Calculating RMS Current from I(t)=3+4sinwt

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To calculate the RMS current from the function I(t)=3+4sin(wt), the correct approach involves using the definition of RMS, which is I_rms = √((1/T)∫_0^T I(t)² dt). The initial misunderstanding was that the RMS value could be derived simply by dividing the maximum value by 1.414, leading to an incorrect answer of 7/√2. The correct RMS value, after proper calculation, is √17, which can be derived by squaring the current function, integrating, and applying the RMS formula. Understanding the derivation and applying trigonometric identities is crucial for accurate results.
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Homework Statement



If the instantaneous current is given by I(t)=3+4sinwt, what is the value of rms current?


The Attempt at a Solution



I thought rms value was max val/1.414. The answer I got was \frac{7}{\sqrt{2}}. The answer given is \sqrt{17}. Why?
 
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chaoseverlasting said:

Homework Statement



If the instantaneous current is given by I(t)=3+4sinwt, what is the value of rms current?


The Attempt at a Solution



I thought rms value was max val/1.414. The answer I got was \frac{7}{\sqrt{2}}. The answer given is \sqrt{17}. Why?

Actually, the first principles definition of RMS value of a current I(t) with periodicity T is

I_{rms} = \sqrt{(\frac{1}{T})\int_0^T{{I(t)}^2}dt}

RMS is Root-Mean-Square. That's exactly what that expression represents, written from left to right. When you evaluate it, you work from inside outward, i.e. square the current, take the mean, then take the root.

Review the derivation of the expression for the RMS of a simple sinusoidal waveform e.g. I(t) = I_0\sin{(\omega t)} and use that method to work out the RMS here. You will need trig identities to simplify then integrate (everything becomes simple at the end because of the bounds of integration).
 
Thank you. I knew I was being pigheaded here. I got it now.
 
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