Problem regarding periodic current functions

[diredragon]

Homework Statement

Three periodic currents have the same ##f=100 Hz##. The amplitude of the second current is ##4 A##. and is equal to half of the amplitude of the third current. Effective value of the third current is 5 times that of the first current. At time ##t_1=2ms## third current increases and changes signs. The first current reached it's amplitude half a period after the second current and the second current is "ahead" in phase to third current by ##\frac{2 \pi}{3}##. Write the functions for the current intensities of the currents.

Homework Equations

3. The Attempt at a Solution [/B]
My course of alternating currents has just started and this is my first problem in the book.
What i know:
##f = 100 Hz##
##Im_1 = 4A##, the amplitude denoted by ##Im##
##Im_3 = 8A##
##\frac{Im_3}{\sqrt 2} = 5*\frac{Im_1}{\sqrt 2} \Rightarrow Im_1 = \frac{8}{5} A##
##t_1 = 2ms##, moment when ##i_3## changes sign and start to increase.
Looking for: ##i_1(t)=?, i_2(t)=?, i_3(t)=?##

The work:
The phase of the current ##i_3## is ##wt+ψ_3##. If at time ##t_1## the current changes sign than at that time the phase must be ##\frac{3\pi}{2}## since the canonical function of the current deals with a cosine function and the cosine increases at that point.
##wt_1+Ψ_3=\frac{3\pi}{2}##
##2\pi *100*2+Ψ_3=\frac{3\pi}{2}##
##ψ_3 = \frac{11\pi}{10}##
This is where i disagree with the book. It says it should be ##ψ_3 = -\frac{9}{10} \pi##
That answer i could obtain by subtracting ##2\pi## from my result but i understand the reason for such a thing. What is wrong here?

 

[ehild]

Homework Statement

Three periodic currents have the same ##f=100 Hz##. The amplitude of the second current is ##4 A##. and is equal to half of the amplitude of the third current. Effective value of the third current is 5 times that of the first current. At time ##t_1=2ms## third current increases and changes signs. The first current reached it's amplitude half a period after the second current and the second current is "ahead" in phase to third current by ##\frac{2 \pi}{3}##. Write the functions for the current intensities of the currents.

Homework Equations

3. The Attempt at a Solution [/B]
My course of alternating currents has just started and this is my first problem in the book.
What i know:
##f = 100 Hz##
##Im_1 = 4A##, the amplitude denoted by ##Im##
##Im_3 = 8A##
##\frac{Im_3}{\sqrt 2} = 5*\frac{Im_1}{\sqrt 2} \Rightarrow Im_1 = \frac{8}{5} A##
##t_1 = 2ms##, moment when ##i_3## changes sign and start to increase.
Looking for: ##i_1(t)=?, i_2(t)=?, i_3(t)=?##

The work:
The phase of the current ##i_3## is ##wt+ψ_3##. If at time ##t_1## the current changes sign than at that time the phase must be ##\frac{3\pi}{2}## since the canonical function of the current deals with a cosine function and the cosine increases at that point.
##wt_1+Ψ_3=\frac{3\pi}{2}##
##2\pi *100*2+Ψ_3=\frac{3\pi}{2}##

It must be a misprint: the time is 2 ms. ##2\pi *100*0.002+Ψ_3=\frac{3\pi}{2}##

##ψ_3 = \frac{11\pi}{10}##
This is where i disagree with the book. It says it should be ##ψ_3 = -\frac{9}{10} \pi##
That answer i could obtain by subtracting ##2\pi## from my result but i understand the reason for such a thing. What is wrong here?

Nothing is wrong. cos(θ+2π)=cos(θ). Sometimes the phase constant is required between -pi and pi. Look in your notes.

 

[diredragon]

It must be a misprint: the time is 2 ms. ##2\pi *100*0.002+Ψ_3=\frac{3\pi}{2}##

Nothing is wrong. cos(θ+2π)=cos(θ). Sometimes the phase constant is required between -pi and pi. Look in your notes.

It doesn't state it specifically but i guess that's how they want it.
I continued solving the problem and i found the phase of the second current with the fact that:
##ψ_2-ψ_3 = \frac{2}{3}\pi \Rightarrow ψ_2 = -\frac{7}{30}\pi##
The phase of the first current was a bit of a problem. I knew only that the first current reaches it's amplitude half the period after the second and i didn't know how to use that. I helped myself with the book solution and was wondering if you could help me understand it. It goes:
##wt_2 + ψ_2 = 0##, i think this means that at time ##t_2## the second current reaches it's amplitude as the cosine is 0.
##wt_1+ψ_1 = 0##
##t_1 = t_2 + T/2##, the time ##t_1## is half a period later
##wt_2 + \frac{wT}{2} = 0##, i don't fully understand this part. Shouldn't this be ##wt_1##? And if so how can it be 0 when it's missing ##ψ_1## term?
It continued
##ψ_1 - ψ_2 + \frac{wT}{2}= 0##, so ##wt_2 = ψ_1 - ψ_2##? Why is this so?
##ψ_1 - ψ_2 = \pi \Rightarrow ψ_1 = \frac{23}{30} \pi##
These are the correct results but i don't understand the reasoning behind the first current's calculations.

 

[ehild]

It doesn't state it specifically but i guess that's how they want it.
I continued solving the problem and i found the phase of the second current with the fact that:
##ψ_2-ψ_3 = \frac{2}{3}\pi \Rightarrow ψ_2 = -\frac{7}{30}\pi##
The phase of the first current was a bit of a problem. I knew only that the first current reaches it's amplitude half the period after the second and i didn't know how to use that. I helped myself with the book solution and was wondering if you could help me understand it. It goes:
##wt_2 + ψ_2 = 0##, i think this means that at time ##t_2## the second current reaches it's amplitude as the cosine is 0.
##wt_1+ψ_1 = 0##
##t_1 = t_2 + T/2##, the time ##t_1## is half a period later
##wt_2 + \frac{wT}{2} = 0##, i don't fully understand this part. Shouldn't this be ##wt_1##? And if so how can it be 0 when it's missing ##ψ_1## term?

There can be misprints and errors in any book. You are right. ##t_1 = t_2 + T/2##, substitute it into the equation
##wt_1+ψ_1 = 0## --> ##wt_2+wT/2+ψ_1= 0##.
You know that ##wt_2 + ψ_2 = 0##, that is, ##wt_2 = - ψ_2 ##, so
##ψ_1- ψ_2+wT/2=0##, and wT/2=pi.