half wave rectifier...

[Ricky]

On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote:

On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote:
On Sunday, July 2, 2023 at 6:27:57 PM UTC-4, Fred Bloggs wrote:
On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote:
On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote:
On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote:

It\'s absolutely clear to me, and no doubt to many others on here that
you have no idea. Which part of 3kW heater running off 250V don\'t you
understand ? I want to halve the power in the 3kW heater. Show me the
maths...
If you want to halve the power in a heater, use an electric oven control
knob. Those are available as repair parts, and they cycle the heating
element they control, to give proportional heat control.

Every broken down cooktop has four of those, and three of \'em will work.
I\'ve never seen an oven or stove control that gave proportional control. Every one I\'ve seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?
Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.
Hmmm... I am talking about the sort of burners or ovens in the home used for cooking.
So am I.
Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!
Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously?

1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal \"thermostat\" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I\'ve had it apart before.

So, when you talk about \"throwing sparks\", you are talking about the normal operation of the contacts. I should have known. What you don\'t know fills volumes.

--

Rick C.

-+-+ Get 1,000 miles of free Supercharging
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[Fred Bloggs]

On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote:

On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote:
On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote:
On Sunday, July 2, 2023 at 6:27:57 PM UTC-4, Fred Bloggs wrote:
On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote:
On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote:
On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote:

It\'s absolutely clear to me, and no doubt to many others on here that
you have no idea. Which part of 3kW heater running off 250V don\'t you
understand ? I want to halve the power in the 3kW heater. Show me the
maths...
If you want to halve the power in a heater, use an electric oven control
knob. Those are available as repair parts, and they cycle the heating
element they control, to give proportional heat control.

Every broken down cooktop has four of those, and three of \'em will work.
I\'ve never seen an oven or stove control that gave proportional control. Every one I\'ve seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?
Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.
Hmmm... I am talking about the sort of burners or ovens in the home used for cooking.
So am I.
Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!
Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously?
1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal \"thermostat\" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I\'ve had it apart before.

Those products are totally shielded against throwing sparks outside the enclosure or they couldn\'t be sold. The water heater controls are beneath a sealed cover that\'s screwed down, that\'s what\'s called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise.

> So, when you talk about \"throwing sparks\", you are talking about the normal operation of the contacts. I should have known. What you don\'t know fills volumes.

Can\'t say the same about you, a vacuum doesn\'t fill anything.

--

Rick C.

-+-+ Get 1,000 miles of free Supercharging
-+-+ Tesla referral code - https://ts.la/richard11209

 

[Ricky]

On Monday, July 3, 2023 at 12:05:13 PM UTC-4, Fred Bloggs wrote:

On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote:
On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote:
On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote:
On Sunday, July 2, 2023 at 6:27:57 PM UTC-4, Fred Bloggs wrote:
On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote:
On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote:
On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote:

It\'s absolutely clear to me, and no doubt to many others on here that
you have no idea. Which part of 3kW heater running off 250V don\'t you
understand ? I want to halve the power in the 3kW heater. Show me the
maths...
If you want to halve the power in a heater, use an electric oven control
knob. Those are available as repair parts, and they cycle the heating
element they control, to give proportional heat control.

Every broken down cooktop has four of those, and three of \'em will work.
I\'ve never seen an oven or stove control that gave proportional control. Every one I\'ve seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?
Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.
Hmmm... I am talking about the sort of burners or ovens in the home used for cooking.
So am I.
Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!
Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously?
1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal \"thermostat\" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I\'ve had it apart before.

Those products are totally shielded against throwing sparks outside the enclosure or they couldn\'t be sold. The water heater controls are beneath a sealed cover that\'s screwed down, that\'s what\'s called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise.
So, when you talk about \"throwing sparks\", you are talking about the normal operation of the contacts. I should have known. What you don\'t know fills volumes.
Can\'t say the same about you, a vacuum doesn\'t fill anything.

So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled.

--

Rick C.

