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-   -   I need help getting down some of the basics. (http://www.diychatroom.com/f18/i-need-help-getting-down-some-basics-26029/)

MEd!c 08-30-2008 11:08 PM

I need help getting down some of the basics.
 
I'm trying to add some lights to my car and I'm having trouble with some of the basics. I will be making a line just for these lights. These lights use inverters which the website claims draw 5mAv. On google I found a link that was talking about those lights and someone said that it was a typing error and it should say it's 5mA. I'm bringing this up because I want to know if mAv actually exists, I dont want to trust something I read off a post in google. Anyways, lets assume it is 5mA, The circuit will be running off the car battery, and I think we all know car batteries provide more power than the small amount of power the inverters use. This is where my understanding comes to a limit.. When you turn on the switch does a large amount of amps @ 12volt start going through the wires, and woosh go through the inverter thus burning the inverter? Or does only 5mA go through the wire constantly because the resistor/appliance(inverter) only draws 5mA?

I guess what's causing this confusion is knowing how if you have some electronic that uses x amps @ y volts, and if you provide lets say x amps @ y+20 volts it will fry the thing. And I'm not sure if the same will happen if you provide too much ampere.

Putting that aside, I'm going to be putting a fuse on the circuit. Where would it be best to put the fuse? After or before the inverter?

TazinCR 08-31-2008 06:38 AM

12VDC will vary +/- 2v. When you hook it up to the battery it will not work off of the switch.It is 5mA Put the fuse before the inverter. As far a woosh with you get 12V the amps is dependent on the load. Example is the started draws more amps than anything on the car. How many watts is your inverter? Be careful because with the car not running your inverter can drain your battery very quickly. amps=watts/volts

BigJimmy 08-31-2008 01:55 PM

Quote:

Originally Posted by MEd!c (Post 153428)
I'm trying to add some lights to my car and I'm having trouble with some of the basics. I will be making a line just for these lights. These lights use inverters which the website claims draw 5mAv. On google I found a link that was talking about those lights and someone said that it was a typing error and it should say it's 5mA.

Draw refers to current which is expressed in A (amps). mA (milli-amp) is simply a thousandth of an amp. There is no such thing as a mAv.

Quote:

Originally Posted by MEd!c (Post 153428)
I'm bringing this up because I want to know if mAv actually exists, I dont want to trust something I read off a post in google.

That makes you smarter than most people. Some of the most insanely inaccurate information I've ever seen can be found in Yahoo questions.

Quote:

Originally Posted by MEd!c (Post 153428)
Anyways, lets assume it is 5mA, The circuit will be running off the car battery, and I think we all know car batteries provide more power than the small amount of power the inverters use.

Umm, not really. Car batteries are designed to provide a massive amount of current for a short period to start the engine. An inverter simply converts DC to AC. Their power ratings simply indicate the maximum output current that they can supply at the output voltage (P=I*V). In themselves, they are not sources of power.

Quote:

Originally Posted by MEd!c (Post 153428)
This is where my understanding comes to a limit.. When you turn on the switch does a large amount of amps @ 12volt start going through the wires, and woosh go through the inverter thus burning the inverter? Or does only 5mA go through the wire constantly because the resistor/appliance(inverter) only draws 5mA?

Ok, if I'm understanding correctly, the lights are sold with inverters? I'm not sure why a light would need an inverter in the first place. Even if they were LED's, those can operate off of AC or DC. Anyway, assuming that the inverter is provided with the light, then they should have been designed for the lamp load and you shouldn't need to worry about anything. Although the resistance of a lamp filament changes as it heats up, the inrush in your case would more than likely be negligible.

Quote:

Originally Posted by MEd!c (Post 153428)
I guess what's causing this confusion is knowing how if you have some electronic that uses x amps @ y volts, and if you provide lets say x amps @ y+20 volts it will fry the thing. And I'm not sure if the same will happen if you provide too much ampere.

In the first instance, you're mostly correct. Devices are usu. designed to operate at some nominal voltage. If you subject the device to a significantly higher voltage, that'll usu. let the smoke out. The source doesn't provide too much current (amperes). The current in the circuit is dictated by what the load is drawing. But, for a given load (which has some resistance), when you crank up the voltage, you increase the current. And that is what causes the sparks to fly.

Quote:

Originally Posted by MEd!c (Post 153428)
Putting that aside, I'm going to be putting a fuse on the circuit. Where would it be best to put the fuse? After or before the inverter?

Fuses are always placed immediately downstream from the source (i.e. as close to the positive battery terminal as possible). You can find automotive fuse holders at most auto parts stores as well as Radio Shack.

Jimmy


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