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If power is current times meter shouldn't the overall resistances and inductances and capacitances of my house appliances be known so the power meter could calculate the power. I look at my house as a closed circuit. some people say the amp meter measures the current that is used by the appliances but according to ohm's law I need to know the impedances to measure the current.The voltage is standard 120 rms volts.

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The impedance is just the voltage phasor divided by the current phasor. If you measure both (as energy meters do) then the impedance could be known (it's a complex number in general). The impedance of a pure inductance or pure capacitance load will be imaginary (no real part).

However there is no need to calculate the impedance. The energy (what you pay for) is just the time integral of the instantaneous product of current and voltage. Since the meter measures both current and voltage, that is all that is needed.

In the case of old-school energy meters, it's done cleverly with a motor-like arrangement using eddy currents. Modern electronic energy meters do the calculations digitally after digitizing the current measurement (from a shunt, current transformer or Rogowski coil) and the voltage measurement (from a voltage divider or potential transformer).

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  • \$\begingroup\$ Pfehany So after measuring the instantaneous power, how would they get rid of the reactive power and only charge us for the real part of the apparent power? \$\endgroup\$ – Jack Feb 9 '16 at 4:00
  • \$\begingroup\$ @MaryE The instantanous power is the real power- so it just needs to be integrated to get the total energy per billing period. If you don't measure instantaneous power- measure voltage and current separately and then multiply- then you get apparent power. It's even possible for the instantaneous power to work out negative after you integrate it- if power is being fed back into the grid. Obviously that can never happen with apparent power. \$\endgroup\$ – Spehro Pefhany Feb 9 '16 at 4:10
  • \$\begingroup\$ According to my book the instantanoues power of a sinusoid consist of both reactive and real power \$\endgroup\$ – Jack Feb 9 '16 at 19:45
  • \$\begingroup\$ @MaryE The reactive portion of the power is zero when integrated over an AC cycle (it draws energy from the source, then gives it back again). <Captain and Tennille flashback>. Capacitors take power at the beginning of each half-cycle and return it at the end, inductors do the opposite. \$\endgroup\$ – Spehro Pefhany Feb 9 '16 at 20:03
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    \$\begingroup\$ That's one way of looking at it- but the units are a bit wonky. Power is measured in kW. You pay for kWh (energy), so the average power you used over a month (kW) multiplied by the number of hours in a month (720 in a 30-day month). If you got charged for 150kWh, you used an average of 208W, measured over the month. \$\endgroup\$ – Spehro Pefhany Feb 9 '16 at 21:00
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The power meter measures voltage and current to the appliance. Those two values are multiplied by means of physic phenomena on rotating disc that is spinning with the rate of u * i * cos(phi). If you measure both current and voltage, then the impedance doesn't to be known, because the meter indirectly "senses" the change of impedance.

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