Find the RMS value of current I

In summary, the conversation discussed a problem involving sinusoidal currents in a network. It was mentioned that the current i2 is phase delayed by 3π/4 behind the current i, and in moments when i2 is minimal, the current i1 has a value of sqrt(2) A, which is two times lower than its maximum value. It was also mentioned that the effective value of the current I is equal to Imax/sqrt(2). The conversation then went on to discuss different attempts at solving the problem, including using the sine rule to find the phase difference between i1 and i2.
  • #1

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] [Broken]
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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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  • #2
Did you ask what "effective current" means? I couldn't even guess.
 
  • #3
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°
 
  • #4
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.
 
  • #5
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?
 
  • #6
rude man said:
What is A? It's not related to anything else in the problem.?
A is the abbreviation of "amperes".
 
  • #7
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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  • #8
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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  • #9
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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1. What is RMS value of current?

The RMS (Root Mean Square) value of current is the effective or average value of an alternating current, taking into account both the magnitude and direction of the current.

2. How is the RMS value of current calculated?

The RMS value of current is calculated by taking the square root of the mean of the squared values of the current over a specified time period. This is also known as the "quadratic mean" or "quadratic average".

3. Why is it important to find the RMS value of current?

Finding the RMS value of current is important because it allows us to compare alternating currents with direct currents in terms of their heating and magnetic effects. It is also used in calculations for circuit analysis and power calculations.

4. What is the difference between RMS value and peak value of current?

The peak value of current refers to the maximum instantaneous value of the current, while the RMS value is the average value of the current over a full cycle. The RMS value is always lower than the peak value for an alternating current.

5. Can the RMS value of current be calculated for any type of current?

Yes, the RMS value of current can be calculated for any type of current, whether it is alternating or direct current. However, the calculation method may differ depending on the type of current being analyzed.

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