RLC superposition

Discussion in 'Homework Help' started by Jess_88, Apr 30, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    hey guys:)

    wondering if my analysis procedure is correct for the cct.

    I need to fined io for the following cct using superposition.

    so I gather I'm soposed to analyse the current io with resect to each individual source in the circuit, by zeroing remaining sources.

    I'm a little unsure of the "zeroing" of each source and if it causes a s/c or o/c.

    These are my circuits I determined by zeroing the sources. do they look right?

    This circuit contains a dc source. therefor the inductor =s/c cap = o/c



    Have I misinterpreted the zeroing sources? or dose this look ok?
    thanks a bunch guys.

    I apologise if I am posting so manny questions in this forum. I'm just trying to get as much help over the range of questions I have to do, while people smarter than myself are online (jegues).
    If i'm posting to manny equation, let me know :)
  2. jegues

    Well-Known Member

    Sep 13, 2010
    Everything looks fine except for the first case where you are zeroing all other sources aside from the 24V DC source.

    You still need the 1 ohm branch with the capacitor, as well as the 2 Henry inductor and 4 ohm resistor.

    Are you assuming steady state for the capacitor/inductor? Is that why you aren't indicating that branch?

    I'm not entirely sure if you can do that. (Is there a reason you think you can?)

    Most of what I've learnt about circuits has been from other members on this fourm, and I myself am still learning.
    Jess_88 likes this.
  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    In my lecture notes it assumes steady state when analysing the DC source for a similar problem. The similar question to fine the voltage across a resister, results in three separate voltages which can be added together due three different frequencies (dc w =0), but is expressed at V = V1 + V2 + V3 (eg: V(t) = 1 + 26sin(3t +15) + 7.8cos(5t + 30))

    hope that helps... and thang you again for the help :)
  4. Jess_88

    Thread Starter Member

    Apr 29, 2011
    Just checking if my method of finding Io for each of the circuits I drew is ok.

    first one is easy(24/6)

    for the second I have
    It = ((1-j2) + (2//(4+j6)))/10<-120
    io = It(2/(2+6)+j6)

    third cct
    Zt = [(2//(1-j2))]//(4+j6)
    Io = [(2//(1-j2)) / Zt] x (6<0)

    look ok?