Delta amperage

Status
Not open for further replies.
Seems like a lot of excess if all you want is line current for resistive phase loads. Iceworm provided the basic formula earlier but didn't simplify and put in electrical nomenclature...

IA = sqrt(IAB?+IAC?+IABIAC)
IB = sqrt(IAB?+IBC?+IABIBC)
IC = sqrt(IBC?+IAC?+IBCIAC)
 
Smart $ said:
Seems like a lot of excess if all you want is line current for resistive phase loads. Iceworm provided the basic formula earlier but didn't simplify and put in electrical nomenclature...

IA = sqrt(IAB?+IAC?+IABIAC)
IB = sqrt(IAB?+IBC?+IABIBC)
IC = sqrt(IBC?+IAC?+IBCIAC)
Click to expand...


Thank you!!!:):thumbsup:

I tried it out and Im getting currents exactly as the spread sheet. Is this the exact value or an approximation for pure resistive loads? If the equation gives currents at least within a few amps of the actual value its exactly what I was looking for:D Thanks again!:happyyes:


I will try and brush up on Iceworms vector methods if I ever need to consider PF as well. (I probably will now after saying this:lol:) Fortunately its electric resistance heaters for now.


If anyone is curious about the spread sheet I found it in a previous thread. From the equations you provided the spread sheet seems right on the money. :cool: Post 6 btw: http://forums.mikeholt.com/showthread.php?t=133354&page=9
 
mbrooke said:
Thank you!!!:):thumbsup:

I tried it out and Im getting currents exactly as the spread sheet. Is this the exact value or an approximation for pure resistive loads? If the equation gives currents at least within a few amps of the actual value its exactly what I was looking for:D Thanks again!:happyyes:


I will try and brush up on Iceworms vector methods if I ever need to consider PF as well. (I probably will now after saying this:lol:) Fortunately its electric resistance heaters for now.


If anyone is curious about the spread sheet I found it in a previous thread. From the equations you provided the spread sheet seems right on the money. :cool: Post 6 btw: http://forums.mikeholt.com/showthread.php?t=133354&page=9
Click to expand...
To say the formulas give an exact value would only be true for ideal circuits. In real-world circuits there are several to many considerations which result in deviated values. However, calculated values should be within ?1%. ;)
 
Smart $ said:
To say the formulas give an exact value would only be true for ideal circuits. In real-world circuits there are several to many considerations which result in deviated values. However, calculated values should be within ?1%. ;)
Click to expand...

Ok that's good to know! :) Voltage drop is one that I can think of, but a + or - 1% deviation is still accurate enough in my case.


Thanks again!
 
Last edited:
mbrooke said:
Ok that's good to know! :) Voltage drop is one that I can think of, but a + or - 1% deviation is still accurate enough in my case.


Thanks again!
Click to expand...
BTW.... I forgot to mention the ?1% is using measured values. If you are using calculated values, such as listed kW of heater divided by nominal voltage, the result could easily be up to ?10%... as actual heater current can vary that much from calculated.
 
Smart $ said:
BTW.... I forgot to mention the ?1% is using measured values. If you are using calculated values, such as listed kW of heater divided by nominal voltage, the result could easily be up to ?10%... as actual heater current can vary that much from calculated.
Click to expand...


You bring up a good point. I have seen that over time. Even when the line voltage matches the current draw sometimes differs. I will keep that in mind. Thanks for reminding me:)
 
Status
Not open for further replies.
Top