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Find the load impedance

  • #1
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0

Homework Statement


Determine the load impedance for the circuit depicted in fIG 11.40 if the source is operating at a PF of (a) 0.95 leading: (b) unity: (c) 0.45 lagging.


Homework Equations


uploaded


The Attempt at a Solution


uploaded. I am not really sure what to do please see attachment.
 

Answers and Replies

  • #2
berkeman
Mentor
57,304
7,282

Homework Statement


Determine the load impedance for the circuit depicted in fIG 11.40 if the source is operating at a PF of (a) 0.95 leading: (b) unity: (c) 0.45 lagging.


Homework Equations


uploaded


The Attempt at a Solution


uploaded. I am not really sure what to do please see attachment.
Attachment not showing for me...
 
  • #3
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forgot to upload it sorry.
 

Attachments

  • #4
berkeman
Mentor
57,304
7,282
So what are the relevant equations? How does power factor affect the measured impedances? TBH, it's pretty hard to read your upload. It would be much better if you could learn to type your equations into your question threads
 
  • #5
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well PF=P/((Ieff)(Veff)), Ieff=Im/sqrt(2), Veff=Vm/sqrt(2), apparent power= (Veff)(Ieff), P=IV. PF depending on whether it is leading or lagging it can be purely resistive or reactive.
 
  • #6
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So A impedance value would have a reactive and resistive component. In (a) it is leading so purely resistive correct so it would just be the resistor value for R and I have a current so I can find V.
 
  • #7
818
67
In (a) it is leading so purely resistive correct ...
What does the power factor of the load tell you about the angle between the voltage across it and the current through it?

Anyway, you need more information than you have provided in the problem statement to determine both the angle and magnitude of the 'load' impedance. Are you sure you've included everything you were given?
 
  • #8
672
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yah that's it. I uploaded the circuit diagram if that helps R= 1000 ohm and I= 275(20 degrees) mA. Load, resistor, and current source are all in parallel
 
  • #9
672
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lets see here the current leads voltage here so capacitive load.If I am right then the cosine^-1(.95)=18.19 degrees?
 

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