Do a/c clamp meter's measure true,apparent or total current?

Question

I am using a capacitive bank to correct the power factor of my house, now using an a/c clamp meter i can see that the current required has gone down... will the electric bill go down by the same amount? (this will be the case if the meter measures total current... but if the meter measures true current it may not be so.

Answer 1

Short answer: I don't expect your electric bill to go down. Explanation below.

The utility's electric meter on the side of your house measures power not current. Because AC has both magnitude and phase, you have to multiple vectors to get the right answer for inductive motor (fans, vacuum, tools, etc). Many such motors have just about the same amps under no load as under high load, but the phase angle of the amps shifts such that V x A = much more power under load.

Since the electric utility 1) provides you with a great power meter, and 2) it reads directly what you're paying for; use it to check if your power consumption is less. Turn off all other circuits and run your equipment with and without the capacitive bank. You don't need to run it long enough enough to spin the little numbered dials. You can just time the mark on the serated wheel that spins around. Each spin of that wheel is a constant fraction of a kWh.

But big picture - if the motor is doing the same work, it is using the same power, give or take the heating up (inefficiencies) of the motor and other circuits downstream of the meter.

For light bulbs and electric heaters, simply use RMS volts times RMS amps. Just what your ammeter and voltmeters read.

RMS = root mean square. Ideally, a meter continuously squares the volts or amps, averages those squares and displays the square root of the average. That makes everything equate back to the scalar quantities we can think of more easily. And equates exactly to DC volts and amps (for resistive loads). A meter will clearly state "true RMS" if it does that. Meters under $100 don't. They use a conversion from peak volts to RMS volts. It works for sine waves, but is a little off for other waveforms (like square waves or modified sine waves from invertors).

Hope that helps.

Answer 2

If the current amount goes does, then it will reduce the wattage used, so yes, you should have a lower bill.

Answer 3

A coil on the middle is pushed with a sq. wave AC voltage. The time for the sq. wave to saturate the middle relies upon on the significance of the DC modern-day in the conductor it is being measured.