Ohms law power problem

elektroda.net NewsGroups Forum Index - Electronic for beginners - Ohms law power problem

Bill Bowden
Guest

Sat Jul 24, 2010 5:56 am

On Jul 23, 8:40 am, John Fields <jfie...@austininstruments.com> wrote:

Quote:

On Thu, 22 Jul 2010 18:34:37 -0700 (PDT), Bill Bowden

wrongaddr...@att.net> wrote:
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 ?

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

A:
E² 12V² 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12

12.1V² 146.41
P2 = ------- = -------- = 12.2 watts
12R 12R

Dp1 = P2 - P1 = 0.2 watts

B:
E 12V
I1 = --- = ----- = 1 ampere
R 12R

P1 = EI = 12V * 1A = 12 watts.

12.1V
I2 = ------- = 1.00833... ampere
12R

P2 = 12.1V * 1.00833A = 12.2 watts

DP2 = P2 - P1 = 0.2 watts

C:

P3 = I²R

= 1.00833...A² * 12R

= 1.0167 * 12R = 12.2 watts

DP3 = P3 - P2 = 0.2 watts.

Yes, that makes sense using hard numbers, I was trying to do it using
small changes, thinking I could get the same result, but it didn't
work.

-Bill

Jon Kirwan
Guest

Sat Jul 24, 2010 7:31 am

On Fri, 23 Jul 2010 19:33:26 -0700 (PDT), Bill Bowden
<wrongaddress_at_att.net> wrote:

Quote:

snip
I can't do calculus Jon. I took it one semester in school and got a
"D"
All I remember is the derivative of x^2 is 2x.

Well, as another said, you don't need calculus. It just adds
another perspective here that gets you to the same place as
all the other approaches also do. And it adds insight where
other methods fail. As a counter to this, some methods (such
as hodographs, for example) provide insights where calculus
fails to help nearly as well. So although it is god, it
isn't a panacea.

Anyway, I hope the pictures I drew (and the algebra) helped.
There was no calculus needed for some of what I wrote.

Jon

Jon Kirwan
Guest

Sat Jul 24, 2010 8:50 am

On Fri, 23 Jul 2010 19:53:53 -0700 (PDT), stratus46_at_yahoo.com
wrote:

Quote:

snip
You don't need calculus for this. As my high school physics teacher
would say, "it's just sixth grade arithmetic".

Well, that's a little off-putting. I think most 6th graders
I've been exposed to (and I volunteered 300 hours a year for
about 5 years running at one grade school, some years back,
as an in-class aide) wouldn't be able to properly help Bill
through the problem. A rare one, maybe.

6th graders are supposed to know how to represent rationals
either as fractions or decimals, can use and apply ratios,
solve percentage problems, and hopefully have been exposed to
talking about things as x grams, y apples, and z frogs. Solve
a few simple things like 3X=5, 2+Y=16, maybe. A very few go
much beyond that.

The algebra is this:

V is some voltage
R is some resistance
P is some power
P = V^2/R

dV = 0.1
V0 = 12
V1 = V0 + dV = 12.1
Rx = 12
P0 = V0^2/Rx
P1 = V1^2/Rx = (V0+dV)^2/Rx = (V0^2+2*V0*dV+dV^2)/Rx
= V0^2/Rx + 2*V0*dV/Rx + dV^2/Rx
= P0 + 2*V0*dV/Rx + dV^2/Rx

or, moving P0 to the left side,

P1 - P0 = 2*V0*dV/Rx + dV^2/Rx

This last expression is the power difference, of course. And
it says that for small dV, it is mostly determined by the
first part, or 2*V0*dV/Rx. But I wouldn't expect a 6th
grader or even an 8th grader to get this far with the problem
or to exhibit much insight.

