Ammeter and Wattmeter in ac RLC circuit

[Deathfish]

What type of power rating does a wattmeter measure in an AC RLC circuit? Is the power measured the apparent power or active power, and is this reading the rms value?

Also which current value is taken by an ammeter in an AC rlc circuit? is this also rms value?

 

[es1]

I suppose it depends on the device. For example I know the digital one I use can give the instantaneous power or average power.

However on the traditional wattmeter wikipedia says:

http://en.wikipedia.org/wiki/Wattmeter
On an ac circuit the deflection is proportional to the average instantaneous product of voltage and current, thus measuring true power, and possibly (depending on load characteristics) showing a different reading to that obtained by simply multiplying the readings showing on a stand-alone voltmeter and a stand-alone ammeter in the same circuit.

There is also a large variety of ammeters out there:
http://en.wikipedia.org/wiki/Ammeter

 

[tiny-tim]

Hi Deathfish!

Is the power measured the apparent power or active power, and is this reading the rms value?

(btw, this isn't really my field, so someone who knows had better check this … )

The apparent power is the rather silly name for what you get if you're silly enough to multiply the readings of a voltmeter and ammeter, ie P_apparent = V_r.m.s.I_r.m.s..

The active power (or real power) is the same as the average power, and is measured by the wattmeter, P_active = P_average = V_r.m.s.I_r.m.s.cosφ.

I don't think there's any such thing as r.m.s. of power …

even for a zero-reactance (purely resistive) circuit, the (instantaneous) power is P = V_maxI_maxcos²ωt = V_maxI_max(1 + cos2ωt)/2 …

as a whole it's always positive, so r.m.s.-ing it seems a little silly

in parts, the first part is a constant, while the second part has an average of 0 …

so the first part is P_average (equal to P_active, which in this case equals P_apparent), V_maxI_max/2 = V_r.m.s.I_r.m.s.,

while the second part is s.h.m., and so does have a r.m.s., equal to half the apparent power (but so what? )

(for a reactive circuit, with phase factor cosφ, P = V_maxI_max(cosφ + cos2ωt)/2,

so in that case the first part is P_active = P_average = V_r.m.s.I_r.m.s.cosφ = P_apparentcosφ, and the second part is exactly the same zero-average s.h.m. as before)​