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#43967 - 10/26/04 03:28 AM Hey guys
AndyP Offline
Member

Registered: 08/11/04
Posts: 17
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?

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#43968 - 10/26/04 03:41 AM Re: Hey guys
pauluk Offline
Member

Registered: 08/11/01
Posts: 7693
Loc: Norfolk, England
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.

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#43969 - 10/27/04 09:16 AM Re: Hey guys
Lee Offline
Junior Member

Registered: 09/30/04
Posts: 9
Loc: philadelphia,pa,usa
^^^
you mean 4.5

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#43970 - 10/27/04 10:54 AM Re: Hey guys
Radar Offline
Member

Registered: 04/30/04
Posts: 349
Loc: Los Angeles, CA
I think 4.5 would be R/E (45/10), which isn't anything really.

10/45 comes out to .2222 on my calculator.
_________________________
There are 10 types of people. Those who know binary, and those who don't.

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#43971 - 10/27/04 10:58 AM Re: Hey guys
ElectricAL Offline
Member

Registered: 10/10/01
Posts: 615
Loc: Minneapolis, MN USA
45 √∑ 10 = 4.5

10 √∑ 45 = .222
_________________________
Al Hildenbrand

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#43972 - 10/30/04 08:45 AM Re: Hey guys
AndyP Offline
Member

Registered: 08/11/04
Posts: 17
.22 repteated

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

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#43973 - 10/30/04 02:33 PM Re: Hey guys
Attic Rat Offline
Member

Registered: 12/14/03
Posts: 530
Loc: Bergen Co.,N.J. USA
...Uh,..OK, so which is it?,.. 4.5,..or .222amps?? Now I'm confused...
Russ
_________________________
.."if it ain't fixed,don't break it...call a Licensed Electrician"

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#43974 - 11/01/04 01:46 AM Re: Hey guys
Scott35 Offline

Broom Pusher and
Member

Registered: 10/19/00
Posts: 2724
Loc: Anaheim, CA. USA
Russ;

 Quote:

...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
_________________________
Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!

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#43975 - 11/01/04 05:08 AM Re: Hey guys
Attic Rat Offline
Member

Registered: 12/14/03
Posts: 530
Loc: Bergen Co.,N.J. USA
...Thanx,Scott for clarifying that,...I was getting dizzy!!!
Russ
_________________________
.."if it ain't fixed,don't break it...call a Licensed Electrician"

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