RMS calculation - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 08:34:04 am
content: So I am having a bash at my latest assignment. I have a light dimmer circuit with a 100W bulb on it and they want me to calculate the power used when the dimmer circuit is set to a firing angle of pi/3.Now putting me thinking cap on as the first half of the wave (0-pi) is the same as the second half (pi-2pi) in terms of RMS outcome I only actually need to consider one half cycle to calculate the RMS as the second half is identical but for the polarity and RMS is designed to deal with that anyway. So my calculation is as per attached. Does this look right ?

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 09:12:05 am
content: I thought I made an earlier post, but it seems to have disappeared.My earlier post was that since the power is independent of frequency, you can write your integral in terms of "sin(x)" and "dx" rather than "sin(wt)" and "dt".And then I followed on with:Also, they have asked for the power, not the RMS voltage.Now the power is proportional to V2, so the power over a full half-wave will be proportional to the integral of the voltage squared over that interval, or:Total power is proportional to: Integral of sin2(x) dx from 0 to piIf the firing angle is pi/3, then the power will be proportional to that fraction of the half-wave:Partial power is proportional to: Integral of sin2(x) dx from pi/3 to piIf you take the ratio of these two integrals you will get the fraction of the total power when the lamp is dimmed. Multiply this fraction by 100 W to get the power of the dimmed lamp.

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 09:29:46 am
content: Surely once I have the RMS voltage I can use the lamp resistance that I can calculate to work out the power as RMS volts is equivalent to the DC volts that have the same effect.Sent from my Moto G (4) using Tapatalk

author: orolo
author_rank: Frequent Contributor
timestamp: September 17, 2017, 10:08:59 am
content: Be careful with the integration limits: going from 0 to Pi with frequency w=50 means integrating over 25 cycles. I think you might want to use w=1.If your bulb is 100W RMS, the power over the load has 141.42W of amplitude. Use that amplitude with your integral (w=1), and you should get about 89.7Watts.

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 10:13:06 am
content: Quote from: Simon on September 17, 2017, 09:29:46 amYes, you can do it that way, but it is a longer calculation that involves unnecessary steps. Since they asked for the power and not the RMS voltage, you can directly calculate the power and skip the voltage.Also, suppose you determine the power to be 80.4 W when dimmed, by the method I outlined. Then using the formulaPower = V2/ Ryou can work out the resistance of the bulb to be 2402/ 100 = 576 ohms like you said.Then you can use the same formula to calculate the RMS voltage at 80.4 W as sqrt(576 * 80.4) = 215 V, if you want to know that.

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 10:27:18 am
content: Quote from: orolo on September 17, 2017, 10:08:59 amThe peak power over an AC cycle is twice the average power. So it should be 200 W, not 141.42 W.

author: Andy Watson
author_rank: Super Contributor
timestamp: September 17, 2017, 10:34:00 am
content: I think your original equation is correct - be careful of the signs of the sines - make sure they don't cancel.There is no short-cut. Dividing the power according to the ratio of on to off time only works for PWM waveforms which have constant DC "on" level.

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 10:52:28 am
content: Quote from: Andy Watson on September 17, 2017, 10:34:00 amYou have to be careful here of the units. If you write this,Integral from pi/3 to pi of (sin(wt))2dtthen you are integrating over time (that's what t stands for), and time is measured in seconds. One half cycle of 50 Hz would be 0.01 s or 10 ms, it would not be pi. So the original equation is not exactly correct, although it captures the right idea.Quote from: Andy Watson on September 17, 2017, 10:34:00 amDividing the power by the ratio of the on to off time has not been proposed though. So it is a non-issue.

