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When estimating power consumption with a current sensor, what is the usual precision that can be achieved for home appliances? I'd like to make a simple circuit using such a sensor and a microcontroller to collect the measurements and estimate consumed power.

For an ideal resistor, we can compute power as I²R. But for an ideal capacitor or inductor, this would be very different. For example, an ideal inductor/capacitor connected to an AC would have 0 average power regardless of current, so I²R over-estimates the actual value.

How much can the above ideal resistor estimate be off for devices that contain inductors, such as motors? And can it ever under-estimate?

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    \$\begingroup\$ For example for AC, the average power would be 0 regardless of current Hmm then why does my panel heater on AC still get hot and I still get charged for the power? Modern (electronic) power meters can monitor both current and voltage at the same time and do the multiplication (to get power) in software. So even a bad power factor can be taken into account. How much precision you get depends on how much money you want to spend. \$\endgroup\$ Nov 23 '17 at 14:35
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    \$\begingroup\$ Before you worry about precision, concentrate on power.... \$\endgroup\$
    – Solar Mike
    Nov 23 '17 at 15:36
  • \$\begingroup\$ @Bimpelrekkie Because your heater isn't an ideal capacitor or inductor. Maybe I didn't make it clear enough that I meant 0 average power for a capacitor/inductor. \$\endgroup\$
    – Petr
    Nov 23 '17 at 18:08
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To know power delivered to a load, you need current and voltage.

If your load has a unity power factor, for example a pure resistance, then you can assume the current is in phase with the voltage, and work from the current sensor output, and an estimate for what the RMS supply voltage is. If you have a nominal 230v supply, and you use that figure instead of the actual measured one, you could be up to 6% out, more in countries with a poor mains supply.

If your load power factor is not unity, then you must measure the instantaneous voltage as well as the instantaneous current. Multiplying the RMS values for voltage and current give you the VA of the load, which in the case of the resistor will be the same as the power. Multiplying the instantaneous values of voltage and current, and integrating them, will give you the true power, which is always equal to or less than the VA, never more.

As with a pure reactive load your true power will be zero, and the VA finite, the ideal resistor estimate will be 100% wrong.

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