Mathcad - 02-16

Mathcad - 02-16 - P2-16. A three-phase load consists of...

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θ 21.801deg = F p cos θ ( 29 = F p 0.928 = b. Since this is a delta-load with a series impedance, we can use the total impedance to solve for the phase current. Recall that the phase current leads the line current by 30 deg V ph V AB = V ph 480V = arg V ph ( 29 0deg = I ph V ph Z = I ph 4.138 1.655j - A = arg I ph ( 29 21.801 - deg = magnitude: I ph 4.457A = c. The magnitude of the line current is the square root of three times the magnitude of the phase current I L I ph 3 e j - 30 deg = arg I L ( 29 51.801 - deg = magnitude: I L 7.719A = P2-16. A three-phase load consists of three, identical impedances connected in Delta. The impedances have a value of (100 + j 40) Ù . Line voltage = 480 V a. Find the power factor of the circuit.
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Mathcad - 02-16 - P2-16. A three-phase load consists of...

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