1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Z(in) of circuit help?

  1. Jun 19, 2013 #1
    Z(in) of circuit help??

    1. The problem statement, all variables and given/known data
    qq2wl0.png
    A. 10+j2.5 (Ω)
    B. 10-j0.4 (Ω)
    C. 10-j2.5 (Ω)
    D. 10+j0.4 (Ω)

    2. Relevant equations
    Z(L)=jwl
    Z(C)=-j/wc

    3. The attempt at a solution
    the problem asks for z(in) without any frequency w given? i think i was just confused by the wording of this problem. s=sigma+jw so does that mean, j2 given means w=2?? then answer would be 10+j2.5 (A). but i'm unsure how to do this without any w.
     
  2. jcsd
  3. Jun 19, 2013 #2

    berkeman

    User Avatar

    Staff: Mentor

    Yeah, it looks like they are asking for Z(w=j2), so j2 must be the frequency w. 2Hz shifted by 90 degrees...?
     
  4. Jun 20, 2013 #3
    Remember your transfer function can be expressed in the form H(jω) = Vo(jω)/Vi(jω), where we can denote jω by s.

    Here you are trying to find the input impedance Z(i), or in other words Z(jω), where Z(jω) = Z(2j).

    So you can conclude that ω = 2 rad/s or ω = 2*pi*f = 2 and find the frequency.

    Find the total impedance of the system and substitute your value of ω.
     
  5. Jun 21, 2013 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    The general expression is V(s). To express in the frequency domain, we simply substitute s with , giving V(jω)

    So V(j2) indicates V(jω) with ω=2 radians/sec.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Z(in) of circuit help?
  1. Help , with circuits (Replies: 5)

  2. Circuits Help! (Replies: 3)

  3. Help with Circuits (Replies: 3)

  4. Help with circuit (Replies: 7)

  5. Help with circuit (Replies: 9)

Loading...