R-L-C Series circuit question

Discussion in 'General Electronics Chat' started by jagjit Sehra, Mar 6, 2008.

  1. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    Hi Folks,
    Got a quick question regarding an R-L-C natural overdamped response. I am basing my equation on the voltae across the capacitor (vC).

    vC = A*exp(m1*t) + B*exp(m2*t)

    Now to obtain the values for A and B. I have assumed two intial conditions which is the only way to do it. The two I have chosen are:

    1) vC at time 0 is 2 V

    2) the derivative of vC (dvC/dt) at time 0 is 0.

    I'm a bit concerened about my second condition, the rate of change of voltage with respect to time at time 0. It shouldnt be 0 should it, it should be a non-zero value. Are there any other conditions I can use.

    Any help will be appreciated.
  2. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    In your second equation, if dt = 0, how do you divide by 0?

    i = C*dv/dt

    dv = i / C*dt (if I did my algebra correctly?)

    A current must be injected into the capacitor over some period of time, to create a voltage across the capacitor.
  3. scubasteve_911

    Senior Member

    Dec 27, 2007
    I may be a bit off, but you do not need initial conditions to solve your equation. You should be aware of the difference between a particular solution and a general solution.

  4. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    Hi Folks,
    Thanks for the reply, I dont understand the first reply I am not dividing by 0. if vC has an intial value of 2 V at time 0:

    2 = A*exp(m1*t) + B*exp(m2*t)

    then dvC/dt

    0 = m1*A*exp(m1*t) + m2*B*exp(m2*t)

    I know it doesnt really matter what value I choose I will still get the familar shape of decay. However I wanted to relate it to a real case. I was concerned that if the voltage is already 2 V across the capacitor then at time 0 the voltage would decay to reach 0 and the derivative of vC cant be zero, any ideas of what it should be?.
    Regarding the second reply to my original question, I havnt given all the information again. The transient response I was looking at was the overdamped where vC = A*exp(m1*t) + B*exp(m2*t) because there are two unknown constants there must be two intial values. At least thats my understanding which could be wrong.

  5. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    Hi Folks,
    Sorry more information I left out.

    a= LC = 0.5
    b= RC = 1.5
    c = 1

    The component values I chose were R = 3R , L = 1H and C= 0.5 F.

    I know they are not very realistic values, however it makes the maths easier for me!

    m1 = (-b+ sqrt(b^2 - 4*a*c))/2*a

    m2 = (-b - sqrt(b^2 - 4*a*c))/2*a

    m1 = -1 and m2 = -2
    at time 0 e(m1*t) will equal 1 therefor equations will look as follows:

    2 = 1A + 1B
    0 = -1A -2B

    Simmulatanous equation of two unkowns.