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Ohm's Law

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Old 09-24-2006, 03:20 AM
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Default Ohm's Law

I found this and figured id post it here in case anyone was interested

Ohm's Law defines the relationships between (P) power, (E) voltage, (I) current, and (R) resistance. One ohm is the resistance value through which one volt will maintain a current of one ampere.

( I ) Current is what flows on a wire or conductor like water flowing down a river. Current flows from points of high voltage to points of low voltage on the surface of a conductor. Current is measured in (A) amperes or amps.

( E ) Voltage is the difference in electrical potential between two points in a circuit. It's the push or pressure behind current flow through a circuit, and is measured in (V) volts.

( R ) Resistance determines how much current will flow through a component. Resistors are used to control voltage and current levels. A very high resistance allows a small amount of current to flow. A very low resistance allows a large amount of current to flow. Resistance is measured in ohms.

( P ) Power is the amount of current times the voltage level at a given point measured in wattage or watts

Thank you the12volt.com

here's the link, its got all kinds of stuff on there including a calculator
http://www.the12volt.com/ohm/ohmslaw.asp
 
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Old 09-24-2006, 03:40 AM
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Default RE: Ohm's Law

VERY USEFUL

Nice post jimmy u get a sticky
 
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Old 09-24-2006, 03:43 AM
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Default RE: Ohm's Law

yay i feel special lol[sm=happy046.gif]
 
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Old 10-24-2006, 12:51 AM
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Default RE: Ohm's Law

The name of this thread is "Ohm's Law" but you forgot to actually state the law

V = IR

I = V/R

R = V/I

Voltage through a load multiplied by the current through a load will yeild Power [in Watts].

P = I*V

I'm an electrical engineer by degree and profession, and an audio/video guru by experience. I worked in 12Volt electronics for along time, and this sometimes proved usefull, but continuity meters proved more usefull in the land of 12V.
 
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Old 07-08-2008, 12:29 AM
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Default RE: Ohm's Law

Voltage through a load multiplied by the current through a load will yeild Power [in Watts].

thats correct. but if u dont mutyply the number by 0.707 it wont give u the true RMS. it will give u the max power
please repost if im wrong. thx

 
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Old 12-04-2008, 10:10 AM
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Default RE: Ohm's Law

ORIGINAL: splblazer

Voltage through a load multiplied by the current through a load will yeild Power [in Watts].

thats correct. but if u dont mutyply the number by 0.707 it wont give u the true RMS. it will give u the max power
please repost if im wrong. thx

You're talking AC, not DC. 12V vehicle electrical systems are primarily DC. The .707 factor is the root-mean-square "average" of a symmetrical AC sine wave.

(old post I know...but I just signed up recently, was browsing around and stumbled across this).

Also being a EE...I wrote up a little article on electronics that someone may find handy (or may not, but hey is there a such thing as too much information! )

http://misterbill.homeip.net/misc/ba...lectronics.PDF
 
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Old 12-29-2009, 01:23 AM
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Default Ohms Law

Ok so Im looking at a rockford amp that is one channel. It has 1000rms at 1 ohm, 750rms at 2 ohm and 500rms at 4ohm. If I want 1000rms at 1 ohm what kind of rockford sub will I be looking for? All I see are specs for 2 and 4 ohms. If I run the amp at 1 ohm does that mean that Ill get 1000rms out of a sub that has 1000 watts max?

THANKS
 
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Old 01-25-2010, 11:39 AM
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Default

Originally Posted by coivabs09
Ok so Im looking at a rockford amp that is one channel. It has 1000rms at 1 ohm, 750rms at 2 ohm and 500rms at 4ohm. If I want 1000rms at 1 ohm what kind of rockford sub will I be looking for? All I see are specs for 2 and 4 ohms. If I run the amp at 1 ohm does that mean that Ill get 1000rms out of a sub that has 1000 watts max?

THANKS
what you need to figure out is if the sub you want to get will take 1kw. If it doesnt then you need to either change the impedance of the woofer or get a smaller amp. If you get a dual 2 ohm sub it will give you a final load of either 1 or 4 ohms. If you get a dual 4 ohm sub it will give you a final load of 2 or 8 ohms. It all depends on what you want to do
 
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