Good video. When I think of integrals I like to think of what you called "all instantaneous values" as simply the area between the function and the x-axis. It just seems to make more sense in my eyes.
If you just want the average or 'mean" of a sine wave from 0 to pi there is no reason to square the sine (the value is always positive from 0 to pi versus from 0 to 2pi) . The average value of a sine wave of amplitude 1 from 0 to pi is: 1/pi times the integral of sine from 0 to pi. This yields 2 divide by pi or approximately 0.637. This differs from the RMS value by about 10%. This is because is for most wave forms you cannot square it over a period where the value changes significantly, find the average (of the squared wave form) and take the square root of the result and expect to get the average. The only periodic wave form for which the RMS value is exactly the same as the average is a pure square wave.
This confused me so much previously. I saw so many people explaining that the significance of the RMS value was the average value of the sin wave when it really wasn't.
can you pls elaborate the part "This is because is for most wave forms you cannot square it over a period where the value changes significantly" thank you!
Thank you very much this was perfect. The only think that i'm curious to know it's why to use the root in all integration? Does have any explanation or it's just the rule?
If the peak wasn't 1, and it was derived like this, then would it be (P*sintheta)^2 in the first step? And then the integral, average, square root, would that result in P coming out just as a factor = 0.707*P? Thank you
Oof. 9 months later haha. So, we multiply by 0.707 for a few reasons, but the way I think about it is that the RMS voltage is A) the universal standard for specifying the AC voltage used in electrical systems (or at least residential and low / medium voltage industrial voltages), all the way from your phone charger to 480VAC hydraulic pumps. B) the RMS is the DC equivalent voltage which is useful for power calculations. If you have a motor that runs at 120VAC 1.0A, you can multiply 120 x 1 and say "alright, this thing uses 120W of power to run." If we used peak voltage, you couldn't do that! You see, the power consumption isn't constant in an AC system, right? In a purely resistive load, it's also sinusoidal. Then we're getting into apparent vs. reactive loads and that's a whole other topic of discussion... But the point is, if you look at the power consumed by a simple resistor over hours or days or weeks, it will be the same if it has 120V DC or 170V AC across it. Again, that's because for a lot of the time during the sine wave's period the voltage across the resistor is actually less than 120V. RMS is just sort of a magic tool for getting the V=IR voltage from sinusoidal sources. It just works haha
Why didn't you set the limits of integration to be 0 and 2 pi? You get the same result and I believe the definition of RMS requires you to set the limits to the start and end of a single period.
he chose the peak voltage of the sine wave form to be 1 if you set it to A , answer of the integral will be A over sqrt of 2.and in this case A was 170 v
Sorry for so many questions. I used Matlab to prove this. I created a sine wave 'd', and multiplied its maximum by the 1/sqrt(2) factor --> (max(d)*1/sqrt(2) but it doesn't equal rms(d) for some reason :/
I'm here because I suck at Matlab, given that I started my Matlab for beginners course last week. Not so bad, in my opinion! :D As for you, imbecile, I got from you the reply I looked for. Thanks for your submission. Much appreciated.
This math is so far beyond my grasp. It's like ya'll just always making up random shiz and symbols and never giving in depth explanation ;( Even the math videos I try to watch to understand it lol
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Good video.
When I think of integrals I like to think of what you called "all instantaneous values" as simply the area between the function and the x-axis. It just seems to make more sense in my eyes.
Yea but area under curves just comes from the fact that it happens to be equal to the sum of all instantaneous values.
ขอบคุณมากๆค่ะ เข้าใจขึ้นเยอะเลย👍👍
If you just want the average or 'mean" of a sine wave from 0 to pi there is no reason to square the sine (the value is always positive from 0 to pi versus from 0 to 2pi) . The average value of a sine wave of amplitude 1 from 0 to pi is: 1/pi times the integral of sine from 0 to pi. This yields 2 divide by pi or approximately 0.637. This differs from the RMS value by about 10%. This is because is for most wave forms you cannot square it over a period where the value changes significantly, find the average (of the squared wave form) and take the square root of the result and expect to get the average. The only periodic wave form for which the RMS value is exactly the same as the average is a pure square wave.
