# Problem regarding periodic current functions

Tags:
1. Jun 12, 2017

### diredragon

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
3. The attempt at a solution

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?

2. Jun 12, 2017

### ehild

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.

3. Jun 13, 2017

### diredragon

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 dont 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.

4. Jun 13, 2017

### ehild

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.

Last edited: Jun 13, 2017