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In a wattmeter, the current coil helps in measuring current and the potential coil is used for measuring voltage... but the real power that the wattmeter measures is \$V \times I \times \cos(\phi)\$.

How is the power factor involved in the measurement made by a wattmeter?

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This kind of wattmeter does not directly work from the power factor. It averages the instantaneous voltage times current. The force on the needle is proportional to the voltage times the current, which is the instantaneous power. A spring causes the needle to deflect linearly with applied force. The inertia of the needle averages the instantaneous product of the current and the voltage, so shows averge real power.

If the voltage and current are out of phase with each other by 90° (power factor = 0), then the power would be positive for half the cycle and negative for half the cycle. If the meter is intended for normal power line frequencies, then the mechanical mechanism will not respond to the individual 50 or 60 Hz half-cycles but show you the low pass filtered power with a significantly longer time constant than individual power cycles. At a power factor of 0, this average will be 0, and the meter will show 0.

If you separately measure the RMS voltage and current then multiply them, you get the VA value (magnitude of power including both real and imaginary parts). You can take the real power shown by the meter and divide it by the VA value to get the power factor.

So again, this meter doesn't directly deal with power factor. It measures real power going back to first principles and whether the voltage and current signals are sines or not.

By the way, the electric meter on your house works on the same basic principle. Instead of deflecting a needle against a spring, the force between the voltage and current coils cause a small motor to spin. The total number of rotations is the time-integral of the power, or the total energy delivered. A series of gears rotate various labeled dials so that this integral can be accumulated and displayed so the meter guy can read the total every month. Newer meters send the accumlated reading to the power company automatically via various communication means.

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But the real power that the wattmeter measures is V×I×cos(ϕ)

Not true.

If you were measuring DC power you could take an instantaneous measurement of voltage and current and get power. You could do this several times in succession to get a better average (should the load and hence current be cyclic).

The wattmeter performs "multiplication" (and averaging) in the magnetic fields and the inertia of the mechanics. It doesn't measure RMS values and compute power factor - it just multiplies and averages (and of course this type of wattmeter works on DC)

Where is power factor involved here? It isn't because if the DC voltage is stable and the current is a sinusoidal waveform impressed on a dc level there is no phase relationship at all between volts and amps.

So, if the voltage and current are sinusoidal, does the wattmeter magically take power factor into account? No it doesn't - it continues to do the math and multiply amps by volts.

Power factor is a convenient way of describing the phase angle between voltage and current in an ac circuit. It's an OK method when loads are simple but when loads are more complex and the current waveform is no-longer sinusoidal, power factor is much harder to determine.

And how does the old analogue wattmeter cope with harmonic distortions on the current waveform? It does just fine because power is volts x amps averaged - the harmonics in the current waveform (which make a nonsence of \$V.I. cos(\phi)\$) are just ignored because when Sin(A) is mulitplied by Sin(B) the average level is zero.

This is why the electrickery supply companies don't like harmonic distortions - it's current that they have to supply that doesn't contribute to power used just like when standard loads have a small power factor - harmonics have a "zero power factor" in effect..

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  • \$\begingroup\$ As he mention "current coil and voltage coil" so it means he is talking about the analog wattmeter. so is it true in case of analog wattmeter also?? \$\endgroup\$ – Atom Sep 13 '13 at 12:13
  • \$\begingroup\$ @SHASWAT an analogue wattmeter performs "multiplication" (and averaging) in the magnetic fields. It doesn't measure RMS values and compute power factor - it just multiplies and averages (and of course an analogue wattmeter works on DC) \$\endgroup\$ – Andy aka Sep 13 '13 at 12:24
  • \$\begingroup\$ Thanx for clearing my doubts...... \$\endgroup\$ – Atom Sep 13 '13 at 12:25
  • \$\begingroup\$ @Andyaka so for AC, what you are saying is that the Wattmeter doesn't calculate the power factor. So how would it differentiate from a V an I when they are for example 30 degrees out of phase or 80 degrees? \$\endgroup\$ – Jack Feb 28 '16 at 3:23
  • \$\begingroup\$ Look at the diagrams of waveforms in this answer (also provided by me): electronics.stackexchange.com/questions/214879/… \$\endgroup\$ – Andy aka Feb 28 '16 at 10:00

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