OP\(1V + 2V + 3V) / (0.0001 + 0.0002 + 0.0003) = 10000[/quote said:NO!
These are trapezoids.
The area of a trapezoid is: 1/2 (b1+b2)*h
For a v(t) = f(t), my function and w = 0.0001
we have:
(1/2) *( [0+1]*w + [1+2] * w + (2+3) * w ..... )
(1/2) * (w +3w + 6w)
(1/2) * 10w = 5w
w=0.0001 s so area under curve = 5*0.0001 = 0.0005 for those slices.
You would divide by the total time over one cycle to get the RMS value of V.
My waveform was a very steep triangle wave. Let's hope I did it right.
I believe that gives the average value of V, not the RMS................
You would divide by the total time over one cycle to get the RMS value of V.
Hello guys,
>So, I guess you have to do something a little fancy: http://www.wolframalpha.com/input/?i=rms
Yes, it seems like the same thing as this (just in lamens terms):
http://www.analytictech.com/mb313/rootmean.htm
HOWEVER, something still doesn't make sence. First of all, please take a quick look at the graph in my attachment below:
If we go by the method explained in my latter link above, then let's assume that Voltage at 100% will be 100V peak. Now, let's do this:
Step#1 (square all values of the top part of the AC wave):
V @ 10% = 10 Volts ² = 100
V @ 20% = 20 Volts ² = 400
V @ 30% = 30 Volts ² = 900
V @ 40% = 40 Volts ² = 1600
V @ 50% = 50 Volts ² = 2500
V @ 60% = 60 Volts ² = 3600
V @ 70% = 70 Volts ² = 4900
V @ 80% = 80 Volts ² = 6400
V @ 90% = 90 Volts ² = 8100
V @ 100% = 100 Volts ² = 10000
V @ 90% = 90 Volts ² = 8100
V @ 80% = 80 Volts ² = 6400V
V @ 70% = 70 Volts ² = 4900V
V @ 60% = 60 Volts ² = 3600V
V @ 50% = 50 Volts ² = 2500V
V @ 40% = 40 Volts ² = 1600V
V @ 30% = 30 Volts ² = 900V
V @ 20% = 20 Volts ² = 400V
V @ 10% = 10 Volts ² = 100V
Step#2 (Sum all values):
Sum of all values is: 67000
Step #3(Do the average of all these values):
Average: 67000/19 slices = 3526
Step#4 (Do the square root of the average):
RMS = √3526 = 59.38 Volts ?????
59.38 volts is *NOT* the rms value of 100V peak????
RMS is supposed to be 70.7 Volts ?????
Can someone explaine this discrepentie. Confused!
r
I have been reading some lectures in this site and I have a question about obtaining the apparent power, true power, reactive power and the power factor of an actual AC circuit.
Basically I am wondering if someone can take a quick look and evaluate or validate that what I am doing is okay.
Let's say I have an 120VAC, 60HZ, RL circuit comprising a 60 ohm resistor and a 160 milli-henry coil. This circuit will have two outputs that will connect to a micro controller. Please view attached schematic. One of the ouputs will read the instantaneous current and the other will read the instantaneous voltage (both at the source of the AC circuit). To read the voltage and current, we will build a small circuit that will convert these values to an analog voltage signal so that the micro controller can read them. Therefore, we do not care about *how* the instataneous voltage and current are read... for example purposes, for now, we just know that the micro controller is able to read those values.
So, if we are able to read the instantaneous voltage and current values into the micro controller, then we should be able to calculate the RMS voltage and currents. So for the sake of this example, lets say we are reading an rms current and voltage of:
I = 1.41A
V = 120 VAC
In doing so, I can calculate the total AC impedence by doing:
ZT = 120/1.41A = 85.10 ohms
By having the total impedence, I can calculate the apparent power by doing:
S = I²Z = (1.41²) x 85.10 = 169.256 va
Also, since we are able to read the instantaneous voltage and current of the AC circuit, we are definitely able to detect the pahse angle between the current and voltage. We can the convert this angle to degrees. Assume we have a 45° angle. Refering to the trigonometric triangle, we can then figure out the true and reactive powers by doing:
119.2 watts = 169.256 * cosine 45° <<< true power
119.9 vars = 169.256 * sine 45° <<< reactive power
Also, we can simply calculate the power factor of this circuit by doing:
0.703 = 119.2/169.256
Is what I am doing correct? All feedback is appreciated.
Yes, but I prefer getting the power triangle through the instantaneous V and I. The reason is let's suppose one day I don't only want the power triangle! Suppose one day I need the instantaneous I and V to do something else.... its just a precaution.
>So all you have to do is multiply each instantaneous current and voltage value, sum them up, and divide by the period
Quick question, this gives us the apparent power ... right?
God I am having a hard time with this. I apologize.
First, doing the product for each instataneous values for V and I, and then summing them up and dividing this result by the period is true power. Okay, I did more reading and true power is defined as work done per time by the resistive components in an AC circuit.
Now, going back to my innitial example, if we would multiply the 120V by the 1.41A, this will give us 169 va's... right!
However, the 120 and the 1.14 are the values that a meter (VOM) would read ... in other words the "rms" voltage and current right?
So then, in my example the peak voltage is (120/0.707)= 169 volts and that the peak current is (1.41/0.707)=1.99A, then, what is the units of the porduct of(169x1.99)= 336.8 called? "336.8 peak va" ???
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