jumper
Senior Member
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Not without a load.![]()
I could be wrong and probably am, but doesn't just energizing the conductors create a miniscule load because of the resistance of the wire?
Not without a load.![]()
I could be wrong and probably am, but doesn't just energizing the conductors create a miniscule load because of the resistance of the wire?
Kind of out of my depth but if I is 0 how would there be a loss?
IR^2
I could be wrong and probably am, but doesn't just energizing the conductors create a miniscule load because of the resistance of the wire?
Are you two selling tickets to Engineers Wrestlemanina?Kind of out of my depth but if I is 0 how would there be a loss?
IR^2
is there a limit to how many recpticals can be on a 20a kitchen circuit?
Beats me, let's wake up Gar-betcha he knows.
If there is resistance (load) in the wire and such, and there is a voltage drop across the load, I think you would have minor losses because you now have amperage.
Are you two selling tickets to Engineers Wrestlemanina?
You can't have a load without a complete circuit.
Are you two selling tickets to Engineers Wrestlemanina?
Yes.I could be wrong and probably am, but doesn't just energizing the conductors create a miniscule load because of the resistance of the wire?
I am thinking much, much smaller. Probably measurable but I don't feel like calculating. I'm up early to finish a project but could not resist dropping by.I am talking about mA here.
No-load capacitance gives us a complete circuit, not to mention the atomic-level alignment of surface charges (miniscule losses indeed).You can't have a load without a complete circuit.
Sorry to hear that my friend. Prayers and get-wells for you.Gotta think more when I post at 2:00AM on pain killers, vicoden.
MRI on back tomorrow, fun.