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Sinusoidal Current of .5 Amps (rms) & 5 kHz

  1. Aug 16, 2014 #1
    This is just a quick question:

    A problem I'm working on says "a sinusoidal current of .5 amps (rms) and 5 kHz." Later, in the problem solution, [itex]I(t)[/itex] is written as [itex].5 \sqrt{2} \cos{(10^4 \pi t)}[/itex]. I think I'm simply misunderstanding something about the construction of a current function when given amps and such. Why is the current not: [itex].5 \sin{(10^4 \pi t)}[/itex]? Where does the [itex]\sqrt{2}[/itex] come from? I assume the cosin and sin are interchangeable.

    Also, what is "rms"?
     
    Last edited: Aug 16, 2014
  2. jcsd
  3. Aug 16, 2014 #2

    ehild

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    The time dependence of the sinusoidal current of frequency f and amplitude A is I=Asin(ωt), ω=2πf .

    The rms value of current and voltage is abbreviation of "root mean square" the square root of the time average of the square of the current (or voltage).

    The average of I2 is [tex](I^2)_{av}=\frac{1}{T}\int_0^T{I^2dt}=\frac{A^2}{T}\int_0^T{\sin^2(ωt)dt}=\frac{A^2}{2}[/tex]

    So ##I_{rms}=\sqrt{A^2/2}=A/\sqrt2##. If you are given the rms current, the amplitude is A=√2 Irms.

    ehild
     
  4. Aug 16, 2014 #3
    If I could delete this thread, I would. Instead, I will answer it:

    The [itex]\sqrt{2}[/itex] and "rms" are very related. The RMS value for some periodic current is the DC current that delivers the same average power. For sinusoidal waves of the form [itex]a \sin{(2 \pi f t)}[/itex], the corresponding RMS value is [itex]\frac{a}{\sqrt{2}}[/itex]. So, for a sinusoidal wave of RMS current .5 A, we just solve for [itex] \frac{a}{\sqrt{2}} = 5[/itex], where [itex]a[/itex] is the amplitude of the sinusoidal current.

    Most information taken from: http://en.wikipedia.org/wiki/Root_mean_square#RMS_of_common_waveforms

    Edit: Whoops! I was writing this while you wrote your answer, ehild. Sorry 'bout that.
     
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