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A theorem is a statement, which, when proved, can be used to prove other statements or derive other results. Each of the following basic theorems requires the use of definitions and postulates for its proof. Note: We shall use the term principle to include important geometric statements such as theorems, postulates, and definitions.

CHAP. 7]

EXAMPLE 8.2 Repeat Example 8.1 for C 10 mF, which results in !0 . As in Example 8.1, the initial conditions are used to determine the constants. 0 A1 A2 0 and A1 0. Then, di d A te t A2 te at e t dt dt 2 from which A2 di=dt j0 2000. Hence, i 2000te 10 t (A) (see Fig. 8-5). Once again the polarity is a matter of the choice of direction for the current with respect to the polarity of the initial voltage on the capacitor.

PRINCIPLE 1:

Lo _

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Since i 0 i 0 ,

Thus, j A > j B in Fig. 2-21.

Fig. 8-5

Fig. 7-31

Fig. 2-21

The responses for the overdamped and critically damped cases plotted in Figs. 8-4 and 8-5, respectively, are quite similar. The reader is encouraged to examine the results, selecting several values for t, and comparing the currents. For example, nd the time at which the current in each of the two cases reaches the values of 1.0 mA and 1.0 mA. Also, in each case, nd t1 for the maximum current. Underdamped or Oscillatory Case < !0 When < !0 , s1 and s2 in the solution to the di erential equation suggested in the preceding are q complex conjugates s1 j and s2 j , where is now given by !2 2 . The solution can 0 be written in the exponential form i e t A1 e j t A2 e j t or, in a readily derived sinusoidal form, i e t A3 cos t A4 sin t

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In the circuit of Fig. 7-32, the triodes are identical, RG % 1, and RL rp = 1 ( RK . Show that the circuit is a di erence ampli er, meaning that vo f v1 v2 . Ans: vo RL v1 v2 = 2RL 2rp

PRINCIPLE 2:

EXAMPLE 8.3 Repeat Example 8.1 for C 1 mF. As before, Then, R 1000 s 1 2L !2 0 1 107 s 2 LC p 107 106 3000 rad=s

Thus, j C > j D in Fig. 2-22.

+ RG RG RK _

The constants A3 and A4 are obtained from the initial conditions as before, i 0 0 and vc 0 200 V. From this A3 0 and A4 0:667. Thus, i 0:667e 1000t sin 3000t A See Fig. 8-6. The function 0:667e 1000t , shown dashed in the graph, provides an envelope within which the sine function is con ned. The oscillatory current has a radian frequency of (rad/s), but is damped by the exponential term e t .

Fig. 2-22

The response of the parallel RLC circuit shown in Fig. 8-7 will be similar to that of the series RLC circuit, since a second-order di erential equation can be expected. The node voltage method gives v 1 t dv 0 1 v dt C R L 0 dt Di erentiating and dividing by C yields d 2v 1 dv v 0 dt2 RC dt LC A solution is of the form

Fig. 7-32

PRINCIPLE 3:

CHAP. 8]

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