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					                      RC circuit with voltage source
                                                                            K. Shum
                                                                          Apr, 2010.


         Objective: Model an RC circuit by linear differential equation




The voltage source is sin(t) (measured in volt). The switched is closed at t=0.
The initial voltage across the capacity is -2 V. Suppose that R=1 Ohm, C=3 H and
=1. Describe the voltage across the resistor VR(t) after the switch is closed.



(To help “debugging”, we will work with symbol R, C and  during derivation, and
plug in numerical values at the end)


Let the voltage of the capacitor be Vc(t) and the current (clockwise) be I(t).


From KVL,
                                (   )=       ( )+   ( )


Using the relationship V = I/C for capacitor and Ohm’s law V= IR, we get


                                (   ) = ( ) / + ′( )


We write the differential equation in standard form

                                         1
                                   ( )+                 ( )=            (         )                                     (*)


Note that the coefficient of I(t) is 1/RC. (RC is usually called the time constant)


We solve the differential equation in (*) by integrating factor. That is, we want to
find a function f(t) so that the derivative of f(t) is f(t)/RC. From our knowledge of
exponential function, we see that et/RC is such a function.


Multiply both sides of (*) by et/RC,
                      /
                                            1       /                                                   /
                           ( )+                               ( )=                  (       )


The left hand side is now the derivative of e / I(t)


                                        /                                               /
                                                ( ) =                   (       )               .                      (**)


Integrate both sides of (**),


                                /                                                   /
                                            ( ) =∫                  (       )                       .                 (***)


Using the integral

                                    =      (         +        )+
                                     +
from calculus, (I use the symbol K for the integration constant),
we integrate the right hand side of (***), and get


                 /
                                                                                1
                        ( )=                                  sin           +           cos                 + .
                                                1
                                    +(              )
Here, K is an arbitrary real number.
The voltage across the resistor is
                                                                    1                                        /
             R I(t) =                               sin        +            cos             +                     .
                                    1
                          +
                               (        )

                                                          2
Substitute R,C and  by their numerical values, the general solution is


                                  9       1                   /
                           ( )=      sin + cos      +             .
                                  10      3


The sine and cosine in the above equation can be combined and written in an
amplitude-and-phase form.


                         ( )=                                     /
                                   cos( − (tan     3)) +                                General
                                                                                        solution

The constant K can be any real number. The phase tan          3 is approximately
equal to 1.25.


From the initial condition (0) = 2, we solve for the constant                 , which turns
out to be 1.7 in this case. The answer to the original problem is

                                9                                     /
                         ( )=      cos( − (tan    3)) + 1.7               .
                                10

The solution with initial voltage 2 is plotted in the following figure.

                  2.5


                    2


                  1.5


                    1
          VR(t)




                  0.5


                    0


                  -0.5


                   -1
                     0    5        10      15       20            25          30
                                            t




We observe an initial transient period of about 15 seconds. After 15 seconds, the
voltage oscillates with amplitude (9/10) = 0.95.

                                          3
Discussion: We plot below several solutions corresponding to different initial
conditions.

                5

                4

                3

                2

                1
           VR




                0

                -1

                -2

                -3

                -4
                     0   5        10       15       20        25       30
                                            t

                                                                      Steady state

We see that after about 15 seconds, the five sample solutions, each for a
particular initial condition, are visually indistinguishable in the graph. In the
general solution, the first term is called the stead-state solution.

                         ( )=                                 /
                                   cos( − (tan    3)) +


                                Steady-state solution
The entire solution is called the transient-state solution. It describes how the
system transits from rest to steady state.


The second term K exp(-t/3) is responsible for the transient effect. Its rate of
decay is dictated by the time constant 3 seconds. Also note that the steady-state
solution does not depend on the initial condition. The initial condition only
affects the constant K, which is attached to the second term only.




                                          4

				
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