RMS and Average Value Difference

[Electric1]

For the above waveform i can calculate the Average value and the RMS value as below
Avg Voltage = 1/0.1*(10*0.07 + 0 * 0.03)=0.7/0.1 = 7 V
Avg Current = 1/0.1*(0*0.05 + 4 * 0.05) = 2 A

Rms Voltage = 10*Sqrt(D) = 10*Sqrt(0.7) = 8.3V
RMS Current = 4*Sqrt(0.5) = 2.82A

Now my main question is when to use the average value and when to use rms value for example in power calculations etc. I was always of the impression that RMS values are only calculated for singnals whose Average value is 0 for example periodic sine signals etc. Is my understanding correct?

 

[Electric1]

The power is calcualted as time average of V * I
1/0.1*(10*4*0.02) = 8Watts. ->1

Power using average voltage and current
7V * 2A = 14Watts ->2

Power using the RMS voltage and current
8.3V * 2.82A = 23.4 Watts. ->3

So both 2 and 3 do not match 1, probably i cannot use 2 and 3 for power calculations.

Now if i consider V = Asin(wt), I = Bsin(wt);
Avg V = 0
Avg I = 0;
P = Time integral of Voltage and current;
P = ABsin^2(wt) -> P = AB/2 -> P = A/sqrt(2) * B/sqrt(2) -> Vrms * I rms;

So the power calculation is dependent on the input waveform, am i correct?

 

[crutschow]

The average value of a waveform does not show the true power.

The RMS calculation gives you the true power delivered by the waveform.
You do the calculation for the waveform shape, and it does not depend upon the waveform having an average value of zero.
For example, if the waveform has a DC offset, it just adds to the RMS calculated value.

To do a discrete RMS power measurement, you can take a large number of simultaneous voltage and current measurements during one cycle of the waveform, multiply each voltage and current value to get the power at that measurement point, and then take the average (the mean from RMS) of those power measurements over the cycle.
That will give a true power value, even for waveforms with a voltage/current phase-shift from a reactive power load, a distorted wave, and/or any arbitrary wave-shape.
This is often done with fast A/D converters and a digital processor to do the math.

 

[Delta Prime]

Six of the most important characteristics of a sine wave are;
PEAK TO PEAK value.
INSTANTANEOUS value.
AMPLITUDE.
PEAK value.
PERIODIC TIME.
AVERAGE value.
RMS value.
The link below will answer all your next questions that you will naturally have..

Measuring the Sine Wave

Understanding the sine wave and measuring its characteristics

learnabout-electronics.org

 

[Electric1]

But as you can see in post2 the power calculations are not matching, why?

 

[Electric1]

View attachment 69810
Six of the most important characteristics of a sine wave are;
PEAK TO PEAK value.
INSTANTANEOUS value.
AMPLITUDE.
PEAK value.
PERIODIC TIME.
AVERAGE value.
RMS value.
The link below will answer all your next questions that you will naturally have..

Measuring the Sine Wave

Understanding the sine wave and measuring its characteristics

learnabout-electronics.org

The average value of the above waveform should be 0 as it has equal areas on the positive and negative side. What am i missing here?

 

[Delta Prime]

Unfortunately, this homework help I cannot give you the answer I could be banned for doing so…

 

[Electric1]

Vavg = [math]\frac{\int_0^\pi V\sin(wt) w dt}{\pi} = V*2/\pi =V* 0.637[/math]
Only positive cycle is considered above, it is confusing each book is following different methodologies.

 

[Harald Kapp]

So the power calculation is dependent on the input waveform, am i correct?

Yes to this.

But there's more to it. With sinusoidal waveforms you also have to take into account the pase shift between voltage and current. For example you may have:
[math] V_{rms} = V_0 \cdot sin(wt) [/math][math] I_{rms} = I_0 \cdot sin(wt+ \phi) [/math]Where φ is the phase shift between voltage and current.

For any complex periodic waveforms you have:
[math]P_{rms} = \sqrt{\frac{1}{T}\int_0^T{(v(t)\cdot i(t)) dt}}[/math]

Only positive cycle is considered above,

That's because the sinusoidal waveform is symmetric. So
[math]\sqrt{\frac{1}{T}\int_0^T{(v(t)\cdot i(t)) dt}} = \sqrt{\frac{1}{(\frac{T}{2})}\int_0^\frac{T}{2}{(v(t)\cdot i(t)) dt}}[/math]

 

[Delta Prime]

Only positive cycle is considered above, it is confusing each book is following different methodologies.

Maybe a visual might help…
Obviously for current below.

 

[Alizayy]

The RMS value represents the "DC equivalent" heating power of a waveform, and you must calculate the average of the instantaneous power (the product of [imath]v(t) \cdot i(t)[/imath]) rather than multiplying independent average or RMS values to find true power in non-sinusoidal or phase-shifted systems.

 

[Delta Prime]

Average values are mostly used when you want to know the net or mean value of a waveform over time, like the average current flowing in a half-wave rectifier, or when designing certain DC circuits.

Thank you for your contributions AI.
"Average" is the broader, general term for central tendency, while "mean" is the specific statistical calculation.