Find current from voltage drop?

J

js5895

Guest
Hi,

The voltage drop formula:

K x I x L x 2
E = ----------------
CMA

I know all the values, except "I", so is there a formula to find "I".

Thanks.
 
js5895 wrote:
Hi,

The voltage drop formula:

K x I x L x 2
E = ----------------
CMA

I know all the values, except "I", so is there a formula to find "I".
Click to expand...
Are you saying that you know all the values in your equation except I,
and you want to find I? If so that is simple algebra.

Multiply both sides of the equation by CMA:-

E x CMA = K x I x L x 2

Then divide both sides by K x L x 2:-

E x CMA
---------- = I
K x L x 2



--
-----------------------------------------------------------------------
To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris
 
On Wed, 23 Mar 2005 20:38:00 GMT in sci.electronics.basics, "js5895"
<js5895@tmp.com> wrote msg <Isk0e.152358$nC5.4171@twister.nyroc.rr.com>:

Hi,

The voltage drop formula:

K x I x L x 2
E = ----------------
CMA

I know all the values, except "I", so is there a formula to find "I".
Click to expand...
See Gareth's answer, unless you don't know either E or I, in which case
you cannot find the answer from the known information.

I have a question. What are K, C, M and A? I'm assuming this is a
formula for the drop across an inductor, where E = I * XL, and XL =
2*Pi*f*L, so somehow Pi*f = K/(CMA)? Thanks.

--
Al Brennan
 
On Wed, 23 Mar 2005 16:19:09 -0700, Kitchen Man <nannerbac@yahoo.com>
wrote:

On Wed, 23 Mar 2005 20:38:00 GMT in sci.electronics.basics, "js5895"
js5895@tmp.com> wrote msg <Isk0e.152358$nC5.4171@twister.nyroc.rr.com>:

Hi,

The voltage drop formula:

K x I x L x 2
E = ----------------
CMA

I know all the values, except "I", so is there a formula to find "I".

See Gareth's answer, unless you don't know either E or I, in which case
you cannot find the answer from the known information.

I have a question. What are K, C, M and A? I'm assuming this is a
formula for the drop across an inductor, where E = I * XL, and XL =
2*Pi*f*L, so somehow Pi*f = K/(CMA)? Thanks.
Click to expand...

Probably wire loss E in volts.

L = length
I = current
CMA = wire cross-sectional area, circular mils
K = fudge factor
2 = round trip

John
 
Thank You very much.
If thats simple algebra, I don't know algebra, lol.

K = Approximate/Exact K of wire
I = Current
L = Length of wire, one way
2 = There and back
E = Voltage drop
CMA = Circular Mill Area of wire

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:efv341ppncqc4bgbocfju75m9rqd2k7bp2@4ax.com...
On Wed, 23 Mar 2005 16:19:09 -0700, Kitchen Man <nannerbac@yahoo.com
wrote:

On Wed, 23 Mar 2005 20:38:00 GMT in sci.electronics.basics, "js5895"
js5895@tmp.com> wrote msg <Isk0e.152358$nC5.4171@twister.nyroc.rr.com>:

Hi,

The voltage drop formula:

K x I x L x 2
E = ----------------
CMA

I know all the values, except "I", so is there a formula to find "I".

See Gareth's answer, unless you don't know either E or I, in which case
you cannot find the answer from the known information.

I have a question. What are K, C, M and A? I'm assuming this is a
formula for the drop across an inductor, where E = I * XL, and XL =
2*Pi*f*L, so somehow Pi*f = K/(CMA)? Thanks.


Probably wire loss E in volts.

L = length
I = current
CMA = wire cross-sectional area, circular mils
K = fudge factor
2 = round trip

John
Click to expand...
 
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