ENGR 12
Assignment 12
Due: next wed
Part I. Drills -- 1 point each
1) A sinusoid having a peak amplitude of 500V is applied to the terminals of a 75 Ohm resistor. Find the
average power delivered to the resistor by a) finding the RMS voltage and using Vrms^2/R, b) finding the RMS
current and using Vrms*Irms.
2)
3) Find the power factor angle and the power factor for the four sets of values in problem 2, and indicate
"lagging" power factor (current phase lags or is smaller than voltage phase, i.e. positive pfa) or "leading"
power factor (current leads voltage, i.e. negative pfa).
4)
5) An electrical load operates at 120 Vrms. The load absorbs an average power of 3 kW at a leading power
factor of 0.5. Calculate a) the power factor angle, b) the apparent power, |S|, c) the reactive power |Q| of
the load. d) the impedance of the load (use S = Vrms Irms*, or Irms = (S/Vrms)*, then find Z = Veff/Ieff )
Part II. Assisted Problem Solving – 2 pts
6) a) Find the capacitance C needed to improve the power
factor to 0.95 lagging in the circuit shown, if the effective
voltage of 120V has a frequency of 60 hertz.
b) how much in percent has the apparent power been
reduced by the pf correction ?
1) Plan: find S = P + jQ for the 20<30 load and then
add enough –Qcap to lower power factor to .95
(that would be a pfa of arccos(.95))
2) First find Sload = Veff Ieff* (where Veff = Vrms)
(realizing Ieff into load = Veff/(20<30), and then
take the complex conjugate and multiply by Veff)
3) You should find the pfa = 30 degrees
4) Find Qload = |Sload|sin(30)
5) Power absorbed by added capacitor will be purely
imaginary:
Scap =0 + j Vrms(Vrms/-jXc)* = -j V2rms/Xc, or
Qcap = - V2rms/Xc
6) The total real power of the combined load does
not change. The total reactive power Qt of the
combined load will then be Qload + Qcap. This
needs to be set to give the desired pfa.
7) Once you find Qcap you can then find Xc and from
there find C using Xc = 1/(wC).
8) For b), find Stotal for the combined loads. The %
reduction in apparent power will be
100*(1 -|Stotal|/|Sload| )
7) The load impedance ZL is adjusted for maximum
average power delivered to ZL. Find the maximum average
power to ZL.
PLAN
1) Find the Thevenin equivalent circuit to the left of ZL
2) Set up an equation for I and solve
3) Use your answer to 2 and voltage divider to find Voc
4) Find Isc = 5 I / 1
5) Find Zth = Voc/Isc
6) Apply ZL = Zth* to the thevenin simplified circuit
7) The reactance elements cancel and we have simply a
resistive voltage divider for finding VL and hence Pmax
Part III. Unassisted Problem Solving – 3 points
8) Find the complex power S delivered to the load, and the power factor angle of S, if ig = 125cos500t mA.