-++- Get 1,000 miles of free Supercharging
-++- Tesla referral code - https://ts.la/richard11209

 

[Fred Bloggs]

On Monday, July 3, 2023 at 2:50:09 PM UTC-4, Ricky wrote:

On Monday, July 3, 2023 at 12:05:13 PM UTC-4, Fred Bloggs wrote:
On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote:
On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote:
On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote:
On Sunday, July 2, 2023 at 6:27:57 PM UTC-4, Fred Bloggs wrote:
On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote:
On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote:
On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote:

It\'s absolutely clear to me, and no doubt to many others on here that
you have no idea. Which part of 3kW heater running off 250V don\'t you
understand ? I want to halve the power in the 3kW heater. Show me the
maths...
If you want to halve the power in a heater, use an electric oven control
knob. Those are available as repair parts, and they cycle the heating
element they control, to give proportional heat control.

Every broken down cooktop has four of those, and three of \'em will work.
I\'ve never seen an oven or stove control that gave proportional control. Every one I\'ve seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?
Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.
Hmmm... I am talking about the sort of burners or ovens in the home used for cooking.
So am I.
Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!
Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously?
1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal \"thermostat\" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I\'ve had it apart before.

Those products are totally shielded against throwing sparks outside the enclosure or they couldn\'t be sold. The water heater controls are beneath a sealed cover that\'s screwed down, that\'s what\'s called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise.
So, when you talk about \"throwing sparks\", you are talking about the normal operation of the contacts. I should have known. What you don\'t know fills volumes.
Can\'t say the same about you, a vacuum doesn\'t fill anything.
So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled.

None of them are in enclosures outside the appliance. You wouldn\'t want to use an oven control as they\'re on the pricey side. They don\'t use a bimetal.. They use a diaphragm activated switch with adjustable bias for temperature control. The diaphragm is activated by a calibrated gas pressure in a capillary tube in the oven chamber. The electronic control ovens use an RTD sensor enclosed in copper tube projecting into the oven chamber. It will be tricky adapting either of them to deliver half power to an immersion heater. It\'s a waste of time to even discuss it.

--

Rick C.

-++- Get 1,000 miles of free Supercharging
-++- Tesla referral code - https://ts.la/richard11209

 

[Ricky]

On Monday, July 3, 2023 at 3:35:44 PM UTC-4, Fred Bloggs wrote:

On Monday, July 3, 2023 at 2:50:09 PM UTC-4, Ricky wrote:
On Monday, July 3, 2023 at 12:05:13 PM UTC-4, Fred Bloggs wrote:
On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote:
On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote:
On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote:
On Sunday, July 2, 2023 at 6:27:57 PM UTC-4, Fred Bloggs wrote:
On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote:
On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote:
On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote:

It\'s absolutely clear to me, and no doubt to many others on here that
you have no idea. Which part of 3kW heater running off 250V don\'t you
understand ? I want to halve the power in the 3kW heater. Show me the
maths...
If you want to halve the power in a heater, use an electric oven control
knob. Those are available as repair parts, and they cycle the heating
element they control, to give proportional heat control.

Every broken down cooktop has four of those, and three of \'em will work.
I\'ve never seen an oven or stove control that gave proportional control. Every one I\'ve seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?
Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.
Hmmm... I am talking about the sort of burners or ovens in the home used for cooking.
So am I.
Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!
Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously?
1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal \"thermostat\" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I\'ve had it apart before.

Those products are totally shielded against throwing sparks outside the enclosure or they couldn\'t be sold. The water heater controls are beneath a sealed cover that\'s screwed down, that\'s what\'s called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise.
So, when you talk about \"throwing sparks\", you are talking about the normal operation of the contacts. I should have known. What you don\'t know fills volumes.
Can\'t say the same about you, a vacuum doesn\'t fill anything.
So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled.
None of them are in enclosures outside the appliance. You wouldn\'t want to use an oven control as they\'re on the pricey side. They don\'t use a bimetal. They use a diaphragm activated switch with adjustable bias for temperature control. The diaphragm is activated by a calibrated gas pressure in a capillary tube in the oven chamber. The electronic control ovens use an RTD sensor enclosed in copper tube projecting into the oven chamber. It will be tricky adapting either of them to deliver half power to an immersion heater. It\'s a waste of time to even discuss it.

LOL! I\'ve never seen you to not be happy wasting everyone\'s time.

