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#43967 10/26/04 06:28 AM
Joined: Aug 2004
Posts: 20
A
AndyP Offline OP
Member
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?

#43968 10/26/04 06:41 AM
Joined: Aug 2001
Posts: 7,520
P
Member
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.

#43969 10/27/04 12:16 PM
Joined: Sep 2004
Posts: 9
L
Lee Offline
Junior Member
^^^
you mean 4.5

#43970 10/27/04 01:54 PM
Joined: Apr 2004
Posts: 349
Member
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.
#43971 10/27/04 01:58 PM
Joined: Oct 2001
Posts: 597
E
Member
45 ÷ 10 = 4.5

10 ÷ 45 = .222


Al Hildenbrand
#43972 10/30/04 11:45 AM
Joined: Aug 2004
Posts: 20
A
AndyP Offline OP
Member
.22 repteated

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

#43973 10/30/04 05:33 PM
Joined: Dec 2003
Posts: 524
Member
...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"
#43974 11/01/04 05:46 AM
Joined: Oct 2000
Posts: 2,721
Broom Pusher and
Member
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!
#43975 11/01/04 09:08 AM
Joined: Dec 2003
Posts: 524
Member
...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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