Firstly, I'm not really an electrician or an electrical guy with lots of knowledge about electronics but I'm always interested learning to it.

Yesterday, I asked my wife to help me out determining the wattage (electrical power) of an electrical device such as TV, Reff, PC, Laptop, Tablet, etc, etc anything around the house that uses electricity and I plotted each one of them and log at least 3 wattage and average it.

I have this:

enter image description here

that's 95% of our electrical devices here around the house. I can play with values and see how much it costs if this device is turned on for this hours. But doing it is pretty tough job because we had to turn off everything and turn on and off each device to check its wattage.

But I wonder if an multimeter can do this? Can I check the wattage of an electrical device such as TV, Laptop , etc even if they are plugged in?

I only have this cheap small digital multimeter I carry around the house. enter image description here

Here in Philippines, meters are exposed and some are digital, like ours. That's the way I check the wattage of the device - sort of like this one:

enter image description here

  • \$\begingroup\$ You should check this out: walmart.com/ip/… Although, to use a Kill-A-Watt, you will have to unplug the device and plug it back in through the meter. This website also has a good estimation calculator for checking your power usage: michaelbluejay.com/electricity/howmuch.html \$\endgroup\$ May 8, 2013 at 1:21
  • \$\begingroup\$ The UT20B is a rebadged Uni-T $5 meter. I'd not use it on 220 V outlets. Despite it's Cat II 300V rating it is unlikely to have adequate protection. \$\endgroup\$ May 8, 2013 at 19:43
  • \$\begingroup\$ What the heck is "current wattage". Which one is it, current or power? \$\endgroup\$ Jul 7, 2013 at 12:11
  • 2
    \$\begingroup\$ @OlinLathrop Asking a non-native speaker "what the heck" he's talking about is less helpful than just editing the question towards an obviously correct use of physical terms. He's clearly not talking about current (in amps) but power (in watts), sometimes also referred to as "wattage". \$\endgroup\$
    – zebonaut
    Jul 7, 2013 at 14:36
  • \$\begingroup\$ @zebon: Editing is only possible when it's clear what it being meant. \$\endgroup\$ Jul 7, 2013 at 16:18

3 Answers 3


A multimeter can be used to measure amps (multiply by volts to get volt-amps, which is typically not far off watts). But most multimeters typically lack the range to be useful for things like appliances.

Instead, a product like the "Kill A Watt" meter is particularly well-suited for what you're trying to accomplish.

  • \$\begingroup\$ Hmm... "Kill A Watt" really echoed around the internet. Lot's of forums tell about this device. My concern is, how do I check lightings and wires hidden in the ceiling? Maybe I can hook up some wires on that device. Wasnt watts has formula like P = V * A? \$\endgroup\$ May 8, 2013 at 2:33
  • \$\begingroup\$ @JaysonRagasa yes, given than P = power, V = voltage, A = current and that your measurements are RMS and in phase. \$\endgroup\$
    – Gunnish
    May 8, 2013 at 11:29
  • \$\begingroup\$ @JaysonRagasa The "Kill A Watt" device is popular because it's simple and does exactly what you are asking. You can get electrician-oriented meters that will clip around a mains wire and measure amps and doesn't require you to cut the wires, but that's less consumer-friendly. Watts = V * A for purely resistive loads (e.g. heaters), but inductive loads (e.g. motors) complicate things. See en.wikipedia.org/wiki/Power_factor \$\endgroup\$
    – tylerl
    May 8, 2013 at 18:30

Power is an RMS voltage measurement multiplied by the current. You can get current clamp meters to safely measure the current, and then having measured the voltage you could perform the calculation manually. There is a good open source project here that explores many options of cheap power measurement.

  • 4
    \$\begingroup\$ NO! Power is NOT just the RMS voltage times the RMS current. Power is the integral of the voltage times the current divided by the integration time. The difference becomes important when there is a phase shift between voltage and current. Or put another way, real power is voltage times current times the power factor. Your statement completely ignores the power factor. Take a capacitor as example. It will draw some finite current, but the average power into a ideal capacitor is 0. \$\endgroup\$ May 8, 2013 at 12:37
  • \$\begingroup\$ Olin's right... but for the purposes of this project, a $20 AC clamp multimeter and a rough idea of your mains voltage RMS is close enough. Or a kill-a-watt or similar, which is likely to be about as accurate. \$\endgroup\$
    – John U
    May 8, 2013 at 18:27

There is systems that allows you to continuously monitor the powerusage around your house, this is a bigger investment but is really cool.

There is project called OpenEnergyMonitor that tries to this with Open Source and Open Hardware, so this project can give you some ideas.



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