Mathcad - 02-03

Mathcad - 02-03 - P2-3: Figure 2-29 shows a sinusoidal...

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arg Z load ( 29 60 - deg = Z load 8.501 = Z load 4.25 7.362i - = Z load V ph I ph = d. The impedance is found by dividing the voltage by the current in phasor notation I = 14.14 at -30 degrees V = 120.2 at -90 degrees c. The cosine leads the sine by 90 degrees, so changing the reference to cosine will subtract 90 degrees from each phasor in part b. I ph 14.14 e j 60 deg amp = I = 14.14 at 60 degrees I rms 14.142 = I rms I pk 2 = I pk 20 = V ph 120.2 e j 0 deg volt = V = 120.2 at 0 degrees V rms 120.208 = V rms V pk 2 = V pk 170 = b. Phasors are alway rms for the magnitude. If sin( ϖ t) is the reference, then the phase angles are the same as in part a. So the phasors are: i(t) = 20 sin ( ϖ t + π /3) amps The current has a peak value of 20 amps and is leading the voltage by 60 degrees or
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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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