This confused me so much previously. I saw so many people explaining that the significance of the RMS value was the average value of the sin wave when it really wasn't.
so if you want the rms of a sine wave from 0 to 2pi this is the time should you use the squared?
can you pls elaborate the part "This is because is for most wave forms you cannot square it over a period where the value changes significantly" thank you!
@@Jpres omg same
We need to use square because we are operating with Power value (Power = U^2 / R = I^2 * R). Not pure voltage or current.
Thank you very much this was perfect. The only think that i'm curious to know it's why to use the root in all integration? Does have any explanation or it's just the rule?
Could we have also gone only till Pi/2 to calculate the RMS? Also why do we square sin^2(theta) @ 5:15? Thank you
If the peak wasn't 1, and it was derived like this, then would it be (P*sintheta)^2 in the first step? And then the integral, average, square root, would that result in P coming out just as a factor = 0.707*P? Thank you
Why do we multiply 0.707 by the amplitude value?
Oof. 9 months later haha. So, we multiply by 0.707 for a few reasons, but the way I think about it is that the RMS voltage is A) the universal standard for specifying the AC voltage used in electrical systems (or at least residential and low / medium voltage industrial voltages), all the way from your phone charger to 480VAC hydraulic pumps. B) the RMS is the DC equivalent voltage which is useful for power calculations. If you have a motor that runs at 120VAC 1.0A, you can multiply 120 x 1 and say "alright, this thing uses 120W of power to run." If we used peak voltage, you couldn't do that! You see, the power consumption isn't constant in an AC system, right? In a purely resistive load, it's also sinusoidal. Then we're getting into apparent vs. reactive loads and that's a whole other topic of discussion... But the point is, if you look at the power consumed by a simple resistor over hours or days or weeks, it will be the same if it has 120V DC or 170V AC across it. Again, that's because for a lot of the time during the sine wave's period the voltage across the resistor is actually less than 120V. RMS is just sort of a magic tool for getting the V=IR voltage from sinusoidal sources. It just works haha
Here in 2024 .......after 7years 😌
Thanks for your help man!
Awesome 👍
Why didn't you set the limits of integration to be 0 and 2 pi? You get the same result and I believe the definition of RMS requires you to set the limits to the start and end of a single period.
Helium?
i don't get why you multiply the final answers with sqrt ?
he chose the peak voltage of the sine wave form to be 1 if you set it to A , answer of the integral will be A over sqrt of 2.and in this case A was 170 v
can these formulas feed me to survive in real life? Hhh
Copy and paste this into WolframAlpha: square root of (integral of sin^2(x) from 0 to pi) divided by pi
0 meter 0.707 bar mount Everest!33000 meters altitude!new Aquator 33643 km! End of world!Better Dooms day!neutronstar!10 bar 0 meter!
Mmm, the integration is over theta. I would have thought it should be over pi.
Nice
What is 120v 40 ohms rms valvue
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Sorry for so many questions. I used Matlab to prove this. I created a sine wave 'd', and multiplied its maximum by the 1/sqrt(2) factor --> (max(d)*1/sqrt(2) but it doesn't equal rms(d) for some reason :/
I can't say the same about your brain or genitals. No manifestation of neither.
I'm here because I suck at Matlab, given that I started my Matlab for beginners course last week. Not so bad, in my opinion! :D As for you, imbecile, I got from you the reply I looked for. Thanks for your submission. Much appreciated.
@@Jdonovanford
Lmao he helped you out and you're being so rude 😂
thanks
173 volt in all
This math is so far beyond my grasp. It's like ya'll just always making up random shiz and symbols and never giving in depth explanation ;( Even the math videos I try to watch to understand it lol