KCL for AC Waveforms: Can I Get the Right Answer?”

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SUMMARY

The discussion centers on the application of Kirchhoff's Current Law (KCL) to AC waveforms, specifically at a central node. The user attempted to solve the problem using KCL and derived a current of 13.33 at an angle of 220.89 degrees. There is ambiguity regarding whether the 5 and 10 Ampere values are RMS or maximum, with a consensus that they are likely peak quantities unless otherwise specified. The conversation highlights the importance of understanding the nature of current values in AC analysis.

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  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with AC circuit analysis
  • Knowledge of RMS and peak current values
  • Basic complex number operations in electrical engineering
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Homework Statement


upload_2018-1-23_13-15-56.png
[/B]

Homework Equations


Can I use KCL at the central node?

The Attempt at a Solution


I tried KCL and got: 5cos0 + j 5sin0 + 10 cos 60 + j 10 sin 60 = (-I1)
5 + 5 + j 10 *1.732 / 2 = (-I1)
10 + j 5 * 1.732 = -I1
This gives 13.33 angle (220.89)

Book answer is A
Is the 5 and 10 Ampere in figure, RMS or Maximum?
 

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Instantaneous values have no phase. So in that sense, the options seem meaningless.

jaus tail said:
Is the 5 and 10 Ampere in figure, RMS or Maximum?
In India, I guess they are taken as peak quantities unless specified otherwise.
I agree with your calculations.

Maybe they have given instantaneous values of the magnitudes and phases of the current sources (in an incorrect way) and they want you to find the magnitude and phase of I1 at that instant.
 
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