Ohms law power problem

In article <47hm46l8eaaj2q8glre6v406kta174qsub@4ax.com>, Jon Kirwan
<snip>

In such cases, saying to "not make things more complex than
needed" is not entirely unlike
Sheesh... enough already.

The flame of interest can be extinguished by a flood of verbiage. ---
Joe, MMX (get the water vs, fire imagery, huh? huh? :))

Or, as Einstein supposedly said,

"Everything should be made as simple as possible, but not simpler."

or as Einstein also supposedly said,

"Any intelligent fool can make things bigger, more complex, and more
violent. It takes a touch of genius -- and a lot of courage -- to move in
the opposite direction."

--- Joe

PS: Beware of confusing (A-B)^2 with A^2 - B^2 and give hints to a
neophye instead of a tidal wave of ... .
 
On Sat, 24 Jul 2010 15:49:56 -0700, none@given.now (Joe)
wrote:

In article <47hm46l8eaaj2q8glre6v406kta174qsub@4ax.com>, Jon Kirwan
snip

In such cases, saying to "not make things more complex than
needed" is not entirely unlike

Sheesh... enough already.

The flame of interest can be extinguished by a flood of verbiage.
snip
It sure can be. I suppose it's now up to Bill to say.

Jon
 
"Bill Bowden" <wrongaddress@att.net> wrote in message
news:5b096e94-89b8-416a-9095-004aabb5868e@k19g2000yqc.googlegroups.com...
Using a 12 ohm load and 12 volt supply, what is the power gain when
the voltage is raised to 12.1 volts?

Considering the formula P=E^2/R the power at 12 volts will be 144/12 =
12 watts. And, at 12.1 volts, the power will be 146.41/ 12 = 12.2
watts, for a gain of 200 milliwatts.

But if the current at 12 volts is 1 amp, and the current at 12.1 volts
is 12.1/12= 1.00833 then the increase is 8.33mA and from the power
formula of P=I*E we get .00833*12.1 = 101 milliwatts which is about
half as much as the first number.

And,considering the increase in current, using the formula P=I^2 * R
we get .00833^2 * 12 = 833 microwatts, which ain't much.

So, which is correct A,B C ?

I vote for A

-Bill
After reading all the responses I got dizzy and I have been teaching this
for 40 years.
You made a math error. You seem capable of catching it.
Check your work.

Tom
 
In article <4c4c604d$0$31276$607ed4bc@cv.net>, "Tom Biasi"
<tombiasi@optonline.net> wrote:

After reading all the responses I got dizzy and I have been teaching this
for 40 years.
Did that student finally learn it?

--- Joe
 
"Joe" <none@given.now> wrote in message
news:none-2507100955030001@dialup-4.231.175.183.dial1.losangeles1.level3.net...
In article <4c4c604d$0$31276$607ed4bc@cv.net>, "Tom Biasi"
tombiasi@optonline.net> wrote:

After reading all the responses I got dizzy and I have been teaching this
for 40 years.

Did that student finally learn it?

--- Joe
Good one:)
 
On Jul 25, 9:03 am, "Tom Biasi" <tombi...@optonline.net> wrote:
"Bill Bowden" <wrongaddr...@att.net> wrote in message

news:5b096e94-89b8-416a-9095-004aabb5868e@k19g2000yqc.googlegroups.com...



Using a 12 ohm load and 12 volt supply, what is the power gain when
the voltage is raised to 12.1 volts?

Considering the formula P=E^2/R the power at 12 volts will be 144/12 > > 12 watts. And, at 12.1 volts, the power will be 146.41/ 12 = 12.2
watts, for a gain of 200 milliwatts.

But if the current at 12 volts is 1 amp, and the current at 12.1 volts
is 12.1/12= 1.00833 then the increase is 8.33mA and from the power
formula of P=I*E we get .00833*12.1 = 101 milliwatts which is about
half as much as the first number.

And,considering the increase in current, using the formula P=I^2 * R
we get .00833^2 * 12 = 833 microwatts, which ain't much.

So, which is correct A,B C ?

I vote for A

-Bill

After reading all the responses I got dizzy and I have been teaching this
for 40 years.
You made a math error. You seem capable of catching it.
Check your work.

Tom
I checked the work, but it doesn't add up.

a small change of 8.33mA and 100mV is only 833 microwatts when it
should be 200 milliwatts.

I think an analogy might be a freeway traffic jam where you add one
more car and everybody stops?

-Bill
 
On Sun, 25 Jul 2010 23:37:03 -0700 (PDT), Bill Bowden
<wrongaddress@att.net> wrote:

On Jul 25, 9:03 am, "Tom Biasi" <tombi...@optonline.net> wrote:
"Bill Bowden" <wrongaddr...@att.net> wrote in message

news:5b096e94-89b8-416a-9095-004aabb5868e@k19g2000yqc.googlegroups.com...



Using a 12 ohm load and 12 volt supply, what is the power gain when
the voltage is raised to 12.1 volts?

