Mathcad - 02-05

Mathcad - 02-05 - C ⋅-= Since this is a series circuit we...

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The total current was found above. Since this is a series circuit, the resistor and capacitor currents are the same Since the circuit is capacitive, the reactive power is negative Q 1.694 - kVAR = Q Im S ( ) = reactive power is the imag part of S: P 2.824kW = P Re S ( ) = real power is the real part of S: S 3.293 kVA = the apparent power is the magnitude of S: S 2.824 1.694j - kVA = S V I = kVAR kW kVA kW define a. We can use the complex power formula arg I ( ) 30.964deg = I 13.72A = I 11.765 7.059j + A = I V Z = Z 15 9j - = Z R j X
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Unformatted text preview: C ⋅-= Since this is a series circuit, we need to find the total impedance and then solve for the circuit current Solution: V 240 V ⋅ = X C 9 Ω ⋅ = R 15 Ω ⋅ = P2-5. Consider a series R-C circuit with R = 15 Ω and -jX C = -j9 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 capacitor 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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