Honu wrote: ↑Thu, 02 Apr 2020, 17:32
coulomb wrote: ↑Thu, 02 Apr 2020, 07:16
Ok, so that's rated for V = √(P·R) = √(2000 · 7) = 118 V. Perhaps it's designed for 120 VAC.
For what i've read, a resistor got no working "voltage"... it got 2 parameters, internal resistance and Max power output it can accept, more then this power.. it blow out. (with perhaps a tolerance of .. let's say 10%)
Sure. But if you apply more than 120 V to this element, it will dissipate more than 2000 W. With a fixed resistance, a power limit also implies a voltage limit.
My logic is that .. Ohm Law give me : V = R x I => R = V/I = 7 Ohm => V and I can only be on a straight line cause V/I = constant...
Yes, fixed resistors come out as a straight line on a V versus I graph. But note that panels' graphs are decidedly not straight lines; they are not fixed resistors.
Ok now, i apply 33V to the 7 Ohm element, it flow 4.7 Amp => should i understand that IF i apply from the generator, a 33V and 10A, only 4.7A will flow through the resistor ?
A generator might have a 10 A limit, but if you plug in a small LED lamp, you won't get 10 A to flow. But yes, a 7 Ω resistor across a 33 V source will cause 4.7 A to flow, as long as the source can maintain that voltage at that current.
If i want 2A to flow i need 14V in my 7 Ohm element => Does it mean a resistor element is driven by voltage not by intensity ?
By "intensity" I'll assume you mean "current". Essentially, yes; the voltage establishes an electric field, which then induces the current. You don't usually think of "applying" a current to a resistor.
This would imply with the Ohm Law that V = R x I that all my couples (V,I) SHOULD be on a straight line, linearly depending on eah other to satisfy my resistance... if i go away from this line i lose power...
It's not that you'll lose power, it's that the current you thought would flow just doesn't happen, and the current demanded by Ohms law does.
Allright, what you say seems to help a little, but .. i still got no response to
A : I'm applying 33V and 2A to my 7Ohm resistor, what is the power dissipated by the element ..?
I thought I made it clear that you can't apply 33 V and
2 A to a 7 Ω resistor. Imagine a power supply with voltage and current limit controls. Initially, you set the voltage limit to 33 V, and the current limit to maximum. You connect the 7 Ω resistor; 4.7 A flows. Now you turn the current limit down to 2 A. You will find that the voltage drops to 14 V, to limit the current to 2 A.
A reply could be : A : The resistor is rated for 7 Ohm, under 33V it will draw at most I = V/R = 33/7 = 4.7 A, you throw 2A at him it will take them and generate : P = 7 x 2 x 2 = 28 Watts
Yes. But in the second case, the voltage won't be 33 V any more.
B : I'm applying 33V and 100A to my 7Ohm resistor, what is the power dissipated by the element ..?
Again, you're specifying voltage and current at the same time. It doesn't work that way.
A reply could be : B : The resistor is rated for 7 Ohm, under 33V it will draw at most I = V/R = 33/7 = 4.7 A,
It won't be "at most" 4.7 A, it will be 4.7 A, no more and no less. Otherwise, it won't be 33 V or it won't be 7 Ω.
you throw 100A at him, it will only take 4.7A and generate : P = 7 x 4.7 x 4.7 = 155 Watts
Meaning in this case some power is lost.
Current doesn't work like that. If you get 100 A into that element, it will blow up; the power would be P = I²R = (100)²·7 = 70 kW.
The only way to have 100 A coming in and only 4.7 A going into the resistor is if there is another path around the resistor, taking the other 95.3 A. Electric current (unlike water current) can't just leak out onto the ground without a conductive path.
The only way power can be "lost" in your example is if you had a source that was say 33 V and had the potential
of providing 100 A. Then by loading that source with 7 Ω you are wasting 95.3% of that potential, so "losing" potential power in that sense.
Nissan Leaf 2012 with new battery May 2019.
5650 W solar, 2xPIP-4048MS inverters, 16 kWh battery.
1.4 kW solar with 1.2 kW Latronics inverter and FIT.
160 W solar, 2.5 kWh 24 V battery for lights.
Patching PIP-4048/5048 inverter-chargers.