--- Well, if you want to determine whether the PF meter is accurate or not you'll have to do _something_ , so let's start at the beginning, OK?
Power factor is defined as the cosine of the phase angle between current and voltage, so if we can make a circuit with a predictable phase angle we can then determine what its power factor will be and, if we plug it into the meter, and that meter doesn't affect our circuit, the PF meter should read what we predicted.
So, What circuit? A simple resistor and capacitor wired in series and plugged into the PF meter's output receptacle would be perfect.
Let's say that we choose a power factor of 0.5 just because it's halfway between 0 and 1, 0 corresponding to a perfecty reactive load and 1 corresponding to a perfectly resistive load.
Now we've gotta do some numbers. Sorry...
Since:
PF = cos(phi)
we get the angle like this:
phi = cos-1(PF) = cos-1(0.5) = 60°
Next, since we don't want to blow anything up we'd like to keep the power dissipation in our test circuit down to something reasonable.
Unfortunately, we don't know what the accuracy of the meter is when it's presented with low power loads but, just for grins, let's go for a 10 watt load.
Then, since:
E² P = --- R
we can rearrange that to get the value of a resistor that'll dissipate 10 watts when it's connected across 240V mains like this:
E² 240V² R = --- = ------ = 5760 ohms P 10W
Not bad. Keeping in mind that adjustable resistors are cheap and adjustable capacitors aren't, lets double that resistance and wattage to 10k ohms and 20 watts, so that the resistor's slider will be somewhere near the middle of the resistor (hopefully) when we figure out the capacitor we need to give us that 60° phase shift.
Now, since:
Xc tan(phi) = ---- R
and we know that phi = 60°, in order to get Xc (which will eventually get us to C) we rearrange again to solve for Xc like this:
Xc = R tan(phi) = 5760R * 1.732 ~ 9976 ohms
And then, finally, since:
1 Xc = --------- 2pi f Cwe can rearrange to solve for C like this:
1 1 C = ---------- = --------------------- = 3.19E-7F 2pi f Xc 6.28 * 50Hz * 9976RThat's 0.319µF, which isn't readily available, but 0.33µF is, and that's pretty close, (since we have an adjustable resistor to play with) but what about the cap's voltage rating?
For 240VRMS mains we're talking about 340V peak which will be placed across the series RC, and the voltage will divide between the R and the C depending on the impedance of the circuit, which will be:
Z = sqrt (R² + Xc²) = sqrt (5760² + 9976²) = 11519 ohms.
That means the current in the circuit will be
E 240V I = --- = -------- ~ 0.0208 amperes. Z 11519R
and that current in the reactance of the capacitor will drop:
E = I Xc = 0.0208 A * 9976R ~ 208VRMS
across the capacitor, so a 0.33µF 350V cap would do.
Now, since the cap is 0.33µF instead of 0.319µF, we've got to go back and find what resistance we need to get a phase angle of 60°.
First though, we need to find the actual value of the cap, so you can either measure it if you have a capacitance meter or use the
50Hz mains and a 1% resistor in series with the cap to find out. Use a transformer to step the mains voltage down to something you can work with comfortably if you're scared of the mains.Now, once you've determined the capacitance of the capacitor, (let's say it's exactly 0.33µF) you determine its reactance by solving:
1 1 Xc = ---------- = ----------------------- ~ 9651 ohms, 2pi f C 6.28 * 50Hz * 3.3E-7Fthen, knowing you want a 60° phase angle, rearrange:
Xc tan(phi) = ---- R
to solve for R, like this:
Xc 9651R R = ---------- = ------- = 5572 ohms tan(60°) 1.732
And here's your circuit:
+------+ 240V50Hz>---| |------+ | | | | PF | [5572R] |METER | | | | [0.330µF] | | | 240V50Hz>---| |------+ +------+Just to make sure we're not going to exceed the resistor's wattage rating we need to go back and figure out the current in the circuit.
Since its impedance will be:
Z = sqrt(R² + Xc²) = sqrt(5572² + 9651²) = 11144 ohms
the current through the resistor will be:
E 240V I = --- = -------- ~ 0.022A Z 11144R
and it'll be dissipating:
P = I²R = 0.022² * 5572R ~ 2.7W,
so a 10000 ohm 10 watt adjustable resistor would be fine.
Notice, however, that the original assumption of a 10 watt load won't be true because of the increase in the circuit's impedance due to the capacitive reactance. But, now that you know how to choose the R and C, you ought to be able to figure out how to get whatever load impedance you need.
-- John Fields Professional Circuit Designer