I need formulas explaining Voltage and Amperage

[NitroX]

You have it upsidedown. That's124% efficiency, not 80%; impossible.

You had it flipped man lol

559.5/746= 75% efficient

 

[NitroX]

The point is convention and consistency to make things easier to communicate and understand. You can do things the opposite way of everyone else when you calculate efficiency, and you can use units no one else uses if you really want. As long as you are consistent you'll be ok. But you are more likely to get confused when you run into anything new and other people will have difficultly understanding you. Asking a person if a 1KW per HP motor is a good motor will get you nothing but blank stares and funny looks.

I'm not sure if this is an insult, I do my math differently. If i confuse people, then its my fault, so be it. It takes a person to understand others before making quick and irrational decisions. i made this unit up your correct (maybe?) Because it is easy to calculate my way through tough problems.

 

[russ_watters]

You had it flipped man lol

559.5/746= 75% efficient

Again, that's upside-down. 559.5 is the input, 746 is the output. Efficiency is output/input, not input/output.

 

[russ_watters]

I'm not sure if this is an insult...

It isn't meant as an insult: we are trying to help you.

...I do my math differently. If i confuse people, then its my fault, so be it. It takes a person to understand others before making quick and irrational decisions. i made this unit up your correct (maybe?) Because it is easy to calculate my way through tough problems.

Therein lies the problem: you are here asking for help and we are trying to help you. If you insist on doing things in non-standard ways, we won't be able to be as helpful. And that's even if we assume that your way is equally valid, which it probably isn't.

 

[NitroX]

sorry russ, i felt like I am arguing against MrSparkle, i never meant it on you or anything else... I am glad for all the help, I am just new, and i don't socialize well which I appologize for behaving like a child...

 

[russ_watters]

sorry russ, i felt like I am arguing against MrSparkle, i never meant it on you or anything else... I am glad for all the help, I am just new, and i don't socialize well which I appologize for behaving like a child...

It's ok - you've been upfront about that, so it's fine.

 

[NitroX]

so confused, what do you mean by upfront?

 

[russ_watters]

so confused, what do you mean by upfront?

You have been honest and we appreciate it.

 

[psparky]

Dear Nitro, I scribbled up this 4 page word document a couple years ago.

Give it a quick read...it's electrical terms put in Layman's terms.

Maybe it will help. Maybe it wont. See attached.

 

[NitroX]

do you have an older version of the document, you can do a save as on your document add .rtf at the end and i can look at it? thank u!

 

[NitroX]

If you want to teach yourself, you are going about it the wrong way. Grab a textbook, read it, and understand the concepts behind the formulas. Then you'll understand how to apply the formulas. If you have questions, then you come and ask. Then work through the problems so you know you understand both the concepts and the formulas and the math. We could give you all the formulas in the world, but that wouldn't stop you from misapplying them. As it is, it sounds like you are trying to break the law of conservation of energy. Well... good luck with that.

But since you asked, the basic equation for an inductor is v = L*(di/dt), and a good page for exponential decay is http://www.allaboutcircuits.com/vol_1/chpt_16/4.html.

heres a problem, is there a way to REVERSE the formulas to show decay rather than charging, yes htis is useful, but i need decay too.

 

[Born2bwire]

heres a problem, is there a way to REVERSE the formulas to show decay rather than charging, yes htis is useful, but i need decay too.

It's all apiece, the decay is simply the inverse behavior of the charge. You simply do the same derivation but you change the initial and steady-state conditions to reflect this. The time constants stay the same regardless. Those are only dependent upon the inductance and resistance of the system (barring any capacitive elements).

 

[NitroX]

hopw do i rewite the formula, canu give me an example please?

 

[Born2bwire]

It's given in the Wikipedia article on RL circuits. If you have a steady state RL circuit and turn off the applied voltage, then the decay is simply $I(t) = I_0e^{-\frac{t}{\tau}}$ since this obviously satisfies the initial condition of the current $I_0$ and the steady state condition of zero current.

 

[NitroX]

thank u so much! i will use that and study it!