I never said this was a good idea. I said it is not proportional. whit3rd referred to the on/off thermostat control as \"proportional\" which I corrected. You then jumped in blabbing about limited power. You continue to backpedal until we are finally in agreement. But now you are still talking about using this control. You should bring that up with someone else. It\'s not a point I was ever discussing.

--

Rick C.

-+++ Get 1,000 miles of free Supercharging
-+++ Tesla referral code - https://ts.la/richard11209

 

[TTman]

No.
...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
  If I apply 250 V across 40 Ohms, I get 125V across each resistor.


Correct.  Now for goodness sake do the next step in the math.

125 volts across 20 ohms produces 781.25 watts in the pot
heater and 781.25 watts in the added resistor.
But you said you want 1500 watts in the pot heater.

Ed
That\'s a surprise...

--
This email has been checked for viruses by Avast antivirus software.
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[ehsjr]

On 7/2/2023 7:35 PM, Don wrote:

Ed wrote:
Don wrote:
Ricky wrote:
Don wrote:
Ricky wrote:
TTman wrote:
Ed wrote:

The communication hasn\'t worked so far, perhaps it\'s been a bit
unclear. I\'ll try a bit differently. The key point is that doubling
the resistance does NOT halve the power.

Take the resistance of your heater (HR), which is about 20.8 ohms,
and figure what voltage needs to be applied to it to run it at 1500
watts. Here\'s the math:
Solve for heater resistance (HR):
P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83

Now what voltage must be applied to HR to equal 1500 watts?
1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so
E = sqroot 31250 so E = 176.77volts

Ok, once you know that voltage you can solve for the rest.
We need to add a series resistance (AR) to drop the voltage
across HR from 250 to 176.77volts

Compute the current through the series string of HR plus AR

Solve for I through HR at 1500 watts:
P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps

The total voltage (250) minus VHR (176.77) equals VAR, the
voltage across the (AddedResistance) AR

250-176.77 = 73.223volts

Solve for power dissipated in AR

P=I*E= 8.486*73.223=621.32 watts in AR

I hope this makes it a bit clearer.
Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.

and... you get half the current, so each resistor will dissipate a quarte
I hope this makes it a bit clearer.
Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.

and... you get half the current, so each resistor will dissipate a quarter
of the original power. The total power will be half of the original power,
but if I understand your limitations on the design, the series resistor can
not be used to heat the pot.

I\'m not going to continue to beat you over the heat to get you to pay
attentn to the equations. But here are equations for power, one more time.

P = E * I
P = E^2 / R
P = I^2 * R

Apply them to the appropriate devices, rather than just saying, \"twice the
resistance, half the power\". Also, try solving for the actual current and
the voltage on each device. This isn\'t hard, but sometimes it takes a bit
of work to get used to it.

Thanks guys. You helped me put it all together. If we keep 250VAC
constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind
misrepresents the V^2 constant as a slope in a linear equation.
But V^2 is not a slope because P=V^2/R is not linear. The
multiplicative inverse term, 1/R, makes it non-linear.
Here\'s a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it
looks similar to a 1/x plot.

This (so far unanswered) question was previously asked by me:

Does the -3 dB half power point play a role with kW (eg not kWh)?

Ordinarily the half power point pertains to bandwidth. It\'s a little
eccentric to apply it to a constant frequency. But it can be done. By
inspection, the half power point of 3000 W yields 40 Ohms.

Note.

[1] octave code:

R=linspace(10,50,40);
P=250^2 ./ R;
plot(R, P);
xlabel(\'R (Ohms)\');
ylabel(\'P (Watts)\');

snipped

You are solving an irrelevant problem. No one cares about reducing the
total power to half. That\'s the mistake that TTman is making. I
originally said he could add an equivalent heater in series with the one
he\'s already using and the total power would be half. He said that won\'t
work because only the pot heater is producing useful heat.

If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts,
how much resistance do you need to add in series to make the pot heater
dissipate 1,500 watts?

By inspection, from my plot, you need to add 0 Ohms in series in order
to dissipate 3000 W from 250 VAC. On the other hand, when you add 20
additional Ohms, in series to the original 20 Ohms, you obtain 1500 W.
Again, by inspection.
As stated above, 250 VAC^2 is a constant.

No. You make a resistive voltage divider when you add a series
resistor. The voltage across the original resistor is reduced
(except if the added R is 0 ohms). Whatever leads one to
think the voltage is constant is erroneous or not properly
understood.

