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VOLTS push AMPS though OHMS.
more V -----> more A
less V -----> less A
less R ------> more A
more R -------> less A
That is all you need to know. If you want to learn it mathematically, then use ohm's law: V = I*R.
* or impedance with AC circuits, statement above assuming steady state DC conditions.
Reduce the load resistance applied to the source to one half of its former value.
Of course, that assumes the power supply/etc is capable of delivering twice the original current.
Wow details abound in this post. Are you talking about a power source, voltage regulator, battery or what?
I dunno. Try multiplying I by 2, and uh... multiplying V by unity.
If U is to be constant and I should double, R has to be halved.
No where enough information here.
In general current is drawn FROM a power source so how much is drawn depends as Steve says on the load. (also on the ability of the power supply to provide the current.)
In what context ? Halving the load resistance will do it.
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