wattmeter beginner need help

[MrPlacid]

Okay, I couldn't resist passing up on old gears and found a digital wattmeter model WD-767 on ebay.

So I took my first few measurements and see if they balanced out.

V_RMS=118
A_RMS=1.09
W=55

To tell you the truth, I am still fuzzy with RMS stuffs, so is power still equal to V x A?
From calculation I got P=VA=128.62 watts, but the display is 55 watts.

[alm]

Apparent power = V x I
Real power = integral(V(t)*I(t)*dt, 0..t) / t = V x I x cos(phi) (for sinusoidal signals), where phi is the phase shift between voltage and current.
Calculating reactive power is left as an exercise for the student . For phi = 0, real power = apparent power, and reactive power is zero.

Think of reactive power as the power drawn by an ideal capacitor: there is voltage over the cap, current flows through the cap, but the ideal cap does not get hot at all. This is because voltage and power are 90 degrees out of phase. Inductive or capacitive load tends to cause this. Power factor correction (PFC), often used in power supplies which can deliver large amounts of power (eg. computer PSU), is designed to compensate for this.

Real power is the sum of the instantaneous product voltage and current, i.e. the sum of the apparent power at each moment in time.

Consumers pay for real power to their utility, commercial customers tend to be charged if they draw large amounts of reactive power, because it causes loss for the utility (that current is going to cause I²R losses, which is real power).

I'm sure you'll find a nice explanation on Wikipedia or any random electronics textbook.

[tecman]

What you are seeing is one of the characteristics of AC power.  VA is volts*amps, which is sometimes called apparent power.  Watts, or real power, is volts*amps*power factor.  Power factor is a number from 0 to 1, and is the cosine of the phase angle between the voltage and current.  In other words, the voltage and current waves are often not in phase with each other.  If the load is inductive, the current will typically lag the voltage, and if the load is capacitive, the current will lead the voltage.  Pure resistive loads usually have a power factor if 1 and are in phase.

If the current is not in phase, the current will not be highest when the voltage is highest, so the actual power will be lower.  The phasing is a result of the power factor, and is present in many AC power systems.  The difference between apparent power and real power is usually referred to as imaginary power.  

Google "Power Factor" and you can find a lot of information on the subject.

paul

[MrPlacid]

Thanks for the reply, Tech and Alm. I think I got it.  Well, gonna check out the power factor lecture on youtube ~1 hour long.


Then sleep on this for a few days, brush up on my math, and hopefully everything sinks in. I am glad, I bought this, I am learning something new

PS: I think I got a power factor of 0.42. Am I on the right track?
Also, I was taking more readings and comparing it to the label on the appliances. It seemed the appliances are usually shown with apparent power and using 120 VAC.

[Tony R]

Real power = integral(V(t)*I(t)*dt) = V x I x cos(phi) (for sinusoidal signals), where phi is the phase shift between voltage and current.

That would be energy, you are taking the integral of power over time

[Kiriakos-GR]

I am glad, I bought this, I am learning something new

I share your Joy too .   

I had also the same flame about exploring my Fluke 87V ,
and you helped me, by uploading for me, all the files of the Fluke CD.

I am also an major supporter of the idea :
 I bought this, I am learning something new 

I am very happy for you too.

 

[alm]

Real power = integral(V(t)*I(t)*dt) = V x I x cos(phi) (for sinusoidal signals), where phi is the phase shift between voltage and current.

That would be energy, you are taking the integral of power over time

You're right, it should be the average instantaneous power, which would be the integral over delta t. Fixed.

[MrPlacid]

I am also an major supporter of the idea :
 I bought this, I am learning something new 
I am very happy for you too.

Thanks, Kiriakos. I just got back into electronics again . It's been almost 3 months absence. Hopefully, this is the last time the electronic bugs get to bite me. I got a new secret weapon for the electronic bugs this time. I hope

You're right, it should be the average instantaneous power, which would be the integral over delta t. Fixed.

Hopefully one day, I will finally be able to understand what you guys are talking about

PS: Guys, I couldn't finish that youtube video. Watching Dave's videos is much more enjoyable.