RMS and average current triangle

In summary, The user is seeking help in finding the RMS and average current of a triangle waveform. They searched the internet but could not find a solution as their waveform is different from what they found online. They have provided the maximum and minimum values of their waveform and are looking for assistance in calculating the RMS and average current. The conversation also includes calculations and equations for finding the RMS and average current.
  • #1
billy fok
40
0
I would like to know how to find RMS and average current of a triangle?
i searched the web. but it focus from 0. which i know.
but my waveform is different from the internet. i have attached the waveform..
Maximum = 2.5 , Minimum = 1.8

i need to find out the RMS and average current
 

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  • #3
Following the definition:

"The RMS over all time of a periodic function is equal to the RMS of one period of the function."

And :

frms=sqrt[integral(f(t)^2/T]

f(t)up=yo+(y1-yo)/(x1-xo)*(x-xo) integrated for x=x1 and x=xo

f(t)down=y1-(y1-yo)/(x1-xo)*(x-x1) for x=2*x1 and x=x1

where T=0.05 xo=0;x1=T/2;yo=1.8;y1=2.5

frms=sqrt{[integral(f(t)up^2+integral(f(t)down^2]/T}

integral(f(t)up^2=aup*x+bup*x^2/2+cup*x^3/3 x=x1 and x=xo

integral(f(t)down^2=adown*x+bdown*x^2/2+cdown*x^3/3 x=2*x1 and x=x1

Let's put k= (y1-yo)/(x1-xo).Then:

aup=yo^2-2*k*yo*xo+k^2*xo^2

bup=2*yo*k-2*xo*k^2

cup=k^2

adown=y1^2+2*y1*k*x1+k^2*x1^2

bdown=-2*y1*k-2*k^2*x1

cdown=k^2

frms=2.16[approx.]
upload_2015-3-30_8-20-57.png
 
  • #4
billy fok said:
i need to find out the RMS and average current
What have you calculated as its average value?
 
  • #5
Following:

http://en.wikipedia.org/wiki/Average

AM=1/n*Sum[y(i)]|y(1)=yo to y(n)=y1

Let's take n number of equal distanced points on upright part of the triangle.The number of intervals

will be n-1.Each y[y(i)] value will be y(i)=yo+(i-1)*(yn-yo)/(n-1)

Sum[y(i)]|y(1)=yo to y(n)=y1=n*yo+ (y1-yo)/(n-1)*Sum(1+2+3+..n-1)

Sum(1+2+3+..n-1)=n*(n-1)/2. Then:

Sum[y(i)]|y(1)=yo to y(n)=y1=n*yo+n*(n-1)/2*(y1-yo)/(n-1)=n*yo+n/2*(y1-yo)

AM=1/n*Sum=yo+(y1-yo)/2=(y1+yo)/2

AM=(1.8+2.5)/2=2.15
 

FAQ: RMS and average current triangle

1. What is the difference between RMS and average current in a triangle wave?

RMS (Root Mean Square) current is a measure of the average current over one complete cycle of a waveform. It takes into account the fluctuations in the current and provides a more accurate representation of the actual average current. Average current, on the other hand, simply calculates the average of the instantaneous current values over one cycle without considering the fluctuations.

2. How is RMS current calculated for a triangular waveform?

RMS current for a triangular waveform can be calculated by taking the square root of the sum of the squares of the instantaneous current values over one cycle, divided by the number of samples in the cycle.

3. Can RMS current be higher than average current in a triangular waveform?

Yes, RMS current can be higher than average current in a triangular waveform. This is because RMS current takes into account the fluctuations in the current, while average current does not.

4. Why is RMS current important in electrical circuits?

RMS current is important in electrical circuits because it is used to determine the power dissipation in a circuit. It is also used to calculate the effective heating effect of an alternating current, which is important for safety considerations.

5. Is there a relationship between RMS and average current in a triangular waveform?

Yes, there is a relationship between RMS and average current in a triangular waveform. The RMS current is always equal to the average current multiplied by the square root of 1/3 for a triangular waveform.

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