text: # Complex power/power factor calculations
  
  ## Homework Statement
  
  (old exam question attached)
  
  ## Homework Equations
  
  Complex power: S = VI* / 2 = P + jQ where P is the average power and Q is the reactive power
  Magnitude of complex power |S| = Vmax*Imax/2
  V = ZI
  
  ## The Attempt at a Solution
  
  We can calculate θv - θi for each load by using arccos(power factor).
  
  arccos 0.8 = 36.86 degrees
  arccos 0.75 = 41.4 degrees
  
  Because I'm fairly new to these types of problems I need to make sure I'm understanding this correctly before I can go any further:
  
  - The first load is given in kW and the other is given in kVA. Does this mean that we can assume that the first load is purely resistive?
  - Can I go ahead and express the complex power of the second load as 40/_41.4 -> 7.5 + 6.6j?
  
  I'm thinking that the answers to these questions will indicate what I should do next.
  
  Regarding part b), how exactly is system power defined? Is it the sum of the power delivered to all of the elements? If so, could I simply add the complex power delivered to the impedance Zs to the complex power delivered to both loads?
  
  EDIT: I just realized I have enough information to calculate the current flowing through the kVA load using S = VI* / 2. Alone, this doesn't help much, but by following a similar procedure for the first load (after determining what the complex power is, exactly) I can apply KCL and determine the current through the branch with the impedance (i1 + i2 = i3).
  
  EDIT 2: Another thought - could I find the current through the impedance directly by using the complex power delivered to both loads as Stotal = S1 + S2 = 40 + (7.5 + 6.6j) and S = VI* / 2? This would solve part a), after finding the voltage of the impedance Zs and using KVL in the leftmost mesh.
  
  #### Attachments
  
  • Screen Shot 2012-07-26 at 1.40.23 PM.jpg
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  First load is not pure resisitive as power factor is not $1$:
  $\begin{cases}P_1=40\,kW\\PF=0.8\end{cases} \Rightarrow Q_1=P_1\frac{\sqrt{1-(PF)^2}}{PF}=30\,kVA\\ \begin{cases}Q_2=40\,kVA\\PF=0.75\end{cases} \Rightarrow P_2=Q_2\frac{PF}{\sqrt{1-(PF)^2}}=\frac{120}{\sqrt{7}}\,kW$
  I think $V=220\angle 0$ means $V_{rms}=220$ not $V_{max}=220$ so $S=VI^*$
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