Find the RMS value of current I

AI Thread Summary
The discussion revolves around calculating the RMS value of a sinusoidal current in a network where one current, i2, is phase delayed by 3π/4 behind another current, i. The participants clarify that the effective current, or RMS value, can be derived from the maximum current values using the sine rule, which relates the three currents as forming a closed triangle. Confusion arises regarding the definition of "minimal" for the current i2, with participants concluding it refers to a negative peak. The correct phase delay for current i1 is established as ψ1 = 45°, leading to a final solution for the RMS value of the current I. The conversation highlights the importance of understanding phase relationships in sinusoidal currents for accurate calculations.
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

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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.
 
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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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