\"If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant\" (as
stated above) is my premise. This 250VAC is treated as Vmains by me.
My mind mechanically treats Vmains as a constant - because it is.
Most electric outlets in my world source about 120 VAC regardless of the
load connected.
When Vmains is held constant, a doubling of Rmains halves Pmains.

Correct. He\'ll get half power out of the combination of HR (the heater
resistance) plus AR (the added resistance). The original power was 3000
watts. Half power is 1500 watts. So he\'ll get 750 watts out of AR and
750 watts out of HR.

BUT THE OP WANTS 1500 WATTS OUT OF HR.

Your math solves the wrong problem.
From the op: \"As part of an experiment with solar, I want to drive my
220V 3kW immersion heater with a suitable diode so that the effective
power is ~1.5kW...\"

Ed

<snip>

 

[Ricky]

On Monday, July 3, 2023 at 5:40:51 PM UTC-4, TTman wrote:

No.
...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.


Correct. Now for goodness sake do the next step in the math.

125 volts across 20 ohms produces 781.25 watts in the pot
heater and 781.25 watts in the added resistor.
But you said you want 1500 watts in the pot heater.

Ed
That\'s a surprise...

That\'s the point. You don\'t understand this and seem to be unwilling to learn anything new. You say you don\'t understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

--

Rick C.

+--- Get 1,000 miles of free Supercharging
+--- Tesla referral code - https://ts.la/richard11209

 

[Ricky]

On Monday, July 3, 2023 at 6:54:37 PM UTC-4, ehsjr wrote:

On 7/2/2023 7:35 PM, Don wrote:
Ed wrote:
Don wrote:
Ricky wrote:
Don wrote:
Ricky wrote:
TTman wrote:
Ed wrote:

The communication hasn\'t worked so far, perhaps it\'s been a bit
unclear. I\'ll try a bit differently. The key point is that doubling
the resistance does NOT halve the power.

Take the resistance of your heater (HR), which is about 20.8 ohms,
and figure what voltage needs to be applied to it to run it at 1500
watts. Here\'s the math:
Solve for heater resistance (HR):
P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83

Now what voltage must be applied to HR to equal 1500 watts?
1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so
E = sqroot 31250 so E = 176.77volts

Ok, once you know that voltage you can solve for the rest.
We need to add a series resistance (AR) to drop the voltage
across HR from 250 to 176.77volts

Compute the current through the series string of HR plus AR

Solve for I through HR at 1500 watts:
P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps

The total voltage (250) minus VHR (176.77) equals VAR, the
voltage across the (AddedResistance) AR

250-176.77 = 73.223volts

Solve for power dissipated in AR

P=I*E= 8.486*73.223=621.32 watts in AR

I hope this makes it a bit clearer.
Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.

and... you get half the current, so each resistor will dissipate a quarte
I hope this makes it a bit clearer.
Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.

and... you get half the current, so each resistor will dissipate a quarter
of the original power. The total power will be half of the original power,
but if I understand your limitations on the design, the series resistor can
not be used to heat the pot.

I\'m not going to continue to beat you over the heat to get you to pay
attentn to the equations. But here are equations for power, one more time.

P = E * I
P = E^2 / R
P = I^2 * R

Apply them to the appropriate devices, rather than just saying, \"twice the
resistance, half the power\". Also, try solving for the actual current and
the voltage on each device. This isn\'t hard, but sometimes it takes a bit
of work to get used to it.

Thanks guys. You helped me put it all together. If we keep 250VAC
constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind
misrepresents the V^2 constant as a slope in a linear equation.
But V^2 is not a slope because P=V^2/R is not linear. The
multiplicative inverse term, 1/R, makes it non-linear.
Here\'s a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it
looks similar to a 1/x plot.

This (so far unanswered) question was previously asked by me:

Does the -3 dB half power point play a role with kW (eg not kWh)?

Ordinarily the half power point pertains to bandwidth. It\'s a little
eccentric to apply it to a constant frequency. But it can be done. By
inspection, the half power point of 3000 W yields 40 Ohms.

Note.