author: orolo
author_rank: Frequent Contributor
timestamp: September 17, 2017, 11:16:47 am
content: Quote from: IanB on September 17, 2017, 10:27:18 amI see, I'ts average power, not RMS power. Then the 80.4W figure is correct. To compute it, I'd use the fact that the load is resistive, so current and voltage are in phase, and then:\$P_{avg} \quad =\quad \frac{1}{\pi}\int_{\pi/3}^{\pi} i(t)\cdot v(t)\, dt \quad = \quad \frac{1}{\pi}\int_{\pi/3}^{\pi} 2I_{rms}V_{rms}\sin^2(t)\,dt \quad = \quad \frac{200}{\pi}\int_{\pi/3}^{\pi}\sin^2(t)\, dt \quad =\quad 80.45\mathrm{W}\$

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 03:10:32 pm
content: Quote from: orolo on September 17, 2017, 10:08:59 amI think the 1/pi at the beginning deals with this but as Ian said i should take "w" out as I am not working with time and frequency here, because you multiply by 1/"time frame" it should work for any length that the integrals also define the wave over. So because i know that the wave is symmetrical I can just work it out over one half cycle as it is the same as looking at the full cycle and integrating the same thing twice. Because we are talking RMS the fact that one half is negative does not matter.

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 03:27:47 pm
content: Well the attached now works in Microsoft mathematics and gives a result of 152.22V (pretend I put 240 there not 230 doh!)Bulb resistance is 576i = 152.22/576 = 0.264AW = 40

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 03:30:54 pm
content: I'm confused about the 80 W because at 60 degrees that is taking a fair chunk off, and if you half the volts you quarter the power.

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 03:40:10 pm
content: actually the question is for 230V

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 04:05:16 pm
content: Quote from: Simon on September 17, 2017, 03:30:54 pmFrom 0 to 180 degrees is the whole half cycle and is full power. Now if you do half of that, from 90 to 180 degrees that would be exactly half the power, or 50 W, due to the symmetry of the graph. It follows that from 60 to 180 degrees is more than half of the graph and must be more than 50 W. (Draw a sketch to visualize it.)

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 04:21:35 pm
content: but my period of calculation is for pi or 180 degrees, as far as my calculation is concerned there is no 180-360 because i know it is the same and the RMS voltage from 180-360 degrees is the same as 0-180 degrees.Where I have gone wrong is to use the RMS value of the full sine wave and not it's peak value, putting that right by adding a multiplication by sqrt(2) in gives the 80.4W

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 04:32:01 pm
content: Quote from: Simon on September 17, 2017, 04:21:35 pmRight, but I didn't write 180-360 degrees in my last post, I wrote 0 to 180 degrees because as you say both half cycles will work out the same and you only have to do the positive half cycle. But it is still true that a quarter cycle is still exactly half the power. There is exactly the same power from 0 to 90 degrees as there is from 90 to 180 degrees.QuoteCool

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 09:01:07 pm
content: sorry I thought you were saying that because i was calculating over half a wave cycle I had ended up with only half the value i should have (through miscalculation which is entirely possible).thank you all for your help that one is put to bed, next one coming up soon.

author: IanB
author_rank: Super Contributor
timestamp: September 17, 2017, 10:55:21 pm
content: Since you have solved the problem, I'll just drop here for reference the details of the method I outlined in my first post. The advantage of this method is you do not have to worry about RMS or peak voltages, and so there is less chance of error:Since power is proportional to the square of the voltage, we can integrate the square of the voltage over one half cycle as:$$\int_{0}^{\pi} \sin^2 x \text{ d}x = {\pi \over 2}$$Normalizing this, we have:$${2 \over \pi}\int_{0}^{\pi} \sin^2 x \text{ d}x = 1$$This gives the power over the whole half cycle normalized to unity.Now, to compute the power with a phase angle starting at \$\pi/3\$ we can modify the starting point of the integral accordingly:$${2 \over \pi}\int_{\pi/3}^{\pi} \sin^2 x \text{ d}x \approx 0.8045$$This tells us the reduced power will be 80.4% of the full wave power.

author: Simon
author_rank: Unknown
timestamp: September 18, 2017, 06:52:22 am
content: Ah yes i see.