Understanding Phasor Representation and Calculating Current in AC Circuits

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Discussion Overview

The discussion revolves around calculating current in AC circuits using phasor representation. Participants are examining the correct application of complex impedance and the mathematical steps involved in the calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states their calculation of total impedance as 210+j232.5 and expresses confusion over the resulting current calculation.
  • Another participant suggests that the real part of the impedance should be 310 ohms instead of 210 ohms.
  • A participant clarifies that they wrote 310 ohms on paper and attempted to multiply by the complex conjugate but obtained a different result.
  • One participant asserts that the solution in the posted image is correct and emphasizes the need to multiply by the complex conjugate of the impedance.
  • A participant shares their calculation, which yields a different result, prompting further questioning about the correctness of their approach.
  • Another participant points out a potential error in taking the square root of the denominator instead of squaring the magnitude of the impedance.

Areas of Agreement / Disagreement

Participants express differing views on the correct values for impedance and the steps in the calculation, indicating that multiple competing views remain and the discussion is unresolved.

Contextual Notes

There are limitations in the clarity of the calculations presented, particularly regarding the treatment of complex numbers and the assumptions made about impedance values.

Who May Find This Useful

Individuals interested in AC circuit analysis, phasor representation, and complex impedance calculations may find this discussion relevant.

nhrock3
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http://i47.tinypic.com/2mg2iyf.jpg
i ant to find the current in phasor representation.
i made found the Z total 210+j232.5
the current sourse is 2170
[tex]I=\frac{2170}{310+j232.5}=1454.56-j1610.4[/tex]

but the calculation igives another result
why?
 
Last edited:
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The first problem that comes to mind is that the 210 ohms (real part) should actually be 310 ohms.

Also, do you know how to express a complex number in such a way that the imaginary part appears only in the numerator (i.e. there is no j on the bottom)?
 
sorry on paper i wrote 310
and i did multiply the numerator and denominator by the "opposite"
of the compex number that is on the denominator

and i got a totally different result

who is correct?
 
The solution in the jpg image you posted is correct. You have to multiply the expression in your first post (numerator and denominator) by the complex conjugate of the impedance and then simplify.
 
[tex] I=\frac{2170(310-j232.5)}{387.5}=1176-j1302[/tex]

i did that as you see
why i still get the wrong result ?
 
It looks like you took the square root of the bottom. You should have zz*=|z|2 there, not |z|.
 
thanks:)
 

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