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Alternating current

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    An inductor has a resistance of 30 Ohm and an inductance of 0,1 H. A current with the peak value of 2 A runs through. Calculate the effect factor and the active effect if it connects to a grid with 230V and a frequency of 50 Hz.

    My book says that the effect factor is 0,69 which I also get, but I can't get the active effect to 60 W. Thereby I believe that I am doing something wrong with the apparent effect (S).

    R (Resistance) = 30 Ohm
    L (Inductance) = 0,1 H
    ipeak (Current peak) = 2 A
    U (Voltage) = 230 V
    f (Freq.) = 50 Hz

    2. The attempt at a solution
    f = 50 Hz ⇒ ω = 2π*f = 100π rad/s
    RL = ω*L = 100π*0,1 = 10π Ohm

    φ = arctan(XL/R) ≈ 46,3 degrees
    Effect factor = cos(φ) ≈ 0,69

    S = U*I
    I = ipeak/sqrt(2)
    ⇒ S = (U*ipeak)/sqrt(2) ≈ 325,27 VA

    Active effect = P = S*cos(φ) ≈ 224 W


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  3. Mar 15, 2015 #2


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    Staff: Mentor

    Hi 9robban6, Welcome to Physics Forums.

    I'm a bit confused by the specification of a peak current of 2 A and a grid connection of 230 V. The load described would carry more current than 2 A peak if connected to a 230 V RMS source at 50 Hz.

    Also, where are you getting the 60 W "active effect" value? I figure that the described load would sink in excess of 800 W both real and reactive, combining for an apparent power in excess of 1200 VA or so (I won't give exact values here).

    Your power factor ("effective value"?) calculation looks good at 0.69 .
  4. Mar 16, 2015 #3
    Sorry for bad description, it's originally written in Swedish. I try again.

    Through an inductor (with the resistance of 30 Ohm and inductance of 0,1 H) runs a current with the peak value of 2 A. Calculate the effect factor and the real power if it is connected to the electricity grid. (Which here in Sweden is 230 V 50 Hz).

    Maybe it shouldn't be 50 Hz? I don't know. But I know that U = 230 V.

    The task is in my book and the book says that the answer to the task is:
    Effect factor = 0.69
    Real power = P = 60W
  5. Mar 16, 2015 #4


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    Staff: Mentor

    Okay, it seems that the important points to take from the problem statement are the current value, the frequency of supply (50 Hz), and the component values. The voltage of the grid is irrelevant since you're given the current through just the circuit fragment consisting of an inductance and a resistance:

    Use a different expression for the power (one that doesn't involve the voltage), or first calculate the voltage across the relevant component due to the current it's passing.
  6. Mar 16, 2015 #5


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    Staff: Mentor

    I think that will solve it!! http://imagizer.imageshack.us/a/img29/6853/xn4n.gif [Broken]
    Last edited by a moderator: May 7, 2017
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