Exercises I solved (asking to see if they're right)

Discussion in 'Homework Help' started by newzed, Jan 29, 2012.

  1. newzed

    Thread Starter New Member

    Jan 11, 2012
    16
    0
    Pleaseeee help me! I already solved them, I'm only asking to you if they're right or not :) :) :)

    So, First page:
    1) Is the only one I wasn't able to solve.. I know I have to attach to the terminals 1V source but then? what am i supposed to do?
    2) Laplace -> 1/2 e^-2Ts - 2/s e^-3Ts + 1/s e^-4Ts
    3) T= L/(Req) where Req= (R // (R+R))+R)
    4) No current right? I mean, it's an ideal opamp so no current in inverting input and no current in noninverting input so everything has 0 current! So 0!

    Second page:
    1) if part one has been open circuited we're looking for impedance parameter, so, for example Z22 will be= v2/i2 while i1=0
    2) in order to compute phasor I
    First i find, after summing in parallel 8-j2 with j6 I
    Then, by current division i find it in the branch i'm looking for
    I= 0.72 (phase 76.35°)

    3) Having apparent power that is Vrms*Irms= 50
    and pf=0.8 --> since V= 20 --> vrms = 20/√2 = 14.14
    so Irms= 2.28
    So z= 6.20 phase of 36.86
    4) 1/(s^2CL + sRC + 1)

    Last Page is about First Order Circuit
    for t<0 vc(0) = 14/3
    for t>0 what is going on?? I mean, the switch closes so the 7A current passes all on the right... so??

    Thank you so much for your time, I really really appreciate it
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    Ix is current flowing from outside into terminal A (and from terminal B out to external circuit)
    Test voltage is same as Vx, and then equivalent resistance of the circuit is
    Rab=Vx/Ix

    Assuming that Ix flows from A to B through R, and assuming that Iy flows from +beta*Vx through 3R and R to negative terminal of beta*Vx, we can write Kirchoff equations:
    Iy=(beta*Vx-Vx)/3R
    or
    Iy=Vx(beta-1)/3R

    and
    Vx=R(Ix+Iy)
    or
    Vx/R=Ix+Vx(beta-1)/3R
    Vx([(1/R)+ (1/3R) - (beta/3R)]=Ix

    finally
    Vx/Ix=1/[(1/R)+ (1/3R) - (beta/3R)]
    Vx/Ix=1/[(3/3R)+ (1/3R) - (beta/3R)]
    Vx/Ix=3/(3+1-beta)
    Vx/Ix=3/(4-beta)
     
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