# Homework Help: AC Circuit Sinusiodal

1. Dec 8, 2012

### 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.

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2. Dec 8, 2012

### Staff: Mentor

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.

3. Dec 8, 2012

### freshbox

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

Effective RMS= Vm x √2

Rms=Vm/√2

Am I correct?

4. Dec 8, 2012

### Staff: Mentor

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.

5. Dec 8, 2012

### freshbox

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.

6. Dec 8, 2012

### Staff: Mentor

Yup.
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.

7. Dec 8, 2012

### 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?

8. Dec 8, 2012

### Staff: Mentor

Yes, they're the same thing.

9. Dec 8, 2012

### 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

10. Dec 8, 2012

### Staff: Mentor

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.

11. Dec 8, 2012

### 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?

12. Dec 8, 2012

### Staff: Mentor

Yes, that's right.
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.

13. Dec 8, 2012

### 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.

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14. Dec 8, 2012

### Staff: Mentor

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).

15. Dec 8, 2012

### 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)

16. Dec 8, 2012

### Staff: Mentor

Sure, looks good.

17. Dec 8, 2012

### 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)

Last edited: Dec 8, 2012
18. Dec 8, 2012

### Staff: Mentor

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.
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.

19. Dec 8, 2012

### freshbox

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

Didn't the formula put area[v2(t)

20. Dec 8, 2012

### Staff: Mentor

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