0
\$\begingroup\$

How does the Emporia Vue 2 work?

It uses a current measurement: magnetic core with coil around it (that’s my understanding). However, without knowing voltage waveform, active power can't be calculated (only apparent power).

For example, if voltage were to be 90 degree out of phase with the current, then active power would be zero (all is reactive power). But the device does not know the voltage waveform; only current.

Do I miss anything?

\$\endgroup\$
4
  • 1
    \$\begingroup\$ 180 phase would be generating rather than receiving power. 90 out of phase is all reactive. \$\endgroup\$
    – Neil_UK
    Nov 12, 2022 at 12:41
  • 1
    \$\begingroup\$ Only 1 voltage need to be known, guessing that others phases are the "same" values... if not, calculus should be a little more complicated. \$\endgroup\$
    – Antonio51
    Nov 12, 2022 at 13:42
  • 1
    \$\begingroup\$ Re, "...coil around electrical conductor..." FYI: The sense coil isn't wrapped around the conductor. It's wrapped around a magnetic core that the conductor passes through. Similar to drawings shown here except that the Vue2 looks like it uses split cores that can be snapped around the conductor instead of requiring the conductor to be disconnected, threaded through, and then re-connected. \$\endgroup\$ Nov 12, 2022 at 14:58
  • \$\begingroup\$ Of course, 90 degrees, fixed my post. Thanks for clarifying on the coil. Answer was given by @RoyC; also measure the phase voltage which is the same everywhere (to first order) \$\endgroup\$
    – divB
    Nov 13, 2022 at 19:40

1 Answer 1

2
\$\begingroup\$

The unit is connected to the phase voltages so will be measuring those voltages. Current is measured by the current transformers on individual loads but the voltage on each of these loads will be the same allowing the meter to calculate power.

\$\endgroup\$
1
  • \$\begingroup\$ Oh I didn’t think of that! Right... instead of immediately rectifying, could do A/D to get voltage waveform. Voltage waveform to first order is the same every where (low voltage drops) while the different phase shifts are in the current \$\endgroup\$
    – divB
    Nov 13, 2022 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.