ohms law formulas confusing me

ok so i bought the mecp book and read it front to back and its pretty simple only the the formulas are kinda easy and hard if that makes sense? ill give an example and use my system as the variables if i want to find power? every formulas gives me a differant answer and i dont know which one to use p=i(squared) x R ok so my amp has 60 fuse rating. 60 squared is 3600. well times R which is 1 because its at 1 ohm now and i get 3600 watts??? its a 600 watt amp p=e x i ok so e is 14.4 and i is 60 once again. now i get 864??? p= e(squared) / R ok so e is 14.4 squared which is 207.36 and divide that by r, 1, i get 207 watts??? im so lost on this and will take any help i can get to understand these formulas. thanks Pioneer DEH-400UB

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Those formulas work on the output of the amplifier. You are trying to use them on the input of the amp. To accurately use either of the formulas, you will need an Oscilloscope, a signal generator and an impedance bridge. The impedance bridge will tell you what the impedance of the speaker is at that frequency. A 4 ohm speaker is not 4 ohms at most frequencies. The Oscilloscope is needed to determine where the amp is clipping and the output voltage. The signal generator is your source.

oh ok so how would i be able to tell what my amp really puts out then? impedance bridge gives me R right? oscilloscope gives me E? wow i thought i could just plug the specs in lol. thanks I AM. i am trying to get certtified to start my career and am having little trouble. now i get it a little better. is there anywhere i can take a practice test?Pioneer DEH-400UB

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I have no idea.

Since the input side is already known, as far as maximum input current, as well as input voltage, you don't have to plug anything in... It's a straight-forward calculation... You don't use the output side's LOAD in that formula, because the power supply is "in the way", as it were... Input 60A * 14.4V = 864W. This is how much power the amplifier could TRANSFER to a load, if it were 100% efficient, meaning zero loss. We all know they AREN'T, though, right? 864w in / 600w out = 70% efficient.   A reasonable number for a mono Class D amplifier, and one I would tend to believe without much question. Cheezy amplifier brand names will raise questions, so I'll just toss that out there right now! LOL Now, if the amplifier is rated 600W per CHANNEL, no way. 600W per channel, RMS means 1200W out. 1200 watts out, divided by the known battery voltage (14.4), divided by the same (hypothetical) efficieny number (70%) = 120A in. That basically addresses the input side. You just always have to remember, that the output side is invisible to, but dependent upon the input side... Meaning, if there is no power out, there is no (effectively) power in. The same applies the other direction, too. Just always remember to forget the specifiactions (and running load, too) of the output side when figuring power input demands. They are separate power delivery and demand circuits.It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

i may have to read that post couple more times to fully understand but thanks so much for the info. so how come that one power formula worked only? 864 at 70% efficieny is like 600 watts yea but how did i get 207 and 3600 on the others? if u answered it in the other post just tell me and ill try to fully understand it lol.Pioneer DEH-400UB

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I know you've seen the pie chart... https://www.the12volt.com/ohm/ohmslaw.asp When you are trying to FIGURE for the inside circle (the unknown) you use the quarter that is adjacent to it, then plug your numbers into the formula.

bumpinexploder wrote: ok so i bought the mecp book and read it front to back and its pretty simple only the the formulas are kinda easy and hard if that makes sense? ill give an example and use my system as the variables if i want to find power? every formulas gives me a differant answer and i dont know which one to use p=i(squared) x R ok so my amp has 60 fuse rating. 60 squared is 3600. well times R which is 1 because its at 1 ohm now and i get 3600 watts??? its a 600 watt amp

Your confusion with thias one is that you are figuring input power with an output load... remember what I said - separate and invisible to one another. The one ohm rating of your amp... What is it? Is THAT the 600 watt rating? If so, then you are using the wrong formula. 600 watts through 1 ohm to figure the current, and then the voltage, you use this formula (and I've already plugged the numbers in) 600 = 24.5^2 X 1 24.5A to figure the RMS voltage across that same 1 ohm load 600 = 24.5^2 X 1 24.5V 24.5V X 24.5A = 600W (One ohm loads are REALLY easy, because voltage and current will be the same results.)

bumpinexploder wrote: p=e x i ok so e is 14.4 and i is 60 once again. now i get 864??? p= e(squared) / R ok so e is 14.4 squared which is 207.36 and divide that by r, 1, i get 207 watts???

The first one worked, because you were using BATTERY voltage, which is NOT output voltage. That's your INPUT power. The second one DIDN'T work, because you were using battery voltage, which is NOT output voltage. You need to plug your OUTPUT voltage in there, which I have already figured for you, in the first portion of my reply.

bumpinexploder wrote: im so lost on this and will take any help i can get to understand these formulas. thanks

Just ask... We'll get you through this! Basically, when figuring input numbers, you ignore the woofer's impedance, because the impedance you need to figure is COMPLETELY unknown... It's the input impedance of the power supply of the amplifier, that's the R you are trying to calculate for. (For your 600W @ 1 ohm amp, by the way, that input resistance is .24 ohms) When figuring output output numbers, you ignore everything you know about the INPUT side (14.4V at 60A - those DON'T apply anymore), and pull out the trusty RMS voltmeter and SINE WAVE generator - you cannot use music, and GET the voltage, so you can plug that number in to the formula. Once you have two of the numbers (power, resistance, current, voltage) in ANY formula, you can figure ALL of the others.It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

NIce tackle, haemph.Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.

wow what an explanation. thanks a ton!! ok so R is always going to be .24 or just in this case? and i need to buy AND LEARN how to use the vom and dmm. 600=24.5^2 x 1 24.5A <<<<<how did u get 24.5 as I or current? 600=24.5^2 x 1 24.5V <<<<< same formula but u just changed the A to a V? i am starting to understand but i thought i would comprehend this a little easier. thanks alot for helping and if u get tired of my questions just reccomend another book or something. wont hurt my feelings lol.Pioneer DEH-400UB

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double post lol am i supposed to get differant answers on each formula in the SAME catagory like power, resistance ect...??? my power formulas are better now than 207 and 3600. with the plug in formula on the website here i got, in order, 864,720, and 600 watts. at least im a little closer.Pioneer DEH-400UB

Kicker 150.2 mids/highs amp

MB Quart fsa 216 front stage

AudioPipe gd-6001 mono amp

AUDIOQUE SD 15S PREFAB BOX!

C&D Technologies Battery, 1500 CCA, 134 amp storage

149.5 DB's