Page 174 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 174

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
               CHAP. 8]
               Overdamped Case ð >! 0 Þ                                                              163
                   In this case, both   and   are real positive numbers.
                                                                        t
                                       i ¼ A 1 e ð  þ Þt  þ A 2 e ð    Þt  ¼ e   t ðA 1 e þ A 2 e   t Þ

               EXAMPLE 8.1 A series RLC circuit, with R ¼ 200 
, L ¼ 0:10 H, and C ¼ 13:33 mF, has an initial charge on the
               capacitor of Q 0 ¼ 2:67   10  3  C.  A switch is closed at t ¼ 0, allowing the capacitor to discharge.  Obtain the
               current transient. (See Fig. 8-4.)
                   For this circuit,
                                                                              q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
                              R                   1
                                   3  1
                                                             5  2
                                              2
                                                                                     2
                                                                                 2
                            ¼   ¼ 10 s ;     ! 0 ¼  ¼ 7:5   10 s ;  and      ¼      ! ¼ 500 s  1
                                                                                     0
                             2L                  LC
               Then,                            i ¼ e  1000t ðA 1 e 500t  þ A 2 e  500t Þ
               The values of the constants A 1 and A 2 are obtained from the initial conditions.  The inductance requires that


                 þ
               ið0 Þ¼ ið0 Þ.  Also the charge and voltage on the capacitor at t ¼ 0 þ  must be the same as at t ¼ 0 , and

               v C ð0 Þ¼ Q 0 =C ¼ 200 V.  Applying these two conditions,
                                                    and
                                       0 ¼ A 1 þ A 2         2000 ¼ 500A 1   1500A 2
               from which A 1 ¼ 2; A 2 ¼ 2, and, taking A 1 positive,
                                                       500t   1500t
                                                 i ¼ 2e     2e     ðAÞ
               If the negative value is taken for A 1 , the function has simply flipped downward but it has the same shape.  The signs
               of A 1 and A 2 are fixed by the polarity of the initial voltage on the capacitor and its relationship to the assumed
               positive direction for the current.









                                                         Fig. 8-4



               Critically Damped Case ð  ¼ ! 0 Þ
                   With   ¼ ! 0 , the differential equation takes on a different form and the two exponential terms
               suggested in the preceding will no longer provide a solution.  The equation becomes
                                                    2
                                                   d i    di    2
                                                      þ 2   þ   i ¼ 0
                                                    2
                                                   dt     dt
               and the solution takes the form i ¼ e   t ðA 1 þ A 2 tÞ.
               EXAMPLE 8.2 Repeat Example 8.1 for C ¼ 10 mF, which results in   ¼ ! 0 .
                                                                                                      þ

                   As in Example 8.1, the initial conditions are used to determine the constants.  Since ið0 Þ¼ ið0 Þ,
               0 ¼½A 1 þ A 2 ð0ފ and A 1 ¼ 0. Then,
                                             di  d       t         at    t
                                               ¼   ðA 2 te  Þ¼ A 2 ð  te  þ e  Þ
                                             dt  dt
                                                               3
                                                              10 t
               from which A 2 ¼ðdi=dtÞj 0 ¼ 2000. Hence, i ¼ 2000te  (A) (see Fig. 8-5).
                                   þ
                   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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