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

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The discussion centers on the application of Kirchhoff's Current Law (KCL) at a central node for AC waveforms. The calculations presented yield a result of 13.33 at an angle of 220.89, which differs from the book's answer. There is confusion regarding whether the given current values of 5 and 10 Amperes are RMS or peak values, with some participants suggesting they are peak unless specified otherwise. The possibility that the values represent instantaneous quantities is also raised, indicating a potential error in the problem's presentation. Clarification on the nature of the current values is necessary for accurate analysis.
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Homework Statement


upload_2018-1-23_13-15-56.png
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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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