[1] octave code:

R=linspace(10,50,40);
P=250^2 ./ R;
plot(R, P);
xlabel(\'R (Ohms)\');
ylabel(\'P (Watts)\');

snipped

You are solving an irrelevant problem. No one cares about reducing the
total power to half. That\'s the mistake that TTman is making. I
originally said he could add an equivalent heater in series with the one
he\'s already using and the total power would be half. He said that won\'t
work because only the pot heater is producing useful heat.

If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts,
how much resistance do you need to add in series to make the pot heater
dissipate 1,500 watts?

By inspection, from my plot, you need to add 0 Ohms in series in order
to dissipate 3000 W from 250 VAC. On the other hand, when you add 20
additional Ohms, in series to the original 20 Ohms, you obtain 1500 W..
Again, by inspection.
As stated above, 250 VAC^2 is a constant.

No. You make a resistive voltage divider when you add a series
resistor. The voltage across the original resistor is reduced
(except if the added R is 0 ohms). Whatever leads one to
think the voltage is constant is erroneous or not properly
understood.

\"If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant\" (as
stated above) is my premise. This 250VAC is treated as Vmains by me.
My mind mechanically treats Vmains as a constant - because it is.
Most electric outlets in my world source about 120 VAC regardless of the
load connected.
When Vmains is held constant, a doubling of Rmains halves Pmains.
Correct. He\'ll get half power out of the combination of HR (the heater
resistance) plus AR (the added resistance). The original power was 3000
watts. Half power is 1500 watts. So he\'ll get 750 watts out of AR and
750 watts out of HR.

BUT THE OP WANTS 1500 WATTS OUT OF HR.

Your math solves the wrong problem.
From the op: \"As part of an experiment with solar, I want to drive my
220V 3kW immersion heater with a suitable diode so that the effective
power is ~1.5kW...\"

The OP\'s problem has been solved many times in this thread, in different ways. But the OP doesn\'t understand any of them. Now, people are just trying to educate the OP.

--

Rick C.

+--+ Get 1,000 miles of free Supercharging
+--+ Tesla referral code - https://ts.la/richard11209

 

[Don]

Ed wrote:
> Don wrote:

<snip>

\"If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant\" (as
stated above) is my premise. This 250VAC is treated as Vmains by me.
My mind mechanically treats Vmains as a constant - because it is.
Most electric outlets in my world source about 120 VAC regardless of the
load connected.
When Vmains is held constant, a doubling of Rmains halves Pmains.

Correct. He\'ll get half power out of the combination of HR (the heater
resistance) plus AR (the added resistance). The original power was 3000
watts. Half power is 1500 watts. So he\'ll get 750 watts out of AR and
750 watts out of HR.

BUT THE OP WANTS 1500 WATTS OUT OF HR.

Your math solves the wrong problem.
From the op: \"As part of an experiment with solar, I want to drive my
220V 3kW immersion heater with a suitable diode so that the effective
power is ~1.5kW...\"

There\'s actually a couple of problems. One problem is people who try to
shout down wrong thought of how doubling a resistor halves the power. My
math solves this problem by illustrating an exception.
If you keep reading my previous followup (the part you snipped)
you\'ll discover how my math eventually illustrates a solution to the
OP\'s 1500 W problem. But, you need to read my followup in its entirety
to see it.

In the end, sloppy nomenclature caused this thread\'s miscommunication.

\"When I use [the word volt],\" Humpty Dumpty said in rather a
scornful tone, \"it means just what I choose it to mean -
neither more nor less.\"

Danke,

--
Don, KB7RPU, https://www.qsl.net/kb7rpu
There was a young lady named Bright Whose speed was far faster than light;
She set out one day In a relative way And returned on the previous night.

 

[TTman]

On 04/07/2023 01:05, Ricky wrote:

On Monday, July 3, 2023 at 5:40:51 PM UTC-4, TTman wrote:
No.
...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.


Correct. Now for goodness sake do the next step in the math.

125 volts across 20 ohms produces 781.25 watts in the pot
heater and 781.25 watts in the added resistor.
But you said you want 1500 watts in the pot heater.

Ed
That\'s a surprise...

That\'s the point. You don\'t understand this and seem to be unwilling to learn anything new. You say you don\'t understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

Yes, I see it, but it defies apparent simplistic logic.I see it\'s V xV
or IxI for power, hence the quarter/sq root. No wonder nothing I
designed/built worked...

