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I have a heater running at 240VAC. I'm trying to measure the power by datalogging the voltage and current then multiply both inorder to get the power reading.

I know that this method works well with DC but is there a more precise method of measuring AC power other than this or this is acceptable?

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  • \$\begingroup\$ Thanks for your feedback! \$\endgroup\$
    – Ybx Nop
    Jul 5, 2023 at 12:18

5 Answers 5

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Much simpler: Buy a Kill-a-Watt and plug the heater into it.

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    \$\begingroup\$ Thanks for your feedback! \$\endgroup\$
    – Ybx Nop
    Jul 6, 2023 at 17:40
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A heater is resistive, so Ohm's law applies on a moment by moment basis. Even logging the voltage is superfluous – the current and the voltage are inherently linked proportionally (by the resistance, which is not constant as it changes with temperature).

So, yes, your logging works of current and voltage at each instant works, both for DC and AC, assuming you log fast enough. It's even more than you need to do.

Seeing your hipot measurement questions: Ohm's law, and how it relays to powers, is really among the very first things you've learned about electricity. Maybe go back and refresh that!

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is there a more precise method of measuring AC power other than this or this is acceptable?

This method is acceptable provided that the measurements for voltage and current are root-mean-square (rms).

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  • \$\begingroup\$ Thanks for your feedback! \$\endgroup\$
    – Ybx Nop
    Jul 6, 2023 at 17:41
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Your solution is fine because you dont havea shift between voltage and current (heater is only resistor load) therefore you can use this method to reliably measure the power of your system. The only inconvenience is the added shunt resisotr that adds up additional power dissipation to your system.

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  • \$\begingroup\$ Thanks for your feedback! \$\endgroup\$
    – Ybx Nop
    Jul 5, 2023 at 12:18
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datalogging the voltage and current

Many heating elements are not simple resistors, and have a significant temperature coefficient. If you know the resistance of the heating element when it is at its operating temperature, then you need only the RMS voltage to calculate the power.

Back to your method, one way to reduce the amount of logging and computation is to use true-RMS circuits or devices to measure the voltage and current before multiplying. This also reduces the effects of line noise on your data.

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  • \$\begingroup\$ Thanks for your feedback! \$\endgroup\$
    – Ybx Nop
    Jul 5, 2023 at 12:18

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