Considering the formula P=E^2/R the power at 12 volts will be 144/12 =
12 watts. And, at 12.1 volts, the power will be 146.41/ 12 = 12.2
watts, for a gain of 200 milliwatts.

But if the current at 12 volts is 1 amp, and the current at 12.1 volts
is 12.1/12= 1.00833 then the increase is 8.33mA and from the power
formula of P=I*E we get .00833*12.1 = 101 milliwatts which is about
half as much as the first number.

And,considering the increase in current, using the formula P=I^2 * R
we get .00833^2 * 12 = 833 microwatts, which ain't much.

So, which is correct A,B C ?

I vote for A

-Bill

After reading all the responses I got dizzy and I have been teaching this
for 40 years.
You made a math error. You seem capable of catching it.
Check your work.

Tom

I checked the work, but it doesn't add up.

a small change of 8.33mA and 100mV is only 833 microwatts when it
should be 200 milliwatts.
---
Algebraically, you can't just use the change in quantity, you have to
use the whole new quantity.

JF
 
"Bill Bowden" <wrongaddress@att.net> wrote in message
news:606dd3ae-7b34-4da6-a306-32caba98a2c3@t2g2000yqe.googlegroups.com...
On Jul 25, 9:03 am, "Tom Biasi" <tombi...@optonline.net> wrote:
"Bill Bowden" <wrongaddr...@att.net> wrote in message

news:5b096e94-89b8-416a-9095-004aabb5868e@k19g2000yqc.googlegroups.com...



Using a 12 ohm load and 12 volt supply, what is the power gain when
the voltage is raised to 12.1 volts?

Considering the formula P=E^2/R the power at 12 volts will be 144/12 =
12 watts. And, at 12.1 volts, the power will be 146.41/ 12 = 12.2
watts, for a gain of 200 milliwatts.

But if the current at 12 volts is 1 amp, and the current at 12.1 volts
is 12.1/12= 1.00833 then the increase is 8.33mA and from the power
formula of P=I*E we get .00833*12.1 = 101 milliwatts which is about
half as much as the first number.

And,considering the increase in current, using the formula P=I^2 * R
we get .00833^2 * 12 = 833 microwatts, which ain't much.

So, which is correct A,B C ?

I vote for A

-Bill

After reading all the responses I got dizzy and I have been teaching this
for 40 years.
You made a math error. You seem capable of catching it.
Check your work.

Tom
I checked the work, but it doesn't add up.

a small change of 8.33mA and 100mV is only 833 microwatts when it
should be 200 milliwatts.

I think an analogy might be a freeway traffic jam where you add one
more car and everybody stops?

-Bill

You can't just take pieces of what you want. The current produced was
1.00833 Amps not .00833 Amps.
The 8.33mA was in addition to the one Amp.
Concept of math error.
Keep pluggin' its fun.

Tom
 
In article
<606dd3ae-7b34-4da6-a306-32caba98a2c3@t2g2000yqe.googlegroups.com>, Bill
Bowden <wrongaddress@att.net> wrote:
I checked the work, but it doesn't add up.

a small change of 8.33mA and 100mV is only 833 microwatts when it
should be 200 milliwatts.
Okay, let's do this as I would in a decent algebra class:

Power is current x voltage.

Let's make this as a formula P = I*V

Better, from an algebraic understanding, is to write this in FUNCTION
NOTATION (dammit!).

P(I, V) = I*V

Where you are confused is thinking that this FUNCTION has the property

<BOGUS> P(I+i, V+v) = P(I,V) + P(i,v) </BOGUS>

<CORRECT> P(I+i, V+v) = (I+i)*(V+v) </CORRECT>

Now, compare <BOGUS> with <CORRECT>:

<CORRECT> Gives P(I+i, V+v) = (I+i)*(V+v) = (remember "FOIL") IV + Iv + iV + iv

<BOGUS> Gives P(I+i, V+v) = (BOGUS) P(I,V) + P(i,v) = IV + iv which is WRONG!

<CORRECT> gives two additional terms: Iv and iV

** NOTE: I never heard of "FOIL" as an acronym until a younger relative
told me about it. Some of my students seem to cling to a rule rather than
an understanding of what is happening (the distributive law). I sometimes
mocked mindless rules by referring to the "FOIL" situation as "Leo Rio".
Which actually stood for "LIORIO" (my own "Left Inner Outer Right Inner
Outer" Roole). :)

--- Joe
 
Joe wrote:
In article
606dd3ae-7b34-4da6-a306-32caba98a2c3@t2g2000yqe.googlegroups.com>, Bill
Bowden <wrongaddress@att.net> wrote:

I checked the work, but it doesn't add up.

a small change of 8.33mA and 100mV is only 833 microwatts when it
should be 200 milliwatts.


Okay, let's do this as I would in a decent algebra class:

Power is current x voltage.

Let's make this as a formula P = I*V

Better, from an algebraic understanding, is to write this in FUNCTION
NOTATION (dammit!).

