# Homework Help: AC current, conceptual question, the difference between Iaverage and Irms?

1. Apr 20, 2012

### connor02

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

This is more of a conceptual problem I'm having difficulty getting my head around.

What is the difference between Iaverage and Irms for a rectified and non rectified AC circuit?

I know Iaverage=(2/π)Imax while Irms=(1/√2)Imax but I don't understand why they are different given that both are essentially measuring the area under the same curve, rectified or not.

2. Relevant equations

Iaverage=(2/π)Imax and Irms=(1/√2)Imax

3. The attempt at a solution

I don't understand why they are different given that both are essentially measuring the area under the same curve.

2. Apr 20, 2012

### collinsmark

Boldface mine.

Not the same curve.

In the average power, we integrate the absolute value of the waveform to find the area under the curve. Then we divide that by the integration time.

$$I_{ave} = {\frac{\int_0^T | \sin \left(\frac{2 \pi}{T}t \right) | dt}{T}}$$

We can get rid of the absolute value by noting that second half of the period is just the negative of the first (before the absolute value), thus we really only have to integrate over the first half, and divide by the first half.

$$I_{ave} = {\frac{\int_0^{T/2} \sin \left(\frac{2 \pi}{T}t \right) dt}{T/2}}$$

RMS current is quite a bit different though. RMS (root mean square) is exactly what it sounds like. First we square the function. Then we take the mean (same method that we did for the average except this time we start with the signal's square -- and because of the square, we know it will always be positive so we don't need to worry about the absolute value). Finally, we take the square root of the whole thing.

$$I_{rms} = \sqrt{\frac{\int_0^T \sin^2 \left(\frac{2 \pi}{T}t \right) dt}{T}}$$

3. Apr 21, 2012

### connor02

Ah I see. it makes sense now. I overlooked that it is root mean square and not mean root square so you that the mean of the square value and not the square root of the square values. TY!