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[Ricky]

On Tuesday, July 4, 2023 at 7:02:34 AM UTC-4, TTman wrote:

On 04/07/2023 01:05, Ricky wrote:
On Monday, July 3, 2023 at 5:40:51 PM UTC-4, TTman wrote:
No.
...HR= 20 Ohms. Add another resistor in series = 20 Ohms
making Rt - 40 Ohms.
If I apply 250 V across 40 Ohms, I get 125V across each resistor.


Correct. Now for goodness sake do the next step in the math.

125 volts across 20 ohms produces 781.25 watts in the pot
heater and 781.25 watts in the added resistor.
But you said you want 1500 watts in the pot heater.

Ed
That\'s a surprise...

That\'s the point. You don\'t understand this and seem to be unwilling to learn anything new. You say you don\'t understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

Yes, I see it, but it defies apparent simplistic logic.

Yes, that\'s what we\'ve tried to tell you. You can\'t just say, double resistance, so half power in the pot.

I see it\'s V xV
or IxI for power, hence the quarter/sq root. No wonder nothing I
designed/built worked...

Yeah, if you ignore the math, things don\'t go well.

What made the lightbulb come on?

--

Rick C.

+-+- Get 1,000 miles of free Supercharging
+-+- Tesla referral code - https://ts.la/richard11209

 

[TTman]

That\'s a surprise...

That\'s the point. You don\'t understand this and seem to be unwilling to learn anything new. You say you don\'t understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

Yes, I see it, but it defies apparent simplistic logic.

Yes, that\'s what we\'ve tried to tell you. You can\'t just say, double resistance, so half power in the pot.


I see it\'s V xV
or IxI for power, hence the quarter/sq root. No wonder nothing I
designed/built worked...

Yeah, if you ignore the math, things don\'t go well.

What made the lightbulb come on?

I went back to basics. I admit I just didn\'t think about this properly
at all. I\'m 72 and clearly my brain is screwed and incapable of clear
thinking. Never mind, I won\'t be long for this world and I won\'t have to
think about things like this.

--
This email has been checked for viruses by Avast antivirus software.
www.avast.com

 

[Ricky]

On Tuesday, July 4, 2023 at 12:36:10 PM UTC-4, TTman wrote:

That\'s a surprise...

That\'s the point. You don\'t understand this and seem to be unwilling to learn anything new. You say you don\'t understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

Yes, I see it, but it defies apparent simplistic logic.

Yes, that\'s what we\'ve tried to tell you. You can\'t just say, double resistance, so half power in the pot.


I see it\'s V xV
or IxI for power, hence the quarter/sq root. No wonder nothing I
designed/built worked...

Yeah, if you ignore the math, things don\'t go well.

What made the lightbulb come on?

I went back to basics. I admit I just didn\'t think about this properly
at all. I\'m 72 and clearly my brain is screwed and incapable of clear
thinking. Never mind, I won\'t be long for this world and I won\'t have to
think about things like this.

So, that is what I have to look forward to in three years? Crap!

I nearly stopped working for some four or five years, and I the next time I had a serious task, I could tell I had lost something. But then I got a sizeable job and had to work intensely for months. It pretty much all came back. I\'m not as quick with numbers, not being able to do as much in my head now. But the logic is all there.

One of the things I do to get to sleep is to recite the alphabet backwards. So, I\'m ready for the intoxication tests now.

--

Rick C.

+-++ Get 1,000 miles of free Supercharging
+-++ Tesla referral code - https://ts.la/richard11209

 

[TTman]

I went back to basics. I admit I just didn\'t think about this properly
at all. I\'m 72 and clearly my brain is screwed and incapable of clear
thinking. Never mind, I won\'t be long for this world and I won\'t have to
think about things like this.

So, that is what I have to look forward to in three years? Crap!

I nearly stopped working for some four or five years, and I the next time I had a serious task, I could tell I had lost something. But then I got a sizeable job and had to work intensely for months. It pretty much all came back. I\'m not as quick with numbers, not being able to do as much in my head now. But the logic is all there.

One of the things I do to get to sleep is to recite the alphabet backwards. So, I\'m ready for the intoxication tests now.

I count / visualise binary up to 127...It\'s interesting at 40,80 and
120... utter nerd

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