Power Measuring!! Need Assistance

Thanks my friend, I just wanted to verify something... but correct me if im wrong? Apparent Power = S=VxI, Active Power=P=VxIxCos(phi) and Reactive Power Q=Vxixsin(phi)? Thanks for the help....

I do have another question though,what would it be best to sense the current a shunt or a CT?(current transformer).

That's only true if the line voltages are symmetric to neutral, which they never really are. Some domestic electric meters assume this: they wind both current-sensing coils around a common laminated core, and use a single potential coil from line-to-line.

It's better to compute power properly,

P = Van * Ia + Vbn * Ib

where the multiplies are instantaneous products, of course.

John

I think thats the only way the OP's going to get the correct answer

NT

ok

Scary stuff!!!

(hint, that is wrong)

Be careful applying this rule when you have alternating current. It is incorrect to multiply average voltage (RMS) by average current (RMS), in case of any but resistive loads.

That is because current may be out of phase with voltage, as it happens with motors.

I am not sure if you should really be designing a power meter, if you are weak on concepts of power etc.

If you have 3 currents, I1, I2, and In (In stands for neutral)

And resistive load

The current In = I1 - I2

110V power is 110*In = 110*(I1-I2) 220V power is 220*(I1-In) = 220*I2

The sum is Power = 110*(I1-I2) + 220*I2 = 110*(I1+I2)

i

Google "Blondel's Theorem" for details.

Of course, using a DSP to do a 2-element power meter at 60 Hz is like clubbing baby seals. Much more sporting is a 90-cent HC05 processor, using the on-chip multiplexed 8-bit ADC.

John

Phil as usual you write only crap. His formulae were way nearer to the truth then what you wrote.

Totally wrong! The average value of a sinusoidal waveform is Zero ... so according to your answer the True Power = 0!

Help me .

AD chip ....... help!!!!!!!!!!

"Go easy with the whisky"

theJackal

Umm......

The product, Vinst*Iinst, is a sine wave which has an average dc-value proportional to the true power.

The product only has an average value of Zero when Iinst is exactly 90 degrees away from Vinst. ie, When the load is a pure reactance, = zero power.

--
Tony Williams.

ha ha ha Prove your assertions instead of dictating them ... Science is based on facts and proof not on dreamers ... learn that ...you wannabe engineer. Either you are capable of intelligent debate or sink yourself in the Pacific . "Go easy with the whisky"

theJackal

LMAO Thats your proof. Pretty cheap stuff Mr. Alison

"Go easy with the whisky"

theJackal

LMAO ... Still no basis of argument but cheap crap .. must make you feel like a genius . HAHAHAHAHA

"Go easy on the whisky"

theJackal

It doesn't appear that anyone answered your actual questions

I believe you need to measure current in L1 x Volts L1 to Neutral and add it to current in L2 x Volts from L2 to neutral to get both the 120V power and the 240V power

For power measurement a CT is most commonly used in order to provide isolation and of course a PT (potential transformer) is used for voltage input.

Designing accurate power measurement devices is a skill - see

for examples of commercially available devices.

Dan

--
Dan Hollands
1120 S Creek Dr
Webster NY 14580
585-872-2606
dhollan3@rochester.rr.com
www.QuickScoreRace.com

I know its quite easy to see but this Phil chap has been bugging me ever since i started writing here ... so I was just teasing him.

Let me give you my version of the facts . To actually talk about the average value of a sine wave over ONE period is senseless as its zero. Thats why when we mention 115V or

250V we are actually talking about a different kind of average the rms which is the square root of the sum of the squares of the averages (over 1 period) of the instantaneous voltages . Square the Sine function and its positive all over 1 period ... and so it makes sense finding its average which is called the rms. Its also interesting to note that this rms value would actually give the same power dissipation as a direct current or voltage of the same value through a resistive load.

"Go easy on the whisky"

theJackal

"@donis"

** You are wrong.

True Power = average value of V x I

where V = instantaneous voltage and I = instantaneous current.

That is what the AD chip computes.

.......... Phil

Quite a simple problem . Being an engineer you know from analysis of linear systems that if you have 2 simultaneous equations (KVLs in the

2 meshes ) with 2 unknown mesh currents you can easily solve for these. If you need details on how thats done let me know.

Power in 2 loads is (I2-I1)^2 * Resistance Load1 + I2 ^2 * Resistance Load 1 .

"Go easy with the whisky"

theJackal

WRONG.

--
 Thanks,
    - Win

"theJackal"

** This Jackass pile of *Eurotrash* is a veritable font of misinformation.

........ Phil

"theJackal"

** This Jackass pile of verminous *Eurotrash* is a veritable font of insane misinformation.

........ Phil