Find the RMS value of current I

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Discussion Overview

The discussion revolves around calculating the RMS (Root Mean Square) value of current in a network of sinusoidal currents, specifically focusing on the relationship between different phase angles and current values at specific moments. Participants explore the implications of phase delays and the definitions of effective current in the context of the problem.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants express confusion over the term "effective current," with one noting it refers to the RMS value.
  • One participant attempts to derive the phase angle ψ1 based on given current values and phase delays, but acknowledges an error in their calculations.
  • Another participant questions the interpretation of "minimal" in relation to sinusoidal currents, suggesting it could refer to a negative peak.
  • Clarification is provided that "A" stands for amperes, addressing a point of confusion among participants.
  • One participant proposes that the phase delay for current i2 should be negative due to its delayed nature.
  • Another participant mentions that the solution they found appears simpler than their approach, prompting curiosity about the method used.
  • There is a suggestion that the Sine Rule may have been applied to relate the three currents forming a closed triangle.

Areas of Agreement / Disagreement

Participants express varying interpretations of the problem statement and the definitions involved, indicating that multiple competing views remain. There is no consensus on the correct approach or interpretation of certain terms.

Contextual Notes

Some participants note confusion regarding the problem statement and the definitions of terms like "effective current" and "minimal." There are unresolved mathematical steps and assumptions regarding phase angles and their implications for current values.

Ivan Antunovic
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Homework Statement


In the network of sinusoidal current, as shown in the figure, the current i2
is phase delayed for the angle of 3π/4 behind the current i.
In the moments in which the current i2 is minimal,
current value of current i1 is sqrt(2) A. This value is two times lower than the maximum
value of the current i1, and in those moments current i1
and growing.
Calculate the effective value of the current I.
Captureelektrijada.jpg[/PLAIN]
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Homework Equations

The Attempt at a Solution


osnovi_elektrijada.png

capture screen
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Is there any way to find the angle Φ2 since I know that in the moments when i2 is minimal value of the current i1 is sqrt(2)?
 
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Did you ask what "effective current" means? I couldn't even guess.
 
rude man said:
Did you ask what "effective current" means? I couldn't even guess.
It's RMS value( Imax/ sqrt(2) ) .

I tried to solve it this way now:
i = Imax *sin(wt)
i1 = I1max *sin(wt + ψ1)
i2 = I2max *sin(wt+ 3*pi/4 )

at time t = t1 when current i2 is minimal.
-I2max = I2max * sin(w*t1 + 3*pi/4) -------> wt1 = -5*pi /4
2*i(t1) = I1max ------> I1max = 2*sqrt(2)

sqrt(2) = 2sqrt(2) * sin (w*t2 + ψ1) / arcsin
ψ1 = 17*pi / 12

which is incorrect the the correct phase delay of the current i1 is ψ1 = 45°
 
Ivan Antunovic said:

Homework Statement


In the network of sinusoidal current, as shown in the figure, the current i2
is phase delayed for the angle of 3π/4 behind the current i.
In the moments in which the current i2 is minimal,
current value of current i1 is sqrt(2) A.
What is A? It's not related to anything else in the problem.
and in those moments current i1
and growing.
?
Totally confusing problem statement.
 
Ivan Antunovic said:
the current i2
is phase delayed for the angle of 3π/4 behind the current i.
In the moments in which the current i2 is minimal,
current value of current i1 is sqrt(2) A. This value is two times lower than the maximum
value of the current i1, and in those moments current i1
and growing.
By "is minimal" for a sinusoid should we understand it to indicate a zero-crossing or a negative peak?
 
rude man said:
What is A? It's not related to anything else in the problem.?
A is the abbreviation of "amperes".
 
Ivan Antunovic said:
I tried to solve it this way now:
i = Imax *sin(wt)
i1 = I1max *sin(wt + ψ1)
i2 = I2max *sin(wt+ 3*pi/4 ) ⇐
i2 is delayed, so phase should be negative.
at time t = t1 when current i2 is minimal.
-I2max = I2max * sin(w*t1 + 3*pi/4) -------> wt1 = -5*pi /4
2*i(t1) = I1max ------> I1max = 2*sqrt(2)

sqrt(2) = 2sqrt(2) * sin (w*t2 + ψ1) / arcsin
ψ1 = 17*pi / 12

which is incorrect the the correct phase delay of the current i1 is ψ1 = 45°
 
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Sorry for the late replay guys,I had exam this week.
NascentOxygen said:
By "is minimal" for a sinusoid should we understand it to indicate a zero-crossing or a negative peak?
I think that they mean for minimal as value when it is a negative peak of the current.
NascentOxygen said:
A is the abbreviation of "amperes".
Correct.
NascentOxygen said:
i2 is delayed, so phase should be negative.
Tried calculations now with negative sign of ψ2 = -3π/ 4 and the result that I got is ψ1 = -π / 12 , which gives ψ1 - ψ2 = 2π/3 as it is correct.Now need to figure out how to find RMS value of the current I.
 
Last edited:
Got the right solution now:

elektrijada_osnovi_sinusoida.png


Their solution
osnovi_elektrijada.jpg


How did they get this expression ? It looks way simpler than my approach.
 
  • #10
Looks like they applied The Sine Rule. The 3 currents form a closed triangle.
 
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  • #11
NascentOxygen said:
Looks like they applied The Sine Rule. The 3 currents form a closed triangle.
Yes you are right
osnovi_elektrijada2.png


Thank you for your help.
 
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