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14.11: RLC Series Transient

A battery of constant EMF V is connected to a switch, and an R, L and C in series.  The switch is closed at time t = 0.  We'll first solve this problem by "conventional" methods; then by Laplace transforms.  The reader who is familiar with the mechanics of damped oscillatory motion, such as is dealt with in Chapter 11 of the Classical Mechanics notes of this series, may have an advantage over the reader for whom this topic is new – though not necessarily so!


"Ohm's law" is                        File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif                                                        14.11.1


or                                             File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image004.gif                                                   14.11.2


Those who are familiar with this type of equation will recognize that the general solution (complementary function plus particular integral) is


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image006.gif                                                    14.11.3


where              File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image008.gif                     14.11.4


(Those who are not familiar with the solution of differential equations of this type should not give up here.  Just go on to the part where we do this by Laplace transforms.  You'll soon be streaking ahead of your more learned colleagues, who will be struggling for a while.)



   Case I.   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image010.gif  is positive.  For short I'm going to write equations 14.11.4 as


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image012.gif                                         14.11.5


Then                                        File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image014.gif                                           14.11.6


and, by differentiation with respect to time,


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image016.gif                             14.11.7


At t = 0, Q and I are both zero, from which we find that


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image018.gif                           14.11.8


Thus                            File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image020.gif                        14.11.9


and                              File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image022.gif                                         14.11.10


On recalling the meanings of a and k and the sinh function, and a little algebra, we obtain


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image024.gif                                                               14.11.11


Exercise: Verify that this equation is dimensionally correct.  Draw a graph of I : t.  The current is, of course, zero at t = 0 and ¥.  What is the maximum current, and when does it occur?



 Case II.   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image010.gif is zero.   In this case, those who are in practice with differential equations will obtain for the general solution


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image026.gif                                                    14.11.12


where                                      File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image028.gif                                                                     14.11.13


from which                              File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image030.gif                                                 14.11.14


After applying the initial conditions that Q and I are initially zero, we obtain


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image032.gif                                         14.11.15


and                                          File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image034.gif                                                                     14.11.16


As in case II, this starts and ends at zero and goes through a maximum, and you may wish to calculate what the maximum current is and when it occurs.


 Case III.   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image010.gif  is negative.   In this case, I am going to write equations 14.11.4 as


                        File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image036.gif                14.11.17


where                                      File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image038.gif                                        14.11.18


All that is necessary, then, is to repeat the analysis for Case I, but to substitute -w2 for k2 and  jw for k, and, provided that you know that File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image040.gif you finish with


                                                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image042.gif                                                    14.11.19


This is lightly damped oscillatory motion.



Now let us try the same problem using Laplace transforms.   Recall that we have a V in series with an R , L and C, and that initially File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image044.gif are all zero.  (The circuit contains capacitance, so Q cannot change instantaneously; it contains inductance, so I cannot change instantaneously.)


Immediately, automatically and with scarcely a thought, our first line is the generalized Ohm's law, with the Laplace transforms of V and I and the generalized impedance:


                                                            File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image046.gif                                         14.11.20


Since V is constant, reference to the very first entry in your table of transforms shows that File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image048.gif and so


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image050.gif                      14.11.21


where                                      File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image052.gif                                                14.11.22


Case I.   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image054.gif


                                    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image056.gif                    14.11.23


Here, of course,           File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image058.gif                        14.11.24


On taking the inverse transforms, we find that


                                                File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image060.gif                                                  14.11.25


From there it is a matter of routine algebra (do it!) to show that this is exactly the same as equation 14.11.11. 


In order to arrive at this result, it wasn't at all necessary to know how to solve differential equations.  All that was necessary was to understand generalized impedance and to look up a table of Laplace transforms.



 Case II.   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image062.gif


In this case, equation 14.11.21 is of the form


                                                            File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image064.gif                                                       14.11.26


where File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image066.gif   If you have dutifully expanded your original table of Laplace transforms, as suggested, you will probably already have an entry for the inverse transform of the right hand side.  If not, you know that the Laplace transform of t is 1/s2, so you can just apply the shifting theorem to see that the Laplace transform of File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image068.gif  Thus


                                                            File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image070.gif                                                                 14.11.27


which is the same as equation 14.11.16.


[Gosh – what could be quicker and easier than that!?]



 Case III.   File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image072.gif                           


This time, we'll complete the square in the denominator of equation 14.11. 21:                              


                                    File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image074.gif                14.11.28


where I have introduced w with obvious notation.


On taking the inverse transform (from our table, with a little help from the shifting theorem) we obtain

                                                            File:C:/Users/DELMAR~1/AppData/Local/Temp/msohtmlclip1/01/clip_image076.gif                                                 14.11.29


which is the same as equation 14.11.19.



With this brief introductory chapter to the application of Laplace transforms to electrical circuitry, we have just opened a door by a tiny crack to glimpse the potential great power of this method.  With practice, it can be used to solve complicated problems of many sorts with great rapidity.  All we have so far is a tiny glimpse.  I shall end this chapter with just one more example, in the hope that this short introduction will whet the reader's appetite to learn more about this technique.