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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

7 5928 99.4 27.7 15.0 10.5 8.26 6.99 6.18 5.61 5.20 4.89 4.64 4.44 4.28 4.14 4.03 3.93 3.84 3.77 3.70 3.64 3.59 3.54 3.50 3.46 3.42 3.39 3.36 3.33 3.30 3.12 2.95 2.79 2.64

8 5981 99.4 27.5 14.8 10.3 8.10 6.84 6.03 5.47 5.06 4.74 4.50 4.30 4.14 4.00 3.89 3.79 3.71 3.63 3.56 3.51 3.45 3.41 3.36 3.32 3.29 3.26 3.23 3.20 3.17 2.99 2.82 2.66 2.51

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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A partial order is strictly de ned as an order that satis es three axioms: every element is related to itself, the relation cannot hold both ways between two elements, and transitivity. It is not required to be total. That is, not every two elements are related. x is related to y if and only if the mapping of x is related to the mapping of y.

9 6023 99.4 27.3 14.7 10.2 7.98 6.72 5.91 5.35 4.94 4.63 4.39 4.19 4.03 3.89 3.78 3.68 3.60 3.52 3.46 3.40 3.35 3.30 3.26 3.22 3.18 3.15 3.12 3.09 3.07 2.89 2.72 2.56 2.41

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.

10 6056 99.4 27.2 14.5 10.1 7.87 6.62 5.81 5.26 4.85 4.54 4.30 4.10 3.94 3.80 3.69 3.59 3.51 3.43 3.37 3.31 3.26 3.21 3.17 3.13 3.09 3.06 3.03 3.00 2.98 2.80 2.63 2.47 2.32

12 6106 99.4 27.1 14.4 9.89 7.72 6.47 5.67 5.11 4.71 4.40 4.16 3.96 3.80 3.67 3.55 3.46 3.37 3.30 3.23 3.17 3.12 3.07 3.03 2.99 2.96 2.93 2.90 2.87 2.84 2.66 2.50 2.34 2.18

Fig. 8-4

15 6157 99.4 26.9 14.2 9.72 7.56 6.31 5.52 4.96 4.56 4.25 4.01 3.82 3.66 3.52 3.41 3.31 3.23 3.15 3.09 3.03 2.98 2.93 2.89 2.85 2.82 2.78 2.75 2.73 2.70 2.52 2.35 2.19 2.04

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 .

20 6209 99.4 26.7 14.0 9.55 7.40 6.16 5.36 4.81 4.41 4.10 3.86 3.66 3.51 3.37 3.26 3.16 3.08 3.00 2.94 2.88 2.83 2.78 2.74 2.70 2.66 2.63 2.60 2.57 2.55 2.37 2.20 2.03 1.88

Classical measurement theory authors have used both versions. The advantage of the second version is that partial orders in the empirical system can be mapped to integers or reals that are both totally ordered.

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.

24 6235 99.5 26.6 13.9 9.47 7.31 6.07 5.28 4.73 4.33 4.02 3.78 3.59 3.43 3.29 3.18 3.08 3.00 2.92 2.86 2.80 2.75 2.70 2.66 2.62 2.58 2.55 2.52 2.49 2.47 2.29 2.12 1.95 1.79

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