I think Bill attempted to use some intuitive ideas to examine
the problem from different angles and I'm impressed that he
struggled with two additional approaches that otherwise might
have been _powerful_, had he understood their meaning fully.
In fact, the values he computed were dV*V/Rx and dV^2/Rx,
which are important parts had he visualized the picture they
were part of, correctly. I'm also impressed that he exposed
himself to criticism, as that a good way to learn. I hope he
won't take your kicking sand in his face as teaching him the
wrong lesson.

The point that Joe made, that the error Bill made was similar
to the cylindrical conduit problem is apt, but I'm not sure
any light was shed for Bill in saying so. Bill's problem is
more easily seen, not with a small margin perimeter where a
smaller square is centered inside the larger -- which is more
as the conduit case -- but instead with two sides and a
corner of the two squares superimposed, I think. All ways
work, but mimicking the conduit by centering squares would
deviate more from Bill's initial insights, I think.

For someone wanting to "understand" and not just work some
recipe that gets right results, I'd recommend looking again
at P=V^2/R. The changing part, per the stated problem, is V
(and consequently P.)

Ignore the R part for now. P is proportional to V^2, where V
is subjected to change. V^2 is a square, with sides of
length V. Changing V is, in effect, changing the length of
both sides of the square. The difference between one square
and a slightly larger square (aligned at one lower-left
corner) imposed on top of it is a rectangular margin on the
right and a similarly rectangular margin at the top. There
are TWO of these margins, so the change in area is 2 times
the area of one rectangle by itself. And the area of one
rectangle is the change in V, the tiny width from V0 to V1
for example, times V. Which is where the 2*dV*V comes from.
That represents the two rectangles, one on the right and one
at the top, each dV*V in size. The last term comes from the
tiny square in the upper right corner that isn't covered by
either of the two marginal rectangles. And that tiny piece
is dV*dV in size. Which explains the entire algebra equation
using entirely visual, geometric and non-algebra means to
yield the same resulting concept.

Bill tried to use finite differences, which is the beginning
of moving towards very powerful concepts. He wrote,
elsewhere:

Quote:

: Yes, that makes sense using hard numbers, I was trying
: to do it using small changes, thinking I could get the
: same result, but it didn't work.

And I think it means he is grasping very close to calculus
thinking. He may not realize just how close he is to
"getting it" and I'd like to encourage him to take on more,
not make him feel badly about failing at "sixth grade
arithmetic." He's close. Very close.

Jon

Jon Kirwan
Guest

Sat Jul 24, 2010 9:22 pm

On Sat, 24 Jul 2010 12:26:07 -0700 (PDT), stratus46_at_yahoo.com
wrote:

Quote:

snip
Since I know Ohm's law to be a law (and not a suggestion) and I got an
answer _different_ from P=E*I, I would assume I made some kind of
mistake and then hunt it down to maintain my motto : Always make _new_
mistakes.

Hehe. Okay. I think that may actually be part of why Bill
wrote. He got a group of "different answers" and wanted to
understand why, when looking at this from different angles,
he didn't come up with the same result. I think it is valid
to think about the world using finite differences (in short,
moving towards a differential viewpoint) and to try and make
sure that works as well as using traditional finite averages
in equations deduced from basic laws.

I think Bill _did_ assume he'd made some kind of mistake. He
just didn't know how to 'right' himself while hanging onto
the differential mindset at the same time.

Quote:

It wasn't meant as a put-down, more of a don't make things more
complex than they need to be and don't be afraid of it.

I have mixed feelings about that approach. If all one wants
is a correct answer for a particular situation and has no
further interest (or ability) in developing a deeper
understanding, I agree with your comment.

But if someone _is_ interested in precise boundaries and
profound meaning to such laws, and not just one of the many
expressions (facets) of them that just happens to be in front
of them at the time, then I think it is wonderful to try and
take something concrete and immediate and play with it more
in order to help develop a deeper insight into their nature.
That may mean doing what Bill attempted (and slightly failed
at.) It is through such questions and struggles that one
may, at times, gain a higher ground from which to see.

