Ohms law power problem

[Bill Bowden]

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

 

[Jon Kirwan]

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

I don't like the use of 'gain' here. Might be dissipation,
instead.

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.

Yes. 12.1*(12.1/12).

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.

Um. You missed the inclusion of a factor of 2, as you might
surmise from this result. It's actually 200mW.

Here's the problem. You compute dV (I prefer V to E) and
multipled it by the new V and divided by R. That's almost
right, except you need a factor of 2. I'll explain a little
by transposing your work.

Let's call V0=12 and V1=12.1. R=12. First, you computed
V1/R and then subtracted from it V0/R to get the current
difference. This is the same thing as computing (V1-V0)/R or
another way of saying that is that since dV=V1-V0, you
computed dV/R. Then you multipled this by V1. In short,
what you computed was V*dV/R.

But look at the original power equation you used.

P = V^2/R

The derivative of that, assuming R is constant, is:

dP = 2*V*dV/R

The change in power is TWO TIMES V*dV/R and you only computed
half that much when you computed V*dV/R. Double it and you
are back in the saddle, again.

Another way of seeing this is to imagine the voltage as one
side of a rectangle and the current as the other side. Power
is then the rectangle area.


V1 +----------------------------+---+
| |. .|
V0 +----------------------------+---+
| / / / / / / / / / | . |
|/ / / / / / / / / /|. .|
| / / / / / / / / / | . |
| / / / / / / / / / |. .|
|/ / / / / / / / / /| . |
0 +----------------------------+---+
0 V0/R V1/R

When you computed V1/R and subtracted V0/R from it, you
basically took the width of a tiny strip of that area where
I've kind of filled it with "." characters. Multiplying by
V1 got you the area of that strip. But you completely missed
that hollow part near the top there, from V0 to V1 and times
V0/R.

To fix up your calculation, you'd need to add it back in.
That would be (12.1-12)*12/12 or, as is obvious, 0.1. Which
gets you right back to your 200mW.

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

Now you are computing dI^2*R.

Instead,

P = I^2*R

therefore,

dP = 2*I*dI*R

And if you used 2*1.008333*0.008333*12, you would once again
get a much more sensible result.

So, which is correct A,B C ?

I vote for A

Calculus is god, though. Best to use that.

Jon

 

[Shaun]

"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

Your doing your math wrong.

Learn to do math properly!! If you do all three methods come up with the
same answer, 200 mW

Shaun

 

[ehsjr]

Bill Bowden 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.

Correct.

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.

The error above is that you used P=I*E to compute based on
the _change_ in current (dI). P <> dI*E

Same thing, below. You used P=I^2 * R to compute based on
the _change_ in current (dI). P <> dI^2 * R

Ed

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

 

[John O'Flaherty]

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

You correctly figured the power at the two voltage levels, and
subtracted the powers.

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.

You subtracted the current, and then calculated the power based on the
current difference. Do what you did for voltage: calculate the powers
based on the two currents, and then subtract the powers.
P = I^2R
p1 = 1^2*12 = 12 W
p2 = 1.00833^2 * 12 = 12.2 W

p2 - p1 = 200 mW.

--
John

 

[Joe]

In article
<5b096e94-89b8-416a-9095-004aabb5868e@k19g2000yqc.googlegroups.com>, Bill
Bowden <wrongaddress@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 ?

I vote for A

-Bill

(B) The reason that you got the wrong result in (B) is that you took the
*increase* in current, .00833 amp and multiplied by 12.1, but you took the
1 amp and multiplied it by 12, when this current (1 amp) should also be
multiplied by 12.1.

(C) The reason that you got the wrong result in (C) is quite analogous to
an error in calculating the volume of concrete necessary to make a
cylindrical conduit.

The true formula is (R^2 - r^2)*pi*L where R is the outside radius, r is
the inside radius, and L is the length of the conduit. The fairly common
mistake is doing (R-r)^2*pi*L, thus giving the whacky result that a 10'
inside diameter conduit and a 1' inside diameter conduit having the same
wall thickness would use the same amount of concrete.

Calculus is hardly necessary to explain either of the above errors.

--- Joe

 

[Bill Bowden]

On Jul 22, 8:03 pm, "Shaun" <r...@nomail.com> 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

Your doing your math wrong.

