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Homework Help: Average power in series LCR

  1. Sep 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A sinusoidal voltage of peak 283 V and frequency 50 Hz is applied to a series LCR circuit in when R = 3ohm, L is 25.48mH and C is 796 microF. Find the power dissipated in the circuit.

    The book says,
    Power = I^2 R
    Where I = rms current.
    Now, I = 283/5 [itex]\sqrt{2}[/itex] = 40 A
    So, power = 40×40×3 = 4800 W.

    Why didn't they include 'the power factor' while calculating the power?
    Also, isn't what they have calculated is a special case? I mean they've taken Z = R, ( because power = I^2 Z cos θ) which is only possible at resonance. (Cosθ = power factor, which they have taken as 1)
    This can't be a general solution, right?
  2. jcsd
  3. Sep 17, 2014 #2


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    Homework Helper

    The reactive elements (capacitor, inductor) do not dissipate power, as the phase difference between the voltage and current is ±90 degrees on them, so cosθ=0. Only the resistances dissipate power, where a power factor is 1.

    To get the dissipated power, calculate the rms current Irms=Urms/|Z| and P = Irms2R.

  4. Sep 18, 2014 #3
    That means cosθ for a series rlc circuit is always 1?
  5. Sep 18, 2014 #4
    No .

    P = irms2Zcosθ , where Z is the impedance of the circuit and cosθ is the power factor . But Zcosθ = R . So we have P = irms2R .

    This is essentially what ehild has explained that power is dissipated only within the resistor .
    Last edited: Sep 18, 2014
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