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The next step is to nd the resistance looking back into terminals a, b (keeping RL disconnected, as before). The 8-ohm and 12-ohm resistances are in parallel when looking back from terminals a, b and hence combine together (product over sum, eq. (34), Chap. 2) to give 12 8 =20 4:8 ohms. The 4.8 ohms is now in series with the 3-ohm resistance, giving us a total of 7.8 ohms in parallel with the 9-ohm resistance. Therefore, looking into a, b we see 7.8 ohms in parallel with 9 ohms, which, using the product of the two, over the sum, is equal to 7:8 9 =16:8 4:179 ohms, approximately. Thus the equivalent Thevenin generator for Fig. 59 is therefore as shown in the following gure, answer.

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Remember that method overloading is not much more than name reuse. The overloaded method is a completely different method from any other method of the same name. So if you inherit a method but overload it in a subclass, you re not subject to the restrictions of overriding, which means you can declare any return type you like. What you can t do is change just the return type. To overload a method, remember, you must change the argument list. The following code shows an overloaded method:

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The arbitrary dependent source (B element) allows an instantaneous transfer function to be written as a mathematical expression. This B element is a standard Berkeley SPICE 3 element. The expressions, [EXPR], given for V and I may be any function of node voltages, currents through any element, or a variety of traditional math functions. In PSpice, the E- and G-controlled source elements are utilized: Format: Bname N+N [ I = EXPR ][V = EXPR ] B1 0 1 I = sqrt(cos(v(1)/(v(2,3)))) B4 outp outn V = exp(i(vdd) 2) B1 1 0 V = V(2) abs(I(V1)) + V(3) B3 1 2 V = I(R1) B2 2 3 I = {V(7) Sin(Time)}

Legal Return Types (Exam Objective 1.4)

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57. First, Isc ows into a conductance equal to, by eq. (61), GT Gg GL , and thus, by eq. (59), Isc Gg GL VL also, IL GL VL thus VL IL =GL , and putting this value of VL into the rst equation and solving for IL gives the answer, eq. (67). 58. (a) The rst step is to nd the short-circuit current Isc , which is the current that ows into the short-circuited terminals a, b as shown in the gure to the left below, with the two loop-voltage equations to the right.

public class Foo{ void go() { } } public class Bar extends Foo { String go(int x) { return null; } }

thus,

Notice that the Bar version of the method uses a different return type. That s perfectly fine. As long as you ve changed the argument list, you re overloading the method, so the return type doesn t have to match that of the superclass version. What you re not allowed to do is this:

public class Foo{ void go() { } } public class Bar extends Foo { String go() { // Not legal! Can't change only the return type return null; } }

SPICE 3 Examples:

We must next nd the value ofthe conductance component Gg ofthe Norton equivalentgenerator. To do this, we now remove the short-circuit from the terminals a, b in the pre ceding gure; Gg now equals the conductance seen looking to the left into the now open-circuited terminals a, b. From inspection of Fig. 59 (disregard the 9ohm branch for the moment), note that we look into one branch consisting of 3 ohms in series with the parallel combination of 8 ohms and 12 ohms, that is, into 3 8 12 =20 7:8 ohms. Since this resistance is in parallel with the 9ohm resistance, the total resistance seen looking into the open-circuited terminals a, b is Rg 9 7:8 =16:8 4:178571 ohms

When a subclass wants to change the method implementation of an inherited method, the subclass must define a method that matches the inherited version exactly. As we saw earlier in this chapter, an exact match means the arguments and return types must be identical. Other rules apply to overriding, including those for access modifiers and declared exceptions, but those rules aren t relevant to the return type discussion. For the exam, be sure you know that overloaded methods can change the return type, but overriding methods cannot. Just that knowledge alone will help you through a wide range of exam questions.

You have to remember only six rules for returning a value:

thus, Gg 1=Rg 0:239316 mhos; approx: and hence the answer is that the Norton equivalent generator for Fig. 59 is

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