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Topic: working of Inverter (Read 883 times) previous topic - next topic

tjones9163

Hello, I am once again trying to wrap my head around inverters.

Let's say I have an appliance that uses 1000 watts and its maximum current is 10amps.
I also have a 2000w inverter and a 12v battery.
Is the math correct to say 1000w/12v = 83.3A

My question is,
Is it the battery that must be able to output 83.3 amps to the inverter, and it is the inverter job to put out a regulated 110v at 10amps(required by my appliance), so it is not my appliance that needs to handle the 83 amps, but instead my battery to the inverter?

I'm pretty sure this is correct but I just wanted to make sure.
P.S I realize that I didn't include the type of battery or amp-hours, more of just a broad understanding.
Thanks

ReverseEMF

Yup!  That's correct! 
"It's a big galaxy, Mr. Scott"

Please DON'T PM me regarding what should be part of the Public Conversation -- Let it ALL hang out!!
Unless, of course, it's to notify me of a mistake.

tjones9163


ReverseEMF

And, to elaborate: The battery current will be even a bit more, because the efficiency will never be 100%. 

Power out = Efficiency * Power In

So:
Pout = Eout * Iout
and
Pin = Iin * Ein
and e = % Efficiency
Pout = e/100 * Pin
Pout = e/100 * Ein * Iin
Iin = (100 * Pout) / (e * Ein)

so, assume an efficiency of 85%

Iin = (100 * 1000) / (85 * 12) = 98Amps!

"It's a big galaxy, Mr. Scott"

Please DON'T PM me regarding what should be part of the Public Conversation -- Let it ALL hang out!!
Unless, of course, it's to notify me of a mistake.

MarkT

and it is the inverter job to put out a regulated 110v at 10amps(required by my appliance)
The inverter puts out a regulated 110Vac.  The load takes the current it needs, the inverter has no say
in this other than when overloaded.

The inverter is a voltage source.  If you control voltage, you cannot also control current.
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

Arley69

Depends on the load on the inverter, these are rarely run at full power. Say at a load of 1000 watts, the current will be 1000 /12=83 amperes. 400amp hours /83amp =4.8 hours. Multiply by 0. 8 to allow for efficiency and the time is about 3.8 hours. It will be less than that however because the battery voltage will drop as it discharges causing more than 83 amps of current to flow.
inverterreview.com

TomGeorge

Hi,
What is the appliance, if it is inductive then there will need to be a higher starting power.

Tom... :)
Everything runs on smoke, let the smoke out, it stops running....

MarkT

so, assume an efficiency of 85%


High power inverters are a lot more efficient that this, to keep the costs down.  Heat dissipation costs
real money and efficiency is easier to achieve at scale, typical 2kW inverters are more like 95% and up.
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

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