. Then the following equations can be written 20 I1 5 I3 10 20 I1 5 I2 2 8 Solving the three equations (1), (2), and (3) simultaneously gives I1 2 A, I2 1 A, and I3 1 A. 2 3

(Yest # Y)2 # Y)2] 2 (Y # Y) Yest)(Yest # Y) # Y)]

Other directions may be chosen for the branch currents and the answers will simply include the appropriate sign. In a more complex network, the branch current method is di cult to apply because it does not suggest either a starting point or a logical progression through the network to produce the necessary equations. It also results in more independent equations than either the mesh current or node voltage method requires. 37

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