In such cases, saying to "not make things more complex than
needed" is not entirely unlike telling a 5th grader who _may_
be asking a profound question like "Why is the moon round?"
and rather than exploring that with them while they show some
interest, instead answering it with a put down, "What? Do
you think it should be square?" as though they were stupid
for asking in the first place. They will get the clue, of
course. And stop asking. But for exploring minds, that is
destructive.

Jon

Guest

Sat Jul 24, 2010 10:26 pm

On Jul 24, 12:50 am, Jon Kirwan <j...@infinitefactors.org> wrote:

Quote:

On Fri, 23 Jul 2010 19:53:53 -0700 (PDT), stratu...@yahoo.com
wrote:

snip
You don't need calculus for this. As my high school physics teacher
would say, "it's just sixth grade arithmetic".

Well, that's a little off-putting. I think most 6th graders
I've been exposed to (and I volunteered 300 hours a year for
about 5 years running at one grade school, some years back,
as an in-class aide) wouldn't be able to properly help Bill
through the problem. A rare one, maybe.

6th graders are supposed to know how to represent rationals
either as fractions or decimals, can use and apply ratios,
solve percentage problems, and hopefully have been exposed to
talking about things as x grams, y apples, and z frogs. Solve
a few simple things like 3X=5, 2+Y=16, maybe. A very few go
much beyond that.

The algebra is this:

V is some voltage
R is some resistance
P is some power
P = V^2/R

dV = 0.1
V0 = 12
V1 = V0 + dV = 12.1
Rx = 12
P0 = V0^2/Rx
P1 = V1^2/Rx = (V0+dV)^2/Rx = (V0^2+2*V0*dV+dV^2)/Rx
= V0^2/Rx + 2*V0*dV/Rx + dV^2/Rx
= P0 + 2*V0*dV/Rx + dV^2/Rx

or, moving P0 to the left side,

P1 - P0 = 2*V0*dV/Rx + dV^2/Rx

This last expression is the power difference, of course. And
it says that for small dV, it is mostly determined by the
first part, or 2*V0*dV/Rx. But I wouldn't expect a 6th
grader or even an 8th grader to get this far with the problem
or to exhibit much insight.

I think Bill attempted to use some intuitive ideas to examine
the problem from different angles and I'm impressed that he
struggled with two additional approaches that otherwise might
have been _powerful_, had he understood their meaning fully.
In fact, the values he computed were dV*V/Rx and dV^2/Rx,
which are important parts had he visualized the picture they
were part of, correctly. I'm also impressed that he exposed
himself to criticism, as that a good way to learn. I hope he
won't take your kicking sand in his face as teaching him the
wrong lesson.

The point that Joe made, that the error Bill made was similar
to the cylindrical conduit problem is apt, but I'm not sure
any light was shed for Bill in saying so. Bill's problem is
more easily seen, not with a small margin perimeter where a
smaller square is centered inside the larger -- which is more
as the conduit case -- but instead with two sides and a
corner of the two squares superimposed, I think. All ways
work, but mimicking the conduit by centering squares would
deviate more from Bill's initial insights, I think.

For someone wanting to "understand" and not just work some
recipe that gets right results, I'd recommend looking again
at P=V^2/R. The changing part, per the stated problem, is V
(and consequently P.)

Ignore the R part for now. P is proportional to V^2, where V
is subjected to change. V^2 is a square, with sides of
length V. Changing V is, in effect, changing the length of
both sides of the square. The difference between one square
and a slightly larger square (aligned at one lower-left
corner) imposed on top of it is a rectangular margin on the
right and a similarly rectangular margin at the top. There
are TWO of these margins, so the change in area is 2 times
the area of one rectangle by itself. And the area of one
rectangle is the change in V, the tiny width from V0 to V1
for example, times V. Which is where the 2*dV*V comes from.
That represents the two rectangles, one on the right and one
at the top, each dV*V in size. The last term comes from the
tiny square in the upper right corner that isn't covered by
either of the two marginal rectangles. And that tiny piece
is dV*dV in size. Which explains the entire algebra equation
using entirely visual, geometric and non-algebra means to
yield the same resulting concept.

