# Measured and calculated power in AC circuit

Discussion in 'Homework Help' started by fektom, Mar 17, 2013.

1. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
Hello everybody,

I've read the page which describes how to make corrections on power factor (http://www.allaboutcircuits.com/vol_2/chpt_11/4.html) and there is something I don't fully understand.

This page says in the second paragraph that I can measure the true power with a wattmeter, while the apparent power can be calculated by multiplying the voltage and the current.

Based on my knowledge, the wattmeter has to wire coils. One is to measure the voltage accross the load, and a second to measure the current through it. The meters deflection comes from the combined effect of two coils inside the meter.

My question is, if the calculated and the measured power (apparent and true) values are created by the same quantities (ie: E and I) how can I get different results?

According to my impression they shall not be different.
If you have any idea, please share!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
The meter deflection is proportional to the time averaged value of the instantaneous product of the the current and voltage functions - a physical response arsing from magnetic field forces produced by the sensing [V & I] coil arrangements, the meter mechanism's mechanical inertia and the current flowing in each coil.

The VA calculation is simply the mathematical product of the derived [RMS] values of the known or measured current & voltage.

One can show mathematically why these are not the same but I'm unsure if you are interested in that approach.

Last edited: Mar 17, 2013
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3. ### Papabravo Expert

Feb 24, 2006
10,021
1,757
This is an example of of why multidimensional problems are hard to understand intuitively. The two dimensions in this case are amplitude and phase. In a resistor the phase difference between current and voltage is always zero so this case is easy to grasp intuitively. With reactive components the phase difference is almost never zero and this is the hard case.

You either grasp the mathematics or you don't. If you don't, then a switch to "Art History" as a pursuit is highly recommended.

4. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Following up on tnk's description, imagine that the two signals are 90 degrees out of phase. Then when the voltage coil is at its peak, the current coil is zero and so no deflection. When the current coil is at its peak the voltage coil is zero and so no deflection. The rest of the time there is a net force but half the time it is trying to deflect the needle one way and the other half of the time it is trying to deflect the needle the other direction. On average, they cancel out and the needle just sits there at no deflection.

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5. ### Tesla23 Active Member

May 10, 2009
318
67
The physics sometimes helps as well:

The true power and apparent power will differ if the circuit has energy storage elements (e.g. capacitors and inductors).

If you apply a sine wave voltage across a capacitor, then as the voltage rises from zero to the maximum, power flows into the capacitor and energy is stored in the electric field. As the voltage then reduces to zero, this power flows back to the source. So whilst current flows, giving apparent power, there is no net power flowing into the capacitor, hence zero true power.

Mathematically, for a capacitor C

$V=cos(\omega t)$ $I = \omega C sin(\omega t)$

so $\overline{VI}=\overline{\omega C cos(\omega t) sin(\omega t)} = 0$

No power is dissipated in the capacitor

Compare this to a resistor R

$V=cos(\omega t)$ $I = cos(\omega t)/R$

so $\overline{VI}=\overline{cos(\omega t) cos(\omega t)/R} = \frac{1}{2R}$

Power is dissipated in the resistor

A similar situation arises for an inductor, as the current increases to a maximum there is energy stored in the magnetic field which is returned to the source as the current drops to zero.

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6. ### WBahn Moderator

Mar 31, 2012
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I don't think the OP is having a problem understanding that there IS a difference between the two, but rather he doesn't see why a wattmeter's reading would reflect that difference.

7. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
So I use completely the same quantities when I measure or calculate.
The explanation is in physic... I'm happy with that.
I didn't take into account that the phase shift between current and voltage has an influence on wattmeter. It shows the combinated effect with the influence of phase shift. (there are 5times zero result in one period under 90 degree phase shift condition)

However, when I measure either the current or the voltage separately, I get separated effect of them even if there is phase shift.

I think I undertand now much better.