My interest in in electricity is purely hobby/handyman so if this question is not appropriate here please disregard. The forum rules didn't seem to exclued totally amateur questions.
I am looking at an inverter which runs off the cig lighter in car. It is a standard US inverter and takes in the car's 12V dc and outputs 120VC AC and is rated for a maximum 1200 watts.
So I think thats alot of power. That could run my wife's mega hairdryer. So I start thinking about the numbers and wonder what kind of draw this will have on the car's electrical system.
So if the hairdryer is using 1200W from 120V AC the current the hairdrayer is pulling is 10 amps. Right?
I = W/E
I = 1200/120
I = 10 Amps.
Maybe this is wrong and the rest of this post is pointless.
If not, this is what is happening on the output side of the inverter. I want to know about the input side. If Power = Power, then, is this correct?
Idc = (Iac * Vac)/Vdc
Idc = (10 * 120)/12
Idc = 100 amps DC
So, that seems like a lot. The alternator in the car only puts out about 60 amps. Even if the car was running I would be draining the battery.
I can't think of any time I have change an automotive fuse where the fuse was rated larger than 30, maybe 40 amp. I checked the fuse on the circut with the cig lighter and it is a 20 amp fuse.
Won't this fuse just blow if I put any kind of significant draw on the AC side of the inverter? Is my math or understanding of the inverter totally incorrect? Curious minds want to know...
I am looking at an inverter which runs off the cig lighter in car. It is a standard US inverter and takes in the car's 12V dc and outputs 120VC AC and is rated for a maximum 1200 watts.
So I think thats alot of power. That could run my wife's mega hairdryer. So I start thinking about the numbers and wonder what kind of draw this will have on the car's electrical system.
So if the hairdryer is using 1200W from 120V AC the current the hairdrayer is pulling is 10 amps. Right?
I = W/E
I = 1200/120
I = 10 Amps.
Maybe this is wrong and the rest of this post is pointless.
If not, this is what is happening on the output side of the inverter. I want to know about the input side. If Power = Power, then, is this correct?
Idc = (Iac * Vac)/Vdc
Idc = (10 * 120)/12
Idc = 100 amps DC
So, that seems like a lot. The alternator in the car only puts out about 60 amps. Even if the car was running I would be draining the battery.
I can't think of any time I have change an automotive fuse where the fuse was rated larger than 30, maybe 40 amp. I checked the fuse on the circut with the cig lighter and it is a 20 amp fuse.
Won't this fuse just blow if I put any kind of significant draw on the AC side of the inverter? Is my math or understanding of the inverter totally incorrect? Curious minds want to know...
