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AC circuit, voltage expressed by time

  1. May 4, 2014 #1
    1. The problem statement, all variables and given/known data

    A voltage source e(t) = 50 sin(2000t + 30°) V supplies the current i(t) = 0.5 cos(2000t) A in a circuit. Find the impedance in the circuit, in both polar and rectangular form. Is the circuit inductive or capacitive?

    2. Relevant equations



    3. The attempt at a solution

    I have a problem here, right of the bat I'm not used to seeing voltage and current being given as an expression including time. Well, in AC circuits, that is. I've looked through my book, but I can't find any examples on this.

    There isn't much work to show here as I'm confused as to where I should start.

    I RMS involved? I picture a sinus wave with current and voltage waves, that's where that idea came from.

    Any input, at all, would be greatly appreciated.
     
  2. jcsd
  3. May 4, 2014 #2

    gneill

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    Staff: Mentor

    Consider how much simpler life would be if both the voltage and current were expressed in terms of the same trig function...
     
  4. May 4, 2014 #3
    So I can convert the cosine to sine, or vice versa. I'm just confused by the whole expression for some reason.

    The voltage and current is expressed by time. Given that I understand that, which frankly I don't, how can I use that to find the impedance?

    What I understand about the expressions is that you can read the values of the current or voltage at a given time. I assume I can use this to find the RMS value, or average value.
     
  5. May 4, 2014 #4

    gneill

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    Staff: Mentor

    The idea is to convert everything to phasors. Then the usual phasor manipulations will be at your disposal.
     
  6. May 4, 2014 #5
    So e(t) = 50 sin(2000t + 30°) -> e(t) = 50 cos(2000t + (30° - [itex]\frac{∏}{2}[/itex]))
    In phasor form this is 50 ej(1/2 - ∏/2 which is 50V ∠1.07°


    i(t) = 0.5 cos(2000t) in phasor form is 0.5 ej0 which translates to 0.5A∠90°

    Am I onto something here? I'm fully aware there will be mistakes here even if I've gone though my numbers twice. I'm not used to these calculations.
     
  7. May 4, 2014 #6

    gneill

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    Staff: Mentor

    That's the right idea. Check your angle calculations. pi/2 is 90°, so 30° - 90° is not likely to be 1.07°. Also, e0 does not yield an angle of 90°. What's e0 numerically?
     
  8. May 5, 2014 #7
    e(t) = 50 sin(2000t + 30°) -> e(t) = 50 cos(2000t + (30° - ∏/2))
    In phasor form this is 50 ej(1/2 - ∏/2) which is 50V ∠-60°


    i(t) = 0.5 cos(2000t) in phasor form is 0.5 ej0 which translates to 0.5A∠0°

    Z = [itex]\frac{V}{I}[/itex] = [itex]\frac{50V∠-60°}{0.5A∠0°}[/itex] = 100Ω∠-60°

    The circuit is capacitive since the voltage lags the current by 60°.
     
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