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Hey guys #43967
10/26/04 05:28 AM
10/26/04 05:28 AM
A
AndyP  Offline OP
Member
Joined: Aug 2004
Posts: 20
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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Re: Hey guys #43968
10/26/04 05:41 AM
10/26/04 05:41 AM
P
pauluk  Offline
Member
Joined: Aug 2001
Posts: 7,520
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.

Re: Hey guys #43969
10/27/04 11:16 AM
10/27/04 11:16 AM
L
Lee  Offline
Junior Member
Joined: Sep 2004
Posts: 9
philadelphia,pa,usa
^^^
you mean 4.5

Re: Hey guys #43970
10/27/04 12:54 PM
10/27/04 12:54 PM
Radar  Offline
Member
Joined: Apr 2004
Posts: 349
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.
Re: Hey guys #43971
10/27/04 12:58 PM
10/27/04 12:58 PM
E
ElectricAL  Offline
Member
Joined: Oct 2001
Posts: 597
Minneapolis, MN USA
45 ÷ 10 = 4.5

10 ÷ 45 = .222


Al Hildenbrand
Re: Hey guys #43972
10/30/04 10:45 AM
10/30/04 10:45 AM
A
AndyP  Offline OP
Member
Joined: Aug 2004
Posts: 20
.22 repteated

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

Re: Hey guys #43973
10/30/04 04:33 PM
10/30/04 04:33 PM
Attic Rat  Offline
Member
Joined: Dec 2003
Posts: 524
Bergen Co.,N.J. USA
...Uh,..OK, so which is it?,.. 4.5,..or .222amps?? Now I'm confused... [Linked Image] [Linked Image]
Russ


.."if it ain't fixed,don't break it...call a Licensed Electrician"
Re: Hey guys #43974
11/01/04 04:46 AM
11/01/04 04:46 AM
Scott35  Offline

Broom Pusher and
Member
Joined: Oct 2000
Posts: 2,713
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!
Re: Hey guys #43975
11/01/04 08:08 AM
11/01/04 08:08 AM
Attic Rat  Offline
Member
Joined: Dec 2003
Posts: 524
Bergen Co.,N.J. USA
...Thanx,Scott for clarifying that,...I was getting dizzy!!! [Linked Image] [Linked Image] [Linked Image]
Russ


.."if it ain't fixed,don't break it...call a Licensed Electrician"

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