But In my text book answer is different. I don't understand why my answer is wrong. Correct answer is I rms =0.3535a and Iavg = 3.185A. Can someone tell me where I am wrong?You don't define the period of time in which to calculate the average, so assuming over infinite time.....
The first two are correct, but the average value of a sine wave is zero.
The formula you used, and carelessly populated, is for a fully rectified sine wave.
According to definition, effective current is equal to square root of mean square.Apply the definitions of rms current and average current (instead of just plugging numbers into whatever formula happens to be nibbling at your toes at the moment).
So what is the defining concept behind rms (look for "effective" voltage and current)?
But In my text book answer is different. I don't understand why my answer is wrong. Correct answer is I rms =0.3535a and Iavg = 3.185A. Can someone tell me where I am wrong?
Hello,But In my text book answer is different. I don't understand why my answer is wrong. Correct answer is I rms =0.3535a and Iavg = 3.185A. Can someone tell me where I am wrong?
Hi,According to definition, effective current is equal to square root of mean square.
No, that is not the definition of effective current.According to definition, effective current is equal to square root of mean square.
Hi there,No, that is not the definition of effective current.
The definition is that the effective value of a current waveform is the DC current that would dissipate the same average power in a purely resistive load as the actual current waveform does over the time period of interest.
Now turn that definition into a matching mathematical relationship. On one side write the equation that evaluates to the average power the actual waveform would deliver to an arbitrary resistance, R. On the other side do the same for the average power a DC current of magnitude Ieff would deliver to that same resistance.
The "effective" value is what counts. It is defined in terms of what it means, not in terms of whatever math operations end up having to be used to get at the corresponding value.vead said: ↑
According to definition, effective current is equal to square root of mean square.
Hi there,
Are you sure about what you are saying, or am i interpreting you wrong here.
The 'effective' value is the square root of the mean of the square, which is usually abbreviated RMS for "Root-Mean-Square".
RMS is the value that produces the same power in a resistor as a DC value would.
So if VacRMS=3 then that produces the same power in a resistor as Vdc=3 would.
Agree?
Which, I thought, was pretty clearly implied when I asked, "So what is the defining concept behind rms (look for "effective" voltage and current)?"Hi there,
Ok i think what you are doing here is you want to go back to the physical definition of 'effective' voltage, then come up with the expression for RMS.
I never said that calling the effective voltage (or current) the RMS voltage was wrong. I asked the TS what the defining concept behind it was (and specifically pointed him to the term "effective voltage"). He responded that the definition of effective current (not he did not even claim it was the definition of rms current) was the square root of mean square. This is wrong. Worse, it demonstrates that his understanding is based on nothing more than memorized and regurgitated formulas with little to no understanding of the concepts upon which they are based, which is why he is having so much difficulty with problems like this.What seems strange though is to imply that RMS is somehow completely wrong when so many references refer to it as RMS being the effective value, which can be confusing.
This is exactly how old-time true-RMS meters worked -- they measured the temperature of a resistor across which the voltage was applied.The way i understand the physical definition comes from an experiment as follows...
We have two identical resistors in identical environments.
We measure the temperature of the two resistors with two identical thermometers.
We apply the unknown voltage to one resistor, we apply a variable DC voltage to the other resistor.
After a time, we adjust the DC voltage so that both resistors measure the same temperature. After a time when the two temperatures are finally the same, the DC voltage applied has the same value as the effective voltage of the unknown waveform.
So in this experiment we actually use the heat as part of the measurement, and we dont even measure the voltage of the unknown waveform, we just measure the effect of the heat produced. Also, there is no math involved.
Well, that would be a lot better than the more common situation of people thinking that Vrms = Vpk/sqrt(2) is right!So while i agree that the base definition includes things other than taking a root, in the end we do have to take a root to get the result, and stating that RMS is completely wrong is not the best idea because then people might think that Vpk/sqrt(2) is wrong too.
Which is why I said, "Because of this coincidental set of mathematical equations, we commonly refer to the effective voltage by the name "root-mean-square" or RMS voltage."Many references indicate that the effective value is the RMS value, because the two are equal. We do that a lot as humans...we take shortcuts to make things easier.
Hi,@The Electrician : It could certainly be argued that it is more rigorous to have an explicit integrand equal to 1 (and I had actually included it originally but recalled a couple of times when its "unnecessary presence" confused someone that couldn't understand how it magically appeared, so I took it out). But I don't think its absence makes the integral undefined since 1 is the identity element for multiplication and so the differential element dt is equal to 1·dt, thus inferring that the integrand is equal to 1.
Some people might make a notational argument claiming that the dt doesn't multiply the integrand, but rather is just a delimiter for the integral (and some have then claimed that it isn't even needed as long as the integral isn't followed by anything that isn't a part of it). I reject the notion that the differential doesn't multiply the integrand. First, it's very appearance is the result of a limiting process in which a small difference (a delta) that becomes a differential in the limit most certainly DOES multiply an expression that becomes the integral in the same limiting process. Furthermore, the differential carries units that only make sense if the differential multiplies the integrand. Thus, in my opinion, leaving the integrand as an inferred 1 is sufficiently rigorous.
It appears that the post I was replying to was deleted by the author after I had started my response. I didn't use Reply because it had formatting issues that made it a bit unreadable so I figured it would be better to just tag him. I had to step away from my computer for a couple hours and when I got back I saw that my response hadn't posted, so I had to hit post again.Hi,
Just curious what post you are replying to.
If they say that the indefinite integral is just 1 then they are wrong. It is not. Period. The arbitrary constant MUST be there in order for it to be correct. If it is a definite integral, then the K does NOT belong there because it cancels itself out. I'm not aware of any exception to this (but would love to either be reminded of one I've forgotten or to learn about one I never knew).Not only that, but i have seen authors show that the integral of dt is just 1 rather than 1+K which is probably even more correct, then later solve for K which may or may not end up to be zero.
The problem is that the second form is wrong due to order of operations. In fact, neither is correct for that reason, but we can take a much better guess at the first one.Also we usually write:
integral x dt
where 'integral' is the curly looking integral symbol and so would mean "the integral of x with respect to t", but i have also seen at least one physicist use:
integral dt x
which is still "the integral of x with respect to t".
That's a little unusual, but it makes sense because we like to know what the variable of integration is and what could be better than seeing it right away rather than later.
The two forms shown together:
integral a+b+c*x^2+d*x^3 dx
integral dx a+b+c*x^2+d*x^3
with his second form there we dont have to scan through the whole thing there before we know what the variable of integration is. Kinda nice.