Learn to do math properly!!  If you do all three methods come up with the
same answer, 200 mW

Shaun

Really? now would you mind explaining why an increase of 8.33mA in a
12 volt load amounts to 200mW?

8.33mA times 12 volts is only 100mW.

Where is the extra 100mW?

-Bill

 

[Jon Kirwan]

On Thu, 22 Jul 2010 22:22:08 -0700 (PDT), Bill Bowden
<wrongaddress@att.net> wrote:

On Jul 22, 8:03 pm, "Shaun" <r...@nomail.com> 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

Your doing your math wrong.

Learn to do math properly!!  If you do all three methods come up with the
same answer, 200 mW

Shaun

Really? now would you mind explaining why an increase of 8.33mA in a
12 volt load amounts to 200mW?

8.33mA times 12 volts is only 100mW.

Where is the extra 100mW?

-Bill

Bill, we both agree that power is V^2/R, right? Let's look
at this, term by term. It is V*V/R. Now, if you increase V
by some small value... call it dV for now... then the new
estimate is (V+dV)*(V+dV)/R. Right? Multiply it out. It
becomes (V^2+2*V*dV+dV^2)/R, yes? Put another way, it is:

V^2/R + 2*V*dV/R + dV^2/R

Now take a deeper look at the above. There are three terms
there. The first term is just the power computed in the
first place. 12V*12V/12Ohms, or 12W. With dV=0.1V, the
third term is barely noticeable. It is .01/12 or that 833uW
you'd earlier computed (remember it from your 3rd calc?) The
middle term is twice what you'd said in your 2nd calculation
because of the 2 there. Note that this 2nd term is two times
12V times (12.1V-12V) divided by 12Ohms. But two times!
Which accounts for earlier conceptual error. You failed to
include the factor of 2 there.

Again, but slightly differently than I did before, let's draw
out a square of sorts. I'm going to remove the 1/R part of
the power equation so that it just leaves us with V*V and
also (V+dV)*(V+dV). Drawing that geometrically instead of
algebraically provides something like this (use a fixed
spaced font, of course, to "see" it):

+----------------+---+ V+dV
| (B) |(D)|
+----------------+---+ V
| / / / / / | |
|/ / / / / /| |
| / / (A) / / |(C)|
| / / / / / | |
|/ / / / / /| |
0 +----------------+---+
0 V V+dV

Region A represents V^2 at V=12V. Region A+B+C+D represents
(V+dV)^2 at V+dV=12.1V. If you then divide A by R=12 you get
the power before changing the voltage by dV. If you then
divide A+B+C+D by R you get the power afterwards. Note that
the difference in area between A and A+B+C+D is B+C+D. B is
just dV*V; C is just dV*V; and D is just dV*dV or dV^2. If
you add B and C together, you get 2*V*dV, right? Once again,
there is that pesky 2 factor. It's there because there are
two areas that are the same shape and size.

The point here is that when you did your 2nd calculation, you
either neglected to take into account B, or else C, and just
calculated one of them. That's why you were basically off by
a factor of two. You only counted one region instead of two.
And there are two of them. Even then, you missed a tiny bit,
namely D. Which as you calculated elsewhere, was only 833uW.

After dividing each by R=12, A=12W, B=0.1W, C=0.1W, and
D=833uW. The total difference is B+C+D or 0.200833W, no
matter how you cut it.

Your 3rd calculation came up with the area of D because what
you did was to compute (dV/R)*(dV/R)*R. This is just dV^2/R.
And since region D is just dV^2 (with an implicit /R to it),
it's what you were actually getting.

I don't know if any of that helps. But there it is. Algebra
AND geometry. Either way, it gets you to the same place.

Jon

 

[Jon Kirwan]

On Fri, 23 Jul 2010 00:10:38 -0700, none@given.now (Joe)
wrote:

snip
Calculus is hardly necessary to explain either of the above errors.

Completely agreed, as I've already evidenced with both
algebra and simple areas.

But calculus is still god!

Jon

 

[Jasen Betts]

On 2010-07-23, Bill Bowden <wrongaddress@att.net> wrote:

On Jul 22, 8:03 pm, "Shaun" <r...@nomail.com> wrote:

Learn to do math properly!!  If you do all three methods come up with the
same answer, 200 mW

Really? now would you mind explaining why an increase of 8.33mA in a
12 volt load amounts to 200mW?