Bill tried to use finite differences, which is the beginning
of moving towards very powerful concepts. He wrote,
elsewhere:

: Yes, that makes sense using hard numbers, I was trying
: to do it using small changes, thinking I could get the
: same result, but it didn't work.

And I think it means he is grasping very close to calculus
thinking. He may not realize just how close he is to
"getting it" and I'd like to encourage him to take on more,
not make him feel badly about failing at "sixth grade
arithmetic." He's close. Very close.

Jon

Since I know Ohm's law to be a law (and not a suggestion) and I got an
answer _different_ from P=E*I, I would assume I made some kind of
mistake and then hunt it down to maintain my motto : Always make _new_
mistakes.

It wasn't meant as a put-down, more of a don't make things more
complex than they need to be and don't be afraid of it.

G²

Joe
Guest

Sat Jul 24, 2010 11:49 pm

In article <47hm46l8eaaj2q8glre6v406kta174qsub_at_4ax.com>, Jon Kirwan
<snip>

Quote:

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? Smile

)

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 ... .

Jon Kirwan
Guest

Sun Jul 25, 2010 12:11 am

On Sat, 24 Jul 2010 15:49:56 -0700, none_at_given.now (Joe)
wrote:

Quote:

In article <47hm46l8eaaj2q8glre6v406kta174qsub_at_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

Tom Biasi
Guest

Sun Jul 25, 2010 5:03 pm

"Bill Bowden" <wrongaddress_at_att.net> wrote in message
news:5b096e94-89b8-416a-9095-004aabb5868e_at_k19g2000yqc.googlegroups.com...

Quote:

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

Joe
Guest

Sun Jul 25, 2010 5:55 pm

In article <4c4c604d$0$31276$607ed4bc_at_cv.net>, "Tom Biasi"
<tombiasi_at_optonline.net> wrote:

Quote:

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

Did that student finally learn it?

--- Joe

Tom Biasi
Guest

Sun Jul 25, 2010 6:03 pm

"Joe" <none_at_given.now> wrote in message
news:none-2507100955030001_at_dialup-4.231.175.183.dial1.losangeles1.level3.net...

Quote:

In article <4c4c604d$0$31276$607ed4bc_at_cv.net>, "Tom Biasi"
tombiasi_at_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:-)

Bill Bowden
Guest

Mon Jul 26, 2010 9:37 am

On Jul 25, 9:03 am, "Tom Biasi" <tombi...@optonline.net> wrote:

Quote:

"Bill Bowden" <wrongaddr...@att.net> wrote in message

news:5b096e94-89b8-416a-9095-004aabb5868e_at_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

John Fields
Guest

Mon Jul 26, 2010 2:55 pm

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

Quote:

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_at_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

Tom Biasi
Guest

Mon Jul 26, 2010 7:54 pm

"Bill Bowden" <wrongaddress_at_att.net> wrote in message
news:606dd3ae-7b34-4da6-a306-32caba98a2c3_at_t2g2000yqe.googlegroups.com...
On Jul 25, 9:03 am, "Tom Biasi" <tombi...@optonline.net> wrote:

Quote:

"Bill Bowden" <wrongaddr...@att.net> wrote in message

news:5b096e94-89b8-416a-9095-004aabb5868e_at_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

Joe
Guest

Mon Jul 26, 2010 9:02 pm

In article
<606dd3ae-7b34-4da6-a306-32caba98a2c3_at_t2g2000yqe.googlegroups.com>, Bill
Bowden <wrongaddress_at_att.net> wrote:

Quote:

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

ehsjr
Guest

Tue Jul 27, 2010 4:37 am

Joe wrote:

Quote:

In article
606dd3ae-7b34-4da6-a306-32caba98a2c3_at_t2g2000yqe.googlegroups.com>, Bill
Bowden <wrongaddress_at_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

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