NCEES #107

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The amplitude of the current is given as a peak value. For AC power, RMS values are used.

Irms = Ipeak / sqrt2

 
The amplitude of the current is given as a peak value. For AC power, RMS values are used.
Irms = Ipeak / sqrt2

Okay so if the frequencies of the two components were the same then the equation:

(7/SQRT2)^2 * 3 would be true?

However Since the frequencies are different we must use the equation:

3*(5/SQRT2)^2 + 3*(2/SQRT2)^2

 
Okay so if the frequencies of the two components were the same then the equation:
(7/SQRT2)^2 * 3 would be true?

However Since the frequencies are different we must use the equation:

3*(5/SQRT2)^2 + 3*(2/SQRT2)^2
Yes. If the signals have the same frequency and are in phase, then you can add the amplitudes.

 
^I'm not so sure. The problem I have is that if either method can be used, then:

(7/SQRT2)^2 * 3 is equall to 3*(5/SQRT2)^2 + 3*(2/SQRT2)^2?

49*3/2 = 25*3/2 + 4*3/2

73.5 = 43.5

This one puzzles me as my first response is the same as benbo's.

 
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^I'm not so sure. The problem I have is that if either method can be used, then:
(7/SQRT2)^2 * 3 is equall to 3*(5/SQRT2)^2 + 3*(2/SQRT2)^2?

49*3/2 = 25*3/2 + 4*3/2

73.5 = 43.5

This one puzzles me as my first response is the same as benbo's.
Maybe I was unclear.

I agree with you that they aren't interchangeable.

One is true when the frequencies are the same.

The other is true when the frequencies are different.

If you have 5sin(wt) + 2sin(wt) obviously it is just 7sin(wt)

But 5sin(wt) + 2sin(2wt) you obviously can't add without a little more math.

I think we agree.

 
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I concur.

(really I'm just disappointed I didn't get to answer any questions. Darn job.)

 
I concur.(really I'm just disappointed I didn't get to answer any questions. Darn job.)
I'm going out of town for a couple days (although I may have internet access). So you won't have to beat me to it! Just Jim and any other EEs who get in on the fun. Besides, somebody is bound to ask something about power and leave me for dead.

 
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