To find amperage would I use
I=E/R

The resistance of a circuit is 45 ohms. The input voltage is 10v. What is the resulting amperage?

Yes, the formula you quote is correct.

I = E / R, so 10 / 45 = 0.222 amps.

That's for a DC circuit or an AC circuit which is purely resistive. If an AC contains inductance or capacitance, then it gets a little more complex.

^^^
you mean 4.5

I think 4.5 would be R/E (45/10), which isn't anything really.

10/45 comes out to .2222 on my calculator.

45 ÷ 10 = 4.5

10 ÷ 45 = .222

.22 repteated

[This message has been edited by AndyP (edited 10-30-2004).]

...Uh,..OK, so which is it?,.. 4.5,..or .222amps?? Now I'm confused...
Russ

Russ;


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

...Thanx,Scott for clarifying that,...I was getting dizzy!!!
Russ