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Given impedance, find voltage drop across resistor

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data


    Determine the voltage V0 in the following network if the impedance Z contains:

    a. a 0.5 mH inductor

    b.a 10 μF inductor

    [Broken]


    2. Relevant equations

    Voltage division for 2 series resistors: (v in)(resistor/total series resistance)

    ZL = ωLj

    Zc = -1j/ωC

    Z = ZL + ZC + R



    Zmagnitude = (R2 + X2)1/2

    θ = arctan(J/R)

    form Z = R + jx


    3. The attempt at a solution

    ω = 377 ?

    Part a:

    ZL = 377 * (0.5 x 10-3) = 0.1885Ωj

    Z total = 100Ω + 0.1885Ωj


    Not sure what to do from here but I guess now try polar conversion, and doing that, Z basically rounds off to

    Z total = (100 ∠ 89.9 deg)

    Z(just the resistor in polar form) = (100 ∠ 90 deg) because the magnitude is 100 and arctan(100/0.000000001) = 90 degrees

    and I think

    V = 100 ∠ 0 deg, from the given source voltage.




    So I think V0t = (100 ∠ 90 deg)*((00 ∠ 90 deg)/(100 ∠ 90 deg))

    = (100 ∠ 90 deg)

    (would be negative though, because the arrow is the opposite direction).

    but not sure. I know part B would basically be the same thing though.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 21, 2013 #2

    rude man

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    Gold Member

    You have a simple voltage divider: Vo/Vin = R/(R + Z)
    where for the inductor Z = jwL etc.

    Vo is a magnitude and a phase angle.
     
  4. May 21, 2013 #3

    gneill

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    Staff: Mentor

    Rework your angle for the total impedance; I can't understand your calculation: arctan(100/0.000000001) ???

    What formula are you using to find Vo? (In symbols)
     
  5. May 21, 2013 #4
    Well, I meant it just for the real resistor and tried to convert just the resistor to polar form... would the resistor angle by itself have an angle of 0, because there's no imaginary impedance (ie, R = 100Ω + 0j)? *Or you can't / you aren't supposed to do that? *

    Actually checking it again too, I had the values / fraction switched around for the arctan theta formula, sorry.

    I know it's voltage division (V0 = V(R1/R1+R2) even though it's impedance in this case ) but I am wondering if I am really supposed to do all calculation in polar form now (and forgot to say that the main problem here might be with polar form again).


    For Z total it was to use

    Zmagnitude = (R2 + X2)1/2

    θ = arctan(imaginary / real)

    so with just R = 100 + 0j, it is R(total alone) = 100 ∠ 0 deg

    while for total impedance:

    Zmagnitude = (1002 + 0.18852)1/2 = 100.0001 still and now

    θ = arctan(0.1885/100) = 0.1 deg

    And then for the real resistor now it would be 0 since I had the formula switched

    so

    Z(total) = 100 ∠ 0.1 deg

    V0t = (100 ∠ 0 deg)*[ (100 ∠ 0 deg)/(100 ∠ 0.1 deg) ]

    ?
     
  6. May 21, 2013 #5

    gneill

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    Staff: Mentor

    Use polar or rectangular formats, or both; addition and subtraction are easier to accomplish in rectangular form (just like vector components), while multiplication and division are often easier in polar form. You should be able to convert back and forth between the formats with confidence. In fact, I bet your calculator has a build in function that will do it for you!

    What's your final result for part (a)?
     
  7. May 22, 2013 #6
    V0t = (100 ∠ 0 deg)V ?


    I take it I would have to stick with polar coordinates for part B too though right?
     
  8. May 22, 2013 #7

    gneill

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    Staff: Mentor

    The magnitude is good. There should be a small angle, pretty close to zero degrees but not exactly.
    The procedure will be the same, only the particular value of Z has changed.
     
  9. May 27, 2013 #8
    Sorry for bumping this so late, but I ended up getting V0t = (100 ∠ 0.1 deg)V for part a

    and

    V0t = ( 35.3 ∠ 69.3 deg)V for part b
     
  10. May 27, 2013 #9

    NascentOxygen

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    Anyone have any idea what the F stands for in inductance measure? :wink:
     
  11. May 27, 2013 #10

    gneill

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    Staff: Mentor

    I think you'll find that the angle for the first one should be negative, but otherwise your results look good.
     
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