P(I, V) = I*V

Where you are confused is thinking that this FUNCTION has the property

BOGUS> P(I+i, V+v) = P(I,V) + P(i,v) </BOGUS

CORRECT> P(I+i, V+v) = (I+i)*(V+v) </CORRECT

Now, compare <BOGUS> with <CORRECT>:

CORRECT> Gives P(I+i, V+v) = (I+i)*(V+v) = (remember "FOIL") IV + Iv + iV + iv

BOGUS> Gives P(I+i, V+v) = (BOGUS) P(I,V) + P(i,v) = IV + iv which is WRONG!

CORRECT> gives two additional terms: Iv and iV

** NOTE: I never heard of "FOIL" as an acronym until a younger relative
told me about it. Some of my students seem to cling to a rule rather than
an understanding of what is happening (the distributive law). I sometimes
mocked mindless rules by referring to the "FOIL" situation as "Leo Rio".
Which actually stood for "LIORIO" (my own "Left Inner Outer Right Inner
Outer" Roole). :)

--- Joe
In other words, what you are saying is that the increase in
power cannot be determined by multiplying the increase in
current by the increase in voltage. True.

And, even if you don't know that, when you write the equation
using functional notation (correctly) the correct answer comes
out.

Nice.

Ed
 
On 10-07-23 09:40 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12

cough..cough...cough....





mike
 
On Tue, 27 Jul 2010 22:34:12 -0600, m II <c@in.the.hat> wrote:

On 10-07-23 09:40 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough....
---

news:n1g056hqktafku6r4qdplcqus6s0jhgldr@4ax.com

JF
 
On 10-07-28 08:39 AM, John Fields wrote:

On Tue, 27 Jul 2010 22:34:12 -0600, m II<c@in.the.hat> wrote:

On 10-07-23 09:40 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough....

---

news:n1g056hqktafku6r4qdplcqus6s0jhgldr@4ax.com

JF


?



mike
 
On Wed, 28 Jul 2010 09:01:01 -0600, m II <c@in.the.hat> wrote:

On 10-07-28 08:39 AM, John Fields wrote:

On Tue, 27 Jul 2010 22:34:12 -0600, m II<c@in.the.hat> wrote:

On 10-07-23 09:40 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough.... <----------------------+
|
--- |
|
news:n1g056hqktafku6r4qdplcqus6s0jhgldr@4ax.com |
|
JF |
|
|
|
?----------------------------------------------+
mike
JF
 
On Wed, 28 Jul 2010 10:51:52 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 28 Jul 2010 09:01:01 -0600, m II <c@in.the.hat> wrote:

On 10-07-28 08:39 AM, John Fields wrote:

On Tue, 27 Jul 2010 22:34:12 -0600, m II<c@in.the.hat> wrote:

On 10-07-23 09:40 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough.... <----------------------+
|
--- |
|
news:n1g056hqktafku6r4qdplcqus6s0jhgldr@4ax.com |
|
JF |
|
|
|
?----------------------------------------------+
mike

JF
---
Or maybe:

cough..cough...cough.... <----------------------+
|
--- |
|
news:n1g056hqktafku6r4qdplcqus6s0jhgldr@4ax.com -+

JF



?
mike

JF

JF
 
On 10-07-28 10:38 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough....<----------------------+

Or maybe:

cough..cough...cough....<----------------------+

Sorry. I left out the mandatory 'ahem'

E˛ 12V˛ 1441 <<<<<<<<<<<<<<<<<<<<< ahem...cough...cough..
P1 = ---- = ----- = ----- = 12W watts
R 12V 12




mike
 
m II wrote:
On 10-07-28 10:38 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough....<----------------------+


Or maybe:


cough..cough...cough....<----------------------+



Sorry. I left out the mandatory 'ahem'

E˛ 12V˛ 1441 <<<<<<<<<<<<<<<<<<<<< ahem...cough...cough..
P1 = ---- = ----- = ----- = 12W watts
R 12V 12




mike
But his point is clearly made - neither typo detracts
from that. :)

Ed
 
On Wed, 28 Jul 2010 13:43:24 -0600, m II <c@in.the.hat> wrote:

On 10-07-28 10:38 AM, John Fields wrote:

All three, when they're worked out properly. :)


A:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12



cough..cough...cough....<----------------------+

Or maybe:

cough..cough...cough....<----------------------+


Sorry. I left out the mandatory 'ahem'

E˛ 12V˛ 1441 <<<<<<<<<<<<<<<<<<<<< ahem...cough...cough..
P1 = ---- = ----- = ----- = 12W watts
R 12V 12
---
Ah, yes, I see it now.


1441
Clearly, ------ = 120.08
12

Thanks, :)
 
On 10-07-29 03:31 AM, John Fields wrote:

Ah, yes, I see it now.
I screw up like that a lot. My Dad used to tell me that mistakes are
harder to spot when your nose is too close to the work. Taking a step
back gives a wider view.

We're all striving to be better. I note with some interest that one or
two denizens of this fine group have already attained perfection. A
weird thing though..perfection seems to bring with it uncontrolled
profanity.

I think I'll stay imperfect.



mike
 

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