But, it's not a 12V load any more.


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[John Fields]

On Thu, 22 Jul 2010 18:34:37 -0700 (PDT), Bill Bowden
<wrongaddress@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:
E˛ 12V˛ 1441
P1 = ---- = ----- = ----- = 12W watts
R 12V 12


12.1V˛ 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 = I˛R

= 1.00833...A˛ * 12R

= 1.0167 * 12R = 12.2 watts

DP3 = P3 - P2 = 0.2 watts.

 

[Dan Coby]

On 7/22/2010 6:34 PM, Bill Bowden 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 ?

I vote for A

I will stick my version of an explanation for this.

I think that you already know that A is the correct answer.

Power is current time voltage.

Increasing the voltage increases the current. If we call the
the original voltage V0 and the initial current I0 and we call
the increase in voltage dV and the increase in current dI.

Then the original power (at 12 volts) is given by.

P0 = V0 * I0 = 12 volts * 1 amp = 12 watts.

The power after the current is increased is given by

P1 = (V0 + dV) * (I0 + dI)

multiplying out the terms gives us:

P1 = V0 * I0 + dV * I0 + v0 * dI + dV * dI

Now let us look at the terms in this equation.

First we have V0 * I0. This is the original power before the voltage
was increased. It is 12 volts times 1 amp = 12 watts.

Second we have dV * I0. I am going to save this term for a little
later since this is the real cause of your confusion.

The third and fourth terms are V0 * dI plus dV * dI. This is the
power increase that you have mentioned. I.e. they are 12 volts *
0.00833 amps = 0.1 watts and 0.1 volts * 0.00833 amps = 0.000833 watts.
Or you can (and you did) combine these together to be 12.1 volts
times 0.0833 amps. Either way you get 0.100833 watts.


Now lets go back to the dV * I0 term. This power is 0.1 volts times
1 amp which is 0.100 watts. This term represents the increase in power
due to the increase in the voltage combined with the original current.
This is the amount that you are missing.

 

[Bill Bowden]

On Jul 23, 12:10 am, n...@given.now (Joe) wrote:

In article
5b096e94-89b8-416a-9095-004aabb58...@k19g2000yqc.googlegroups.com>, 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 ?

I vote for A

-Bill

(B) The reason that you got the wrong result in (B) is that you took the
*increase* in current, .00833 amp and multiplied by 12.1, but you took the
1 amp and multiplied it by 12, when this current (1 amp) should also be
multiplied by 12.1.

Yes, that looks right. I missed the 1 amp at 12.1V for the extra
100mW.

(C)  The reason that you got the wrong result in (C) is quite analogous to
an error in calculating the volume of concrete necessary to make a
cylindrical conduit.

The true formula is (R^2 - r^2)*pi*L where R is the outside radius, r is
the inside radius, and L is the length of the conduit.  The fairly common
mistake is doing (R-r)^2*pi*L, thus giving the whacky result that a 10'
inside diameter conduit and a 1' inside diameter conduit having the same
wall thickness would use the same amount of concrete.

Calculus is hardly necessary to explain either of the above errors.

---  Joe

 

[Bill Bowden]

On Jul 22, 7:12 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

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?

I don't like the use of 'gain' here.  Might be dissipation,
instead.

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.

Yes.  12.1*(12.1/12).

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.

Um.  You missed the inclusion of a factor of 2, as you might
surmise from this result.  It's actually 200mW.

Here's the problem.  You compute dV (I prefer V to E) and
multipled it by the new V and divided by R.  That's almost
right, except you need a factor of 2.  I'll explain a little
by transposing your work.

Let's call V0=12 and V1=12.1.  R=12.  First, you computed
V1/R and then subtracted from it V0/R to get the current
difference.  This is the same thing as computing (V1-V0)/R or
another way of saying that is that since dV=V1-V0, you
computed dV/R.  Then you multipled this by V1.  In short,
what you computed was V*dV/R.

But look at the original power equation you used.

P = V^2/R

The derivative of that, assuming R is constant, is:

dP = 2*V*dV/R

The change in power is TWO TIMES V*dV/R and you only computed
half that much when you computed V*dV/R.  Double it and you
are back in the saddle, again.

Another way of seeing this is to imagine the voltage as one
side of a rectangle and the current as the other side.  Power
is then the rectangle area.

