Mathcad - 02-04

Mathcad - 02-04 - = P V 2 R = a The real power will only be...

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arg I tot ( 29 56.976deg = I tot 22.019A = I tot 12 18.462j + A = I tot I R I C + = I C 18.462jA = I C V jX C = I R 12A = I R V R = b. Since the resistor and capacitor are in parallel, we can calculate the current through each and add them together S 5.285 kVA = S P jQ + = The apparent power is the magnitude of the complex power jQ 4.431j - kVAR = jQ V 2 jX C = Note: the bar over the right hand side indicates the conjugate of the expression (remember S=VI* ) kVA kW kVAR kW define The reactive power is only associated with the capacitive reactance, so: P 2.88kW
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Unformatted text preview: = P V 2 R = a. The real power will only be disipated in the resistor. Since this is a parallel load, we can write: V 240 V ⋅ = X C 13-Ω ⋅ = R 20 Ω ⋅ = P2-4. Consider a parallel R-C circuit with R = 20 Ω and -jX C = -j13 Ω , with an applied voltage of 240 p o V. a. Calculate the real, reactive, and apparent power b. Calculate the total current from the source and the resistor and inductor currents....
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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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