HV to Lv conversion

Gents, first, thanks for the wealth of knowledge on other questions it is much appreciated. My new question is, I have a HV supply with readings of Watts, VAR and amps into my PLC, can anyone point to a formula that can help me calculate the LV side current.
THanks in advance, Steve D

ok

If you know the HV and LV and amps on the HV. HV/LV*amps on HV = amps on LV. If i understand the question

Steve,

Please tell me your a student and not a controls engineer at Three Mile Island....

Check out this site (great info): http://www.patchn.com/electron.htm

actually, I was wondering if I had to take into account the efficiency (VAR) when calculating my LVA. I can make a PLC take off its pants a waggle its thing but I can't remember 1st year apprenticeship stuff, such is the dumb questions of a specialist.

Steve,

The eff of the device used for voltage conversion (maybe a transformer?) may be negligable. If you explain the app a little more someone may be able to help you with that issue.

BTW, Bookmark the link I sent you. Great reference for thoes times when the brain shuts down.

Also..Join the club & register on the forum.

Best regards, Mike

VA to VA conversions are constant.

KVAR are the vertical components of the two power triangles.

KW are the horizontal components of the two power triangles.

The diagonal are the KVA, and the resultant of the other two vectors.

The efficiency of the transformer is a heat loss issue, and would be accounted for as KW in the HV side, but would still be present in the VA portion of both sides. You won't lose Watts without associated amps. KVAR will not affect the KW at all, except as it relates to additional VA current, which will cause more KW loss, but you will not be able to separate this component of the KW from the "load" component of the KW without specifically trying to isolate it with temperature measurements. Regardless, it will not be KVAR, but KW, anyway, if it causes heat. (Short form of the previous: Extra amps that cause heat are KW, not KVAR, no matter what caused the extra amps to flow).

The KVAR's are needed by the load for magnetization and charge (plus and minus vars), and have absolutely nothing to do with efficiency, but are integral to power factor. Power factor has absolutely nothing to do with efficiency, except as noted above/causing more amps.

If this doesn't help, perhaps you could specify your need for information a little more, and we'll try to help.

Don