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Painting QR-Code in Software A series RLC circuit, with R 200

EXAMPLE 8.1 A series RLC circuit, with R 200
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, L 0:10 H, and C 13:33 mF, has an initial charge on the capacitor of Q0 2:67 10 3 C. A switch is closed at t 0, allowing the capacitor to discharge. Obtain the current transient. (See Fig. 8-4.) For this circuit, q R 1 103 s 1 ; 7:5 105 s 2 ; !2 and 2 !2 500 s 1 0 0 2L LC Then, i e 1000t A1 e500t A2 e 500t
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The values of the constants A1 and A2 are obtained from the initial conditions. The inductance requires that i 0 i 0 . Also the charge and voltage on the capacitor at t 0 must be the same as at t 0 , and vC 0 Q0 =C 200 V. Applying these two conditions, 0 A1 A2 and 2000 500A1 1500A2
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from which A1 2; A2 2, and, taking A1 positive, i 2e 500t 2e 1500t A
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If the negative value is taken for A1 , the function has simply ipped downward but it has the same shape. The signs of A1 and A2 are xed by the polarity of the initial voltage on the capacitor and its relationship to the assumed positive direction for the current.
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Fig. 8-4
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Critically Damped Case !0 With !0 , the di erential equation takes on a di erent form and the two exponential terms suggested in the preceding will no longer provide a solution. The equation becomes d 2i di 2 2 i 0 dt dt2 and the solution takes the form i e t A1 A2 t .
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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.
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Since i 0 i 0 ,
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HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
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[CHAP. 8
Fig. 8-5
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
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
i e 1000t A3 cos 3000t A4 sin 3000t
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 .
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