ohms law formulas confusing me

bumpinexploder wrote: 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.

He got that by using solving for the current in the Ohm's law formula. sqrt( P * R) = I. Therefore the square root of 600 * 1 is 24.5A. Using Ohm's law again. P = I * E. Solving for E (the voltage) P/I = E, so 600 / 24.5 = 24.5V. In that case the current and voltage have the same value of 24.5 (different units though). One more thing, impedance is not the same as resistance. Impedance is going to be the opposition of current flow for AC current while resitance is the opposition of current flow for DC current. Impedance actually changes with frequency so it will not be the same as resitance. The resistance would only be 0.24 ohms if you used 60A for the current and 14.4V for the voltage. If either of those changes your resistance will change.

Only in this case. If you have an amp (obviously rated quite a bit higher than 600 watts) that has a 120A fuse in it, the input impedance COULD drop as low as .12 ohms. And yes, my .24 ohm statement is really the minimum impedance of the power supply input. Whiterob... Your last line IS partly right. My statement IS a drastic oversimplification, as the power supply on the inside is an impedance, it is not as simple as I made it seem to be. The MINIMUM IMPEDANCE of this example power supply can safely drop as low as .24 ohms. This is why the 60A fuse. That is all the devices in the power supply end of the amplifer can safely transfer/convert/switch. This input impedance number will also change, based on the output demands placed on the amplifier. The 60A fuse won't just blow as soon as you hook it up, will it? That MUST mean that at low (or NO) power, the impedance (resistance) is MUCH higher, right? As you request the amp to make more power, the current input goes up. The input voltage isn't changing, is it? Well, for current to go up, with the voltage remaining static, the resistance must go down. Ohm's Law, baby!

stevdart wrote: NIce tackle, haemph.

Thanks! I do try my best!

bumpinexploder wrote: 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.

Well, I don't know... What numbers are you using to figure those results? As your third grade teacher probably used to tell you: "Show your work..." I believe I ended up with 864, so at least ONE of your calculations were correct. The others? Ummm... Not so much.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."

well id show my work but now its pointless cuz its wrong lol. i was using the 60 for current not the 24.5 lol. now i understand about 90 % thank yall so much. i guess now i just need some practice problems. thats why i want to take a pracice mecp before i waist 65 bucks and not pass lolPioneer DEH-400UB

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