Solving AC Circuit Sinusoidal Waveform: Effective (r.m.s) Value

[freshbox]

From the 1st screenshot part ii of the question:

My working:
Vm x √2 = 28.284 x √2 = 40v (Answer Correct)

I would like to ask a question from the 2nd screenshot. "Determine the effective (r.m.s) value of this voltage waveform over one complete cycle.

I don't know how to solve this part. Is the 1st screenshot question which I got the answer the same method as the 2nd ?


I would appreciate if someone can give some guidance to me, thanks.

 

[gneill]

Things are not quite so simple when the waveform is not a sinusoid. Apply the definition of rms and integrate over the period. Here you can "do it manually" since the waveform is composed of simple segments of constant voltage.

 

[freshbox]

Ok, just want to check is the 1st question answer correct 40v?

Effective RMS= Vm x √2

Rms=Vm/√2

Am I correct?

 

[gneill]

Ok, just want to check is the 1st question answer correct 40v?

Effective RMS= Vm x √2

Rms=Vm/√2

Am I correct?

The RMS value of a sinusoid should be less than the peak value. If Vm is your peak, then divide by root 2, not multiply.

 

[freshbox]

So the answer is actually 20v, the book answer is wrong?

And how do you determine if the Vm is at peak or not? Since you say if Vm is at peak, x 2. So I need to determine if the Vm is at peak or not

Thanks gneill.

 

[gneill]

So is actually 20v?

Yup.

And how do you determine if the Vm is at peak or not? Since you say if Vm is at peak, x 2. So I need to determine if the Vm is at peak or not

You are given a waveform:

v(t) = 28.284 sin(...) V

The peak value of the sine function is 1, so the peak value of the voltage function is 28.284 V.

 

[freshbox]

I would like to clarify one thing:

-Calculate the RMS values for voltage
-Determine the effective RMS value of v(t)

Are both question asking the thing? RMS is also effective RMS?

 

[gneill]

I would like to clarify one thing:

-Calculate the RMS values for voltage
-Determine the effective RMS value of v(t)

Are both question asking the thing? RMS is also effective RMS?

Yes, they're the same thing.

 

[freshbox]

Ok. so for the 2nd question, since I'm not given an expression and the waveform is not sinusoidal, do i need to work out an expression for myself and from there mutiply by √2?Omg how do i write an expression that is not sinusoidal o_0

 

[gneill]

Ok. so for the 2nd question, since I'm not given an expression and the waveform is not sinusoidal, do i need to work out an expression for myself and from there mutiply by √2?

Note that the constant factor √2 applies to the relationship between peak and rms for sinusoids. It does not so apply for arbitrary waveforms.

You should apply the definition of RMS to the given waveform.

 

[freshbox]

Do you mean that the concept of x √2 or / √2 does not apply to this question because it is not a sin wave form?

If yes how do i find the rms value for square wave form?

 

[gneill]

Do you mean that the concept of x √2 or / √2 does not apply to this question because it is not a sin wave form?

Yes, that's right.

If yes how do i find the rms value for square wave form?

Apply the definition of RMS. The name gives a hint on how to remember what to do; it's the square Root of the Mean of the Square value of the function over its period.

$$f_{rms} = \sqrt{\frac{1}{T}\int_0^T f(t)^2 dt}$$

Since your voltage signal is comprised of several constant values over the period of 10ms, you can break the sum of the squares into a few pieces that are easy to do by inspection.

 

[freshbox]

I can also use the 2nd formula from the picture? It's the same as the 1st one and also the one from you right?

I managed to find the area which is 45, please correct me if I'm wrong.

But i don't know the meaning of [i²(t)], can you please explain to me?Thanks.

 

[gneill]

I can also use the 2nd formula from the picture? It's the same as the 1st one and also the one from you right?

I managed to find the area which is 45, please correct me if I'm wrong.

But i don't know the meaning of i²(t), can you please explain to me?


Thanks.

i(t)² is the square of the current function. Say the current is 4A at some instant in time. Then its square is 16A² at that instant.

In your attachment showing the voltage waveform over a period the first portion of the period shows a value of 6V for a time of 5ms. Dropping the units for now, integrating the square over that portion of the curve yields (6)²x5 = 180. Do the same for the rest of the segments and sum the results. Take the mean by dividing the sum by the whole period (10).

