Finding the Peak Value of Current: A Puzzling Solution

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Discussion Overview

The discussion revolves around determining the peak value of current from a given expression involving complex impedance. Participants are analyzing the calculations and interpretations of the peak current based on the provided formula.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims the peak value of current is 12.0 A based on their conversion to polar form.
  • Another participant asserts that their professor confirmed the answer of 12.0 A is correct, suggesting the solution key is wrong.
  • Some participants argue that the magnitude of the denominator is SQRT(200) or 14.12, leading to a calculation of 8.485 A when dividing 120 by this value.
  • There is a contention regarding whether the peak voltage and impedance values are being correctly interpreted in the context of calculating peak current.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct peak value of current, with some supporting 12.0 A and others supporting 8.485 A. No consensus is reached on which value is correct.

Contextual Notes

Participants have not fully resolved the assumptions regarding the definitions of peak voltage and impedance, which may affect the calculations presented.

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Homework Statement



Determine the peak value of current of:

[tex]\underline{I} = \frac{120\angle0^0}{10+j10}[/tex]


The Attempt at a Solution



I change this into polar form and got:

[tex]\underline{I} = 12.0 \angle -45^0[/tex]

Therefore, the peak value of the current should be 12.0 A. However, the solution key says 8.4853 A. How can this be so?
 
Last edited:
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I just confirmed with my prof that the solution was wrong. The answer is 12.0 A.

Mods, you can lock this thread.
 
I like the old answer better.

The bottom part of the fraction has a magnitude of SQRT( 200) or 14.12.

Divide 120 by this to get 8.485.
 
vk6kro said:
I like the old answer better.

The bottom part of the fraction has a magnitude of SQRT( 200) or 14.12.

Divide 120 by this to get 8.485.

But you would have to change that to polar form to get the peak current.
 
No,

The 120 is already the peak voltage and the 14.12 ohms is the impedance so the peak current is 120 / 14.142 or 8.485 amps.
 

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