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Instantaneous Current in single phase circuit

  1. Sep 3, 2015 #1
    1. The problem statement, all variables and given/known data

    SOURCE:
    v(t) = (311)cos(314t) [v]

    IN SERIES WITH:
    R = (0.3) [Ω]
    L = j(0.7) [Ω]
    "load" = 2.5 + j(1.0) [Ω]


    2. Relevant equations
    Find the instantaneous current, and the phasor current.

    3. The attempt at a solution
    (1.) I first found the frequency of the source:
    (314) / (2*pi) ≈ 50 Hz.

    (2.) Figure out the value of the inductor:
    jωL = j(0.7) => j(50)L = j(0.7) => L = (0.7/50) = 14 [mH]

    (3.) I started doing KVL around the loop:
    (-311)cos(314t) + (0.3)(i(t)) + (14e-3) (di/dt) ...

    Then I didn't know what to do for the "load." It isn't specified if it's a capacitor/inductor/mix, and I'm not sure how to go from 2.5 + j(1.0) [Ω] to something I can use in the time domain? Is it just the real part of that value?
    So... (-311)cos(314t) + (0.3)(i(t)) + (14e-3) (di/dt) + 2.5 = 0 ?

    ---------------------------------------------------------

    (1.) For the phasor I found the total impedance:
    0.3 + j(0.7) + 2.5 + j(1.0) = 2.8 + j1.7 = 3.27∠31.26 [Ω]

    (2.) Then did:
    I = V/Z = (311∠0) / (3.27∠31.26) = 94.94∠-31.26 [A]

    Does the phasor look correct, and can I back into the instantaneous from the phasor?

    Thanks for any help! I'm just getting back into this and I'm pretty rusty!
     
  2. jcsd
  3. Sep 3, 2015 #2

    NascentOxygen

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    Staff: Mentor

    Hi Robjob. :welcome:

    You haven't provided the question, so it's difficult to say whether your answer is correct. Please include the problem statement.

    One error: XL = jwL where w is 314 not 50
     
  4. Sep 4, 2015 #3
    To put 94.94<-31.26 A into instantaneous form you would do:

    I(t) = 94.94cos(314t-31.26) Amps
     
  5. Sep 5, 2015 #4

    rude man

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    Homework Helper
    Gold Member

    Recompute ω!
     
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