Inverter Current

It depends upon the inverter, but most are better than 90% efficient. So the input wattage generally (roughly) equals the output wattage. Current will depend on the relative voltages, so if battery voltage is less than RMS AC voltage, then the battery amperage will be greater than AC amperage. If battery voltage is higher than AC voltage, then battery current will be lower than the AC current.

The power you can get out cannot be less than the power you put in. If you lower the input voltage you will also increase the input current, likewise for the power out. An inverter generally doesn't care what the input voltage or current is, just that it more than is taken out of it (power= voltage * current).

For example: the input could be 100 amps at 1 volt or 1 amp at 100 volts (100 watts either way) just so long as it is greater than the power out, 90 volts at 1 amp or 90 amps at 1 volt (90 watts) assuming 90% efficiency of the inverter. You have to put in more than you take out and the design of the inverter is what determines what is put in and what is taken out and likewise the output power (p=I*E) must be less than the input power. This is true no matter whether it is a single or three phase input and/or single or three phase output. Power in >Power out.

Good Luck, Old Salt

The power you can get out cannot be less than the power you put in.
I think you meant "The power you can get out cannot be more than the power you put in."

K Fry-

Yes. Thank you for your alertness.

Good Luck, Old Salt

Strange that your answer, which was very good (but for the typo) was marked off topic. I've marked it back on topic.

K Fry-

Thank you for the complement, even with my typo. Fingers become spastic at later hours. They have a mind of their own. I appreciate the removal of the OT. Usually I can see why I get an OT, Experimental Experiments especially, but not on this one.

Your, and several other's, postings are the ones I admire and trust the most just as I am sure others do likewise.

Again, Thanks.

Good Luck, Old Salt

Dear Mr. sujinvipin,

The DC input Voltage and Current cannot be directly compared with output AC Voltage and Current and it is the POWER that can be compared where efficiency of the INVERTER comes in to picture.

For a given power if the Voltage is low, the Current will be high and viseversa. For AC, the Power Factor comes in to picture.

Thanks,

DHAYANANDHAN.S

Hi Sujin

Like Mr Dhan... stated, DC cant be related with AC....

In a single phase inverter you use Opamps to output frequencies from 2 pins and drive mosfet drivers to create AC wave.

The load to the motor will decide the amount of current the battery can discharge.

Like Mr Dhan... stated, DC cant be related with AC....

In fact, there is a relationship between DC and AC. The power is the same (if we exclude losses in the inverter). That means that if you know the power at the motor, you can calculate the power at the battery, and therefore its current, given its voltage.

The load to the motor will decide the amount of current the battery can discharge.

This leaves a lot of room for misinterpretation by the OP, and is possibly a source of the OP's confusion. Motor "load" is generally measured in amps, i.e. current. That is only one part of the story, because the motor is also fed at a particular voltage, so power can be relatively high or low, depending on the product of voltage and current. So a motor loaded to 50 amps will draw something other than 50 amps (higher or lower) from the battery, depending upon the relative voltages (at battery and motor). The only case in which the two currents are the same is if the voltages are the same (ignoring losses) or if they vary by a factor equal to the inverter efficiency (accounting for losses).

time the 2nd like Mr. dhayanandhan . . . said there is no 1:1/static/ non RMS correlation - trying to go faster my shadow - i'm studing this topic - seen various materials - but i don't fully comprehend the GO! yet . . . so anyway it goes like::

say you pull 3x60Hz from batteries = 180Hz(!) there should be a pre-charge of the primary coils withch is likely PWM-d sines (◄ that's the fuzzy part) - anyway the good inverter adjust to your load at the 2-ndary side (making it even more fuzzier) so only the T>60Hz^-1 will start likely to correspond your E conservation more correctly

besides there are the effective values for Ammeters ??? but if their digital then i donno what they show A.RMS or 1/√2·A or A.Max(/peak)

T>(60Hz)^-1>16ms (replaces the error above)

some practical results (R1 is DC (A) , R2 is AC (V) -- so it's the integral 1/¶ · 2 for full half-cycle)