 

[freshbox]

Before I go on to find the rms value, can you help me check the answer for part vii.

My working for area:
(6x5)+(9x3)+(-6x2)=45

Average value:
45/10=4.5v (Ans)

 

[gneill]

Before I go on to find the rms value, can you help me check the answer for part vii.

My working for area:
(6x5)+(9x3)+(-6x2)=45

Average value:
45/10=4.5v (Ans)

Sure, looks good.

 

[freshbox]

Ok thanks, back to finding the value of rms.

I don't understand why (6)2x5 = 180. this working.

i(t)2 is the square of the current function. Say the current is 4A at some instant in time. Then its square is 16A2 at that instant.

I would put 45 in the area, since i found it earlier and T = 10s. So for v(t)² is 36+81+36=153

√45(153)/10 = 26.23V (Ans Wrong)

 

[gneill]

Ok thanks, back to finding the value of rms.

I don't understand why (6)2x5 = 180. this working. Is this v² x T?

That's 6 squared multiplied by 5. 6V is the value of the voltage over a period of 5 milliseconds. So v²t becomes (6)² x 5 = 180.

From the formula √ area[v²(t)/T doesn't tally with your working.

I would put 45 in the area, since i found it earlier and T = 10s But how about v²(t)?


I am confused now and I know I am wrong

You are no longer working with area. It's no longer just voltage x time (which corresponds to an area under the curve), but v² x time.

If you wish, replace each of the unique segments of the curve with a function and do the integration piecewise. So for example, for the second segment it would be:

$$\int_5^8 9^2 dt$$

Sum up the values for each segment and then divide by the total period, yielding the MEAN of the SUM of the SQUARE VALUES.

 

[freshbox]

Why is it not voltage x time which is the area but v² x time?

Didn't the formula put area[v²(t)

 

[gneill]

The definition of RMS squares the voltages and multiplies by time. It's not area for RMS.

 

[freshbox]

#Post 13 it is divide by Time from the formula.

 

[gneill]

It is divide by Time from the formula.

?

Yes, the sum of the squares is divided by total time (the period) to yield the Mean of the Sum of the Squares over time.

$$\frac{1}{T}\int_0^T v(t)^2 dt$$

Then you take the square root of all that and you've got the RMS.

 

[freshbox]

How come the area he got is 495 while I get 45, which result in getting 49.5 as the average value while i got 4.5

Since you say my answer is correct, how did he even get the 2nd part correct?

 

[gneill]

How come the area he got is 495 while I get 45, which result in getting 49.5 as the average value while i got 4.5

Since you say my answer is correct, how did he even get the 2nd part correct?

THAT'S NOT AREA! It's the sum of the squares, which he then divided by the period to find the MEAN of the SUM OF THE SQUARES. Then he took the square root of that to yield the RMS value of 7.04 V.

Edit: Okay, technically, it is an area, but it's the area under the v²(t) curve, not the area under the v(t) curve. For a piecewise continuous function like we're dealing with here, it amounts to summing the squares of the values multiplied by the times over which the values hold. Then divide the sum by the total time to find the mean.

 

[gneill]

#Post 13 it is divide by Time from the formula.

Okay, I think I see where the confusion may be arising. In that image in post #13, he's indicating that he's taking the area under the i²(t) curve, not multiplying the area under the i(t) curve by i²(t).

 

[freshbox]

Can you show me where is the v²(t)curve and v(t) curve?

 

[gneill]

Can you show me where is the v²(t)curve and v(t) curve?

The v(t) curve for one period was in your second attachment. It looked something like this:

The v²(t) curve you'd have to construct on your own if you think you need it, but the v(t) curve gives you all the information you need to work with in this problem; the squares of the values of v can be easily obtained by inspection, along with the periods of time over which they occur.

EDIT: Actually, the curve depicted in the attachment of post #13 is in fact the v²(t) curve for this problem. I hadn't noticed that before; I took it to be something from some other class demonstration problem. Sorry about that!

 

[The Electrician]

Here is the v(t) and v(t)^2 plot. The v(t) is blue and v(t)^2 is red.

Also shown is the calculation of the average and the RMS value.

 

[freshbox]

Thank you gneill & The Electrician for the explanation.