V1 +----------------------------+---+
   |                            |. .|
V0 +----------------------------+---+
   | /  /  /  /  /  /  /  /  /  | . |
   |/  /  /  /  /  /  /  /  /  /|. .|
   |  /  /  /  /  /  /  /  /  / | . |
   | /  /  /  /  /  /  /  /  /  |. .|
   |/  /  /  /  /  /  /  /  /  /| . |
 0 +----------------------------+---+
   0                         V0/R  V1/R

When you computed V1/R and subtracted V0/R from it, you
basically took the width of a tiny strip of that area where
I've kind of filled it with "." characters.  Multiplying by
V1 got you the area of that strip.  But you completely missed
that hollow part near the top there, from V0 to V1 and times
V0/R.

To fix up your calculation, you'd need to add it back in.
That would be (12.1-12)*12/12 or, as is obvious, 0.1.  Which
gets you right back to your 200mW.

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

Now you are computing dI^2*R.

Instead,

P = I^2*R

therefore,

dP = 2*I*dI*R

And if you used 2*1.008333*0.008333*12, you would once again
get a much more sensible result.

So, which is correct A,B C ?

I vote for A

Calculus is god, though.  Best to use that.

Jon

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.

-Bill

 

On Jul 23, 7:33 pm, Bill Bowden <wrongaddr...@att.net> wrote:

On Jul 22, 7:12 pm, Jon Kirwan <j...@infinitefactors.org> wrote:





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?

I don't like the use of 'gain' here.  Might be dissipation,
instead.

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.

Yes.  12.1*(12.1/12).

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.

Um.  You missed the inclusion of a factor of 2, as you might
surmise from this result.  It's actually 200mW.

Here's the problem.  You compute dV (I prefer V to E) and
multipled it by the new V and divided by R.  That's almost
right, except you need a factor of 2.  I'll explain a little
by transposing your work.

Let's call V0=12 and V1=12.1.  R=12.  First, you computed
V1/R and then subtracted from it V0/R to get the current
difference.  This is the same thing as computing (V1-V0)/R or
another way of saying that is that since dV=V1-V0, you
computed dV/R.  Then you multipled this by V1.  In short,
what you computed was V*dV/R.

But look at the original power equation you used.

P = V^2/R

The derivative of that, assuming R is constant, is:

dP = 2*V*dV/R

The change in power is TWO TIMES V*dV/R and you only computed
half that much when you computed V*dV/R.  Double it and you
are back in the saddle, again.

Another way of seeing this is to imagine the voltage as one
side of a rectangle and the current as the other side.  Power
is then the rectangle area.

V1 +----------------------------+---+
   |                            |. .|
V0 +----------------------------+---+
   | /  /  /  /  /  /  /  /  /  | . |
   |/  /  /  /  /  /  /  /  /  /|. .|
   |  /  /  /  /  /  /  /  /  / | . |
   | /  /  /  /  /  /  /  /  /  |. .|
   |/  /  /  /  /  /  /  /  /  /| . |
 0 +----------------------------+---+
   0                         V0/R  V1/R

When you computed V1/R and subtracted V0/R from it, you
basically took the width of a tiny strip of that area where
I've kind of filled it with "." characters.  Multiplying by
V1 got you the area of that strip.  But you completely missed
that hollow part near the top there, from V0 to V1 and times
V0/R.

To fix up your calculation, you'd need to add it back in.
That would be (12.1-12)*12/12 or, as is obvious, 0.1.  Which
gets you right back to your 200mW.

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

Now you are computing dI^2*R.

Instead,

P = I^2*R

therefore,

dP = 2*I*dI*R

And if you used 2*1.008333*0.008333*12, you would once again
get a much more sensible result.

So, which is correct A,B C ?

I vote for A

Calculus is god, though.  Best to use that.

Jon

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.

-Bill

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

G˛

 

[Bill Bowden]

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

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

           12.1V˛    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 = I˛R

        = 1.00833...A˛ * 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]

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

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]

On Fri, 23 Jul 2010 19:53:53 -0700 (PDT), stratus46@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

 

[Jon Kirwan]

On Sat, 24 Jul 2010 12:26:07 -0700 (PDT), stratus46@yahoo.com
wrote:

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.

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

 

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

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.

G˛