# ac waveform question

Discussion in 'Homework Help' started by hotrocks, Jul 24, 2010.

1. ### hotrocks Thread Starter New Member

Jun 29, 2010
8
0
I am currently working my way through Electrical Installation Calculations 7th Edition and have got to exercise 6 regarding AC waveforms. the question is as follows

An alternating current has the following values taken at intervals of 30 degrees over 1 half cycle

angle ° current A
0----------0
30---------10.5
60---------17.5
90---------19.7
120--------150
150--------11.5
180--------0

Determine the rms and average values of this current.

I calculate that for a Sinusoidal waveform 90° (peak) x 0.707=rms so 19.7x0.707= 13.93A
and 90° (peak) x 0.637=average so 19.7x0.637= 12.55A

however the answers in the book are given as 31.1A and 14.1A

Can any of you knowledgeable folk help me out please

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
Did you check the publishers or authors site for any errors in the book?

I don't see any errors in your logic.

3. ### Ghar Active Member

Mar 8, 2010
655
72
That is a significantly distorted sinusoid... it's supposed to be symmetric around 90 degrees.

That rms value of 31 V can't make sense because rms is always lower than the peak...

4. ### hotrocks Thread Starter New Member

Jun 29, 2010
8
0
Thanks for the speedy reply guys you`ve just confirmed what I was thinking however best to check before continuing with the book. Looks like a publication error after all. No doubt ill have more questions for you all as I progress

Thanks again

5. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
Things like that are one of the reasons Tony started the AAC book at the top of the page. Basically being open source we'll be able to nail mistakes like that long term.

6. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Did you notice the 150A value @ 120 degrees? Is that a typo (meant for 15A) of the OP or actual part of the exercise?

Last edited: Jul 25, 2010
7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,284
328
The book answers lead to a form factor (ratio of RMS to average) of more than 2. This can only happen if the waveform is very peaky, which would make one think that the peak value of 150 is intended.

On the other hand, I can't get the book answers from the given values. If I leave that 150 peak value in, I get 34.87 for the average.and 62.5 for the RMS value.

If I change the 150 to 15.0, I get 12.37 for the average and 13.9 for the RMS value.

8. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
The book is an ebook. Now this is not to disparage ebooks, but there should be a way for the people who disagree with something or have inquiries, to get the answers.

Jul 7, 2009
1,585
141
The OP doesn't state whether this waveform is symmetrical or not. If it isn't, we need the data for the other half cycle. If it is, then we can use the definition of RMS to get:

$RMS^2 = \frac {2(0^2 + 10.5^2 + 17.5^2 + 19.7^2 + 150^2 + 11.7^2)} {2(6)} = 3906.91$

or $RMS = 62.51$. I knew Electrician's answer would be right, so I'm repeating him. But I did it because some folks might not know that you can calculate the root mean square from its definition

$RMS = sqrt{\frac{x_1^2 + x_2^2 + ... + x_n^2}{n}}$

for any set of n points.

10. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
I thought it might be a periodic waveform because it's certainly not symetrical, but the math didn't hold out. (sum of squares / n )^.5

Even if you tried working it backwards (answer back to the original data) your not getting what was posted.

on edit ...

The ebook deals with house wiring, so I would imagine its not a periodic wave, but a symeterical wave at 60 or 50 Hz.

Last edited: Jul 24, 2010