Find the RMS value of current I

[Ivan Antunovic]

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.
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Homework Equations

The Attempt at a Solution

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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)?

 

[rude man]

Did you ask what "effective current" means? I couldn't even guess.

 

[Ivan Antunovic]

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°

 

[rude man]

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.

 

[NascentOxygen]

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?

 

[NascentOxygen]

What is A? It's not related to anything else in the problem.?

A is the abbreviation of "amperes".

 

[NascentOxygen]

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°

 

[Ivan Antunovic]

Sorry for the late replay guys,I had exam this week.

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.

A is the abbreviation of "amperes".

Correct.

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.

 

[Ivan Antunovic]

Got the right solution now:

Their solution

How did they get this expression ? It looks way simpler than my approach.

 

[NascentOxygen]

Looks like they applied The Sine Rule. The 3 currents form a closed triangle.

 

[Ivan Antunovic]

Looks like they applied The Sine Rule. The 3 currents form a closed triangle.

Yes you are right

Thank you for your help.