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RL parallel circuits

Discussion in 'Electronic Basics' started by conrad, May 4, 2007.

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  1. conrad

    conrad Guest

    I have an RL parallel circuit.
    The AC voltage is 120 and the frequency
    is 60 Hz. One branch has a resistor (60 ohms)
    and the other branch(the one parallel to
    the resistor) has an inductor with 60 mH.

    The inductive reactance of my inductor is
    22.61(rounded to nearest hundreth) ohms.

    My question is this, to find the impedance
    I use Z = 1/(sqrt([1/60]^2 + 1/[22.61]^2)) ?
  2. Rich Grise

    Rich Grise Guest

    If that's what the textbook says, then, obviously, yes.

    Good Luck!
  3. Bill Bowden

    Bill Bowden Guest

    I don't do it that way. First I decide what's reasonable, and since
    you have a 60 ohm resistor across the source, the total impedance must
    be less than 60 ohms. The currents are 90 degrees out of phase, so the
    solution is to vector the resistor current at 90 degrees from the
    inductor current, get the result, and then convert that to impedance
    at some angle.

    The resistor current will be 120/60 = 2 amps, and the inductor current
    will be 120/(6.28*60*.06) = 5.31 amps. The total current will be
    sqrt(5.31^2 + 2^2) = 5.67 amps. Therefore, the impedance will be
    120/5.67 = 21.15 ohms. And you can figure out the phase angle.

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