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How do you determine Peak current in Sine Wave 1

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drax

Mechanical
Mar 2, 2000
95
I always heard that you multiply 2.2 by the RMS value but how is this derived?

thanks,
mark
 
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I meant to say how do you determine peak to peak, I've been told to mult rms by 2.3. where does 2.3 come from?

thanks,
Drax
 
RMS means root mean squared, i.e. you multiply the RMS value by the square-root of 2, which is 1.41.


 
If you mean, by "peak to peak" the maximum instantaneous current at two peaks in a single cycle sinewave, then it is 2 times the square root of 2. Or in otherwords, 2.8284.
 
Ooops, I didn't get the second sentence finished. It should say -

Or in otherwords, 2.8284 times the RMS current value.
 
Thanks for all of your help. An engineer at work finally explained the 2.3 multiplier. He showed me a graph using a 8.8 X/R ratio that translates to the 2.3 number instead of the 2.82 number.

thanks again
 
Drax,

Like the others said in normal condition the peak of the sinusoidal current is equal the rms current * sqrt (2).
The peak sinusoidal current only depends on the X/R ratio when short circuit happens. During short circuit the first few cycle will have a DC component, which will add to the sinusoidal current. This current is called asymmetrical short circuit current. The peak and the duration of the asymmetry depend on the X/R (Inductive reactance /Resistance) ratio of the AC circuit.
 
thanks 1x1x1x for explaining that, even though he showed me a chart, I still didn't understand until your email. I'm a mechanical engineer trying to learn the electrical side.
thanks a million,

Drax
 
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