Question:
Equation for resistive loss is P = I^2 * R, where P is Wattage, I is Current and R is Resistance.
Since P = I * V (Voltage), wouldn't I divide both sides by I and come up with the equation of V = I * R? Wouldn't that be the same thing?
Question:
Equation for resistive loss is P = I^2 * R, where P is Wattage, I is Current and R is Resistance.
Since P = I * V (Voltage), wouldn't I divide both sides by I and come up with the equation of V = I * R? Wouldn't that be the same thing?
Rest in peace Mike Clements, aka "Yellowbeard"
Rest in peace Joerg Theissen, aka "GI Joe"
Right... because Ohm's Law is I = V / R which is the same as V = I * R. Damn... Why bother with the P = I^2 * R. Sheesh!
Rest in peace Mike Clements, aka "Yellowbeard"
Rest in peace Joerg Theissen, aka "GI Joe"
yes correct, its called algebra, you will find all sorts of relations if you keep on dividing and multiplying. but its been done before and the equations we have I think were all done that way, apart from the 'laws' which were experimentally derived from observation, and perhaps from 'first principles' by the clever people.
why bother? depends what 'variable's' you have available and where you want to get to.
the one's on the right are called the 'independant' variables, and the left one is 'dependant' as its dependant on the others, or sometimes called a 'function' of the others.
your favorite people the arabs, came up with al gabra, when some of them were enlightened.
Last edited by ianm2; 10-02-2007 at 03:42 AM.
Thanks smart ass. I got an A in algebra, perfect score on the exam and was a programmer for some years. That's not the point of my question.
I was just wondering if I could apply it to the electrical equations because the paper I was looking at was clearly trying to differentiate P = I^2 * R and V = I * R and I can't for the life of me understand why.
Rest in peace Mike Clements, aka "Yellowbeard"
Rest in peace Joerg Theissen, aka "GI Joe"
If you don't know the voltage, but just need the power the circuit dissipates then you'd have to use two formulas to find the power without P=I^2 * R. Now, in what circumstance you'd not know or care what the voltage supplied to the circuit was... I haven't a clue. But it's good to have something to cover all bases I guess.
Well, yeah. That's kind of my point. How would you actually know the wattage and not either the voltage or current (of course, if you knew either and the wattage than you can figure out the other.)
I just think some people make explaining electrical theory overly complicated just for the sake of it.
Rest in peace Mike Clements, aka "Yellowbeard"
Rest in peace Joerg Theissen, aka "GI Joe"
Well, say you have a circuit that has to stay under 500 watts (because you're using a shared 110v/15a branch circuit), and you have to need to maintain a constant current of 5 amps (for say a lead acid battery charger). You need to find out what the resistance needs to be so you can choose your components. You'd have to use P=I²R to solve the problem.
500 = 5²R;
500 = 25R;
500/25 = 25R/25;
20 = R.
Your circuit can have a resistance of no more than 20 Ohms.
I think you already wrote the answer to your question in your first post
The P = I² * R formula is nothing else then the short version of P=I*I*R which is nothing else then P=I*V | V = I * R (the trick is to view the V as an variable, where you can put in any equivalent therms).
Basically, this has nothing to do with Ohm`s law, it is pure algebra i would say
I am pretty sure the point of physics/science is to make things more complicated than they need to be, I'm pretty sure one of my old text books actually says that on the cover...