On Saturday, July 1