I have 2 equation describing the alternating amperage $I_1$ and $I_2$. I need to get amount of these amperages.

My equations: $$I_1=10\sin(\omega t+30)$$ $$I_2=20\sin(\omega t-50)$$

How can i make it? I forgot.

Thank you.


closed as too localized by Qmechanic Feb 15 '13 at 15:29

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  • $\begingroup$ Please don't post your homework, if that's what this is. Otherwise, it's unclear to me what you're asking. $\endgroup$ – Mark Eichenlaub Nov 14 '10 at 6:28
  • $\begingroup$ @shk: do you need to add i1 and i2 and reduce to the shortest form? $\endgroup$ – Peter Mortensen Nov 14 '10 at 10:13
  • 4
    $\begingroup$ Better always use "current" instead of "amperage", especially on a physics site. $\endgroup$ – Peter Mortensen Nov 14 '10 at 10:19
  • $\begingroup$ I am not sure what you are really asking here. Do you just want the R.M.S. current? $\endgroup$ – Noldorin Nov 14 '10 at 16:06
  • $\begingroup$ What's wrong with "amperage"? dictionary.reference.com/browse/amperage $\endgroup$ – endolith Nov 15 '10 at 17:01

I am going to assume that you ask R.M.S. current and will answer that. If you want some other form of answer just clarify the question a bit more and I'll adjust the answer.

To calculate the RMS(Root-Mean-Square) value of a current, you do the following calculation:

$\sqrt{\frac{1}{T} \int_0^T i(t) dt }$

where T is the period of the signal and t is the variable of integration.

  • $\begingroup$ Oh let me note that you can apply superposition to add both RMS values up, provided that your circuitry is linear. $\endgroup$ – Cem Nov 17 '10 at 12:35

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