# RC by shitingting

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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

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)) +

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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