RLC Circuit

Discussion in 'Homework Help' started by ahmed95, Sep 3, 2014.

  1. ahmed95

    Thread Starter New Member

    Aug 31, 2014
    3
    0
    Hello,
    I am having problems in determining the capacitor current. I am asked to find the equation for v(t).
    intial conditions are i(0)=40A and v(0)= 40V.
    L is 12.5mH, R=0.1 ohm , C =0.2F
    the thing is I cant figure out the relation for Ic.
    i got v(t)= A1e^-10t+A2e^-40t
    for second equation i have to derivate and use the relation for ic in terms of ir and i as per the diagram.
    i sued ic=-ir-i but thats not right because my answer is wrong.
    please guide me. thanks.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hi,

    If you already calculated v(t) and that v(t) is the voltage directly across the capacitor Vc(t), then you can simply take the derivative of that equation and multiply by C because one of the electrical definitions of C is:
    i=C*dv/dt

    where
    dv/dt is the derivative of the voltage across C with respect to time (in this case the voltage (in volts) at the top is positive, bottom negative),
    i is the current through the cap in Amperes (conventional current flow in the cap from top through to the bottom),
    C is the value of the capacitor in Farads,
    t is time in seconds.

    The derivative of the form of the voltage A*e^(-at)+B*e^(-bt) is pretty simple so this comes out pretty easy to do actually.
    Since you are given all the values in the circuit and all the initial conditions, you should be able to calculate this all out to constants with the only variable left being the time t.
     
    JoyAm likes this.
  3. ahmed95

    Thread Starter New Member

    Aug 31, 2014
    3
    0
    hey thanks for the awaited reply but I have a few corrections to offer.
    i) isnt i the current across the inductor?
    ii) isnt Ic( current across capacitor) be -(Ir+I) ?

    thanks.
     
  4. JoyAm

    Member

    Aug 21, 2014
    112
    4
    I dont think i really understand what your question is .
    i is the current across the inductor, but thats obvious , i mean it is shown in the picture you attached
    also ic=-ir-i where
    ic=c*dVc(t)/dt
    and
    i=Vc(t)/R
    I hope this will help you
     
    ahmed95 likes this.
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    Hi,

    Normally when we talk about current we say that the current flows THROUGH an element, and a voltage appears ACROSS an element.

    In the equation i have:
    i=C*dv/dt

    i explained that 'i' is the capacitor current not the inductor current. The capacitor current is related to the inductor current by:
    iC=-iR-iL

    but let me rewrite the equation again in a more clear form:
    iC=C*d(vC)/dt

    So here you can see that if you know vC (a function of time) you can get the capacitor current iC by taking the derivative of vC with respect to time adn then multiplying by the capacitor value. Written out completely in function form it looks like this:
    iC(t)=C*d(vC(t))/dt

    where we see that iC and vC are functions of time.

    Since vC(t) has the form:
    vC(t)=A*e^(-a*t)+B*e^(-b*t)

    the derivative is:
    d[vC(t)]/dt=-a*A*e^(-a*t)-b*B*e^(-b*t)

    and so the current THROUGH THE CAP is:
    iC(t)=C*(-a*A*e^(-a*t)-b*B*e^(-b*t))

    or rewritten:
    iC(t)= -C*(a*A*e^(-a*t)+b*B*e^(-b*t))

    You should probably solve for A and B first though and then you can get the final solution in the end. A and B are found from the initial conditions.
     
    ahmed95 likes this.
  6. ahmed95

    Thread Starter New Member

    Aug 31, 2014
    3
    0
    thanks a lot for the help i got it :)
     
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