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Parallel RL circuit

  1. Jun 14, 2005 #1

    ng

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    Hi there,

    Can someone help me with parallel RL circuit across ac supply?wats the differential equation .......
    Thanx for all help.

    ng
     

    Attached Files:

    • I.doc
      I.doc
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  2. jcsd
  3. Jun 14, 2005 #2

    Astronuc

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    Staff: Mentor

  4. Jun 14, 2005 #3

    ng

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    Thanx for the response.I have already seen that and din help me much.
    The problem is with getting the transient solution.
    RC parallel -If the input is Vcoswt then current throu R will be V/Rcoswt.throu C is
    -CwV sinwtThis is te steady state i understand.But the transient and getting to it throu te differential equations is my problem for both RL and Rc parallel across ac supply.
     
  5. Jun 14, 2005 #4

    Astronuc

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    Staff: Mentor

    In the doc file, you have a differential equation.

    V = VR(t) = VL(t) = L [itex]\frac{di(t)}{dt} [/itex] = ( I(t) - i(t) )/R,

    but I(t) = V(t)/R + i(t), and let i(t) = VL/ZL where Z is the impedance.

    Also consider i(t) = io cos ([itex]\omega[/itex] t + [itex]\theta[/itex]), where [itex]\theta[/itex] is the phase angle.

    The voltages are in phase, the currents are out of phase.

    Evaluate at time, t = 0.

    See if that helps.
     
  6. Jun 14, 2005 #5

    SGT

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    In your attached file you have already arrived to the differential equation, you just didn't write it in a form to solve it. Write your equation as:
    [tex]L\frac{di}{dt} + Ri = RIcos\omega t[/tex].
    This is a non homogeneous differential equation. Find the solution of the homogeneous equation iH(t) and a particular solution that will be of the form iP(t) = Acos(ωt + φ).
    The general solution is i(t) = iH(t) + iP(t)
    Replacing the initial conditions you have the solution. It contains two terms: the exponential term is the transient and the cosine term is the steady state.
     
  7. Jun 14, 2005 #6

    ng

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    hey sgt,thankyou for that reply.yeah,how stupid of me.....u r right.
    thanx again.
     
  8. Jun 14, 2005 #7
    For your last circuit. I started it out like this,

    Apply KCL at the top supernode and you have

    [tex]I_{in} = i_R + i_L + i_c[/tex]

    Note, that all cicuit components have the same potential and recall that

    [itex]L \frac {di_L}{dt} = v_L[/itex] and [itex]C \frac {dv_c}{dt} = i_c[/itex]
     
  9. Jun 14, 2005 #8

    ng

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    thanx for the reply, corneo
    hmm... okay.but i was wondering why shud i assume the solution
    Acos(wt+Phi)?

    Is this what is assumed for first order circuits subject to ac supply?i read its Acoswt+Bsinwt for second order circuits.am i right?
    :confused:
     
  10. Jun 15, 2005 #9
    You should assume the solution is [tex]A \cos (\omega t + \phi)[/tex] because I think your going to use methods of undetermined coefficients. I think cosines and sines appear in your solution to 2nd order ODE because the circuit is underdamped ([itex]\zeta < 1[/itex]). I just took a class in circuits so a lot of this is still new to me.
     
  11. Jun 15, 2005 #10

    ng

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    Thankyou corneo,
    Okay
    so i get

    Ldi/dt+Ri=RIcoswt

    to get the complementary solution assume
    Ldi/dt+Ri=0

    int di/i=- intR/L dt

    ln i= -R/Lt+K

    i=exp(-t/tau+k)

    at t=0 initial condition i is V/R

    then its..?is it right?

    for particular solution i assume i=Acoswt
    putting in first eqn

    -Lwsinwt+RAcoswt=RIcoswt...

    where am i going wrong?
    please help.
     
  12. Jun 15, 2005 #11

    SGT

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    Both solutions can be used with first and second order circuits (or higher orders). They are equivalent.
    [tex]Acos(\omega t + \phi) = Acos\omega t cos\phi - Asin\omega t sin\phi = A_1 cos\omega t + A_2 sin\omega t[/tex]
    where
    [tex]A_1 = Acos\phi[/tex]
    [tex]A_2 = -Asin\phi[/tex]
     
  13. Jun 15, 2005 #12

    ng

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    Okay i get it.thanx SGT.
    but ...heheh :biggrin: how do u find the value of A?
     
  14. Jun 15, 2005 #13

    SGT

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    You replace the solution and its derivative in the differential equation. When expanding the sine and the cosine of [tex]\omega t + \phi[/tex], you will have in both members of the equation terms in sine and in cosine. Equalling the coefficients of like terms you get the two unknowns A and [tex]\phi[/tex].
     
  15. Jun 15, 2005 #14

    ng

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    SGT,Thanx again ALOT!!!
    So,
    for particular solution i assume i=Acoswt
    putting in first eqn

    Ldi/dt+Ri=RIcoswt

    -Lwsinwt+RAcoswt=RIcoswt
    equating the coefficients on LHS and RHS
    looks like i am going wrong...is it?
    please help.
     
  16. Jun 15, 2005 #15

    SGT

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    No, you put [tex]i = Acos(\omega t + \phi)[/tex]
    or [tex]i = Acos(\omega t) + Bsin(\omega t)[/tex]
    calculate [tex]\frac{di}{dt}[/tex]
    and replace i and [tex]\frac{di}{dt}[/tex] in the equation.
     
  17. Jun 15, 2005 #16

    ng

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    :smile:
    SGT...thanx a million.

    i hope i am right this time....

    -L(Awsinwt+Bwcoswt)+R(Acoswt+Bsinwt)=RIcoswt

    (-LBw+RA)coswt=RIcoswt

    -LBw+RA=RI
    -LAw+RB=0

    solve this to get A and B
    w=2 pi f

    this approach is used for second order also right?

    thankyou once again!!! :approve:
     
  18. Jun 15, 2005 #17

    ng

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    :biggrin: just one more question...
    why cant it be Acoswt?why the phi?
    why shud it be coswt?why not sinwt?

    (am i being really stupid? :rolleyes: )
     
  19. Jun 16, 2005 #18

    SGT

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    When you differentiate cosωt you get -sinωt. If you replace both in the equation you get a term in sinωt in one side that must be equalled to zero on the other side, this is inconsistent. Just try it in your equation and you will see.
    Of course you can use the sine instead of the cosine, they are the same function, only with different phases.
    cosωt = sin(π/2 - ωt)
     
  20. Jun 16, 2005 #19

    ng

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    okay i understand.
    Is this used in second order equations as well?
    hmm.... an underdamped case for instance
    there r two coefficients in it right?
    how is that found?

    i hope i am notgetting onto ur nerves!
    thankyou for all the help.
     
  21. Jun 16, 2005 #20

    SGT

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    Both in first and second order equations the steady state response to a sinusoidal input have two parameters to be evaluated: A and φ or A1 and A2.
    You have two equations: one with the coeficients of the terms in sine and one with the coeficients of the terms in cosine. Two equations, two unknowns ...
     
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