L C and R component of load operation w/ lagging PF

Discussion in 'Homework Help' started by Turbo89, Sep 4, 2014.

  1. Turbo89

    Thread Starter New Member

    Jul 25, 2014

    I'm taking a course in AC power systems, and it's straining my memory of AC circuit analysis.

    Here's a low down of where I am: I have a AC circuit with three loads identified and an AC source. Two of the loads are purely resistive and the voltage source has phase angle of 0. The last component is operating at 373 watts, with a lagging pf of .8.

    Why I'm trying to ultimately calculate current generated by the source and the three powers of the circuit (apparent, reactive and real). My approach at solving for the last component is as follows:

    1) Use the Known E and pf to calculate the component's apparent power (P = S*pf)
    2)Use the S to back into a component current (S=V*I = E*I =>> I=S/V).
    3) Calculate the component's phase angle by taking the arcos(.8), lets call this theta
    4) Calculate the component's reactive power using the E, I and phase angle I previously calculated (Q=E*I*sin(theta).
    5) Then use the reactive power to calculate the imaginary part of impedance (X = E^2 / Q)
    6) Calculate the resistive part of impedance using E and the calculated current (R = E/I)
    7) Combine these two into a complex notation R + jX

    From here I solve for circuit Impedance by summing the 3 parallel impedance's (this is fairly painful).

    I then calculate the circuit P by summing the wattage (r1 + r2 + 373W), use this wattage to back into a equivalient circuit current (P=I^2 * R), calculate apparent power by multiplying the Current by E, then use the power triangle to solve for reactive power of the circuit.

    ... Well that was long Winded, DOES anyone see any holes in this approach?!

    Thanks in advance!!!
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Probably more work than is actually necessary but looks OK. You don't need to find the impedance(s). It can all be done without those related steps.
  3. Turbo89

    Thread Starter New Member

    Jul 25, 2014
    Thanks for taking the time to review it (I know I was long winded)!

    With respect to how is could be done easier, I'm sure you're right but I couldn't think of a way... and pointers as to where to look would be appreciated (I wouldn't mind redoing the problem to check my initial calcs).!