Russ;
...Uh,..OK, so which is it?,.. 4.5,..or .222amps?? Now I'm confused...
The total Amperes will be 0.222 A for this Circuit.
The Calculation
I=E÷R would result in an Amperage of 0.222, if a 45 Ohm Resistance Load was connected to a Power Source with an output Voltage of 10 Volts.
Looks like this:
I=E÷R,
R = 45 Ohms
E = 10 Volts
E (10 Volts) / R (45 Ohms) = 0.222 Amps ("I")
Concequentially (sp?), the 45 Ohm load connected to a 10 Volt System, will draw 2.22 Watts of True Power from the Power Supply.
The "4.5 Amps" thing was incorrect for this scenario (E=10V / R=45 Ohms).
If the values were reversed, then the "4.5 Amps" thing would be correct, meaning if the system's Voltage was 45 Volts and the connected load's Resistance was 10 Ohms, the resultant Current Flow would be 4.5 Amperes - with a resulting True Power (Wattage) of 202.5 Watts drawn from the Power Supply.
Sorry that this was not pointed out clearly, and hope it has been cleared up with this reply.
Scott35
