Mathcad - 03-03 and 04 and 05

Mathcad - 03-03 and 04 and 05 - Problem 3-3: Solution...

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NOTE: THD R = THD 2 or 46.2% THD 2 0.462 = THD 2 I 3 2 I 5 2 + I 7 2 + I 9 2 + I rms := using the rms current to find THD, we find I rms 41.72 amp = I rms I 1 2 I 3 2 + I 5 2 + I 7 2 + I 9 2 + := The rms of the current is the square root of the sum of the squares of the rms magnitudes of the components: Problem 3-5: NOTE: THD F = THD 1 or 52.1% THD 1 0.521 = THD 1 I 3 2 I 5 2 + I 7 2 + I 9 2 + I 1 := Based on the fundamental component, the total harmonic distortion is Problem 3-4: Note that the neutral current is larger than the phase current for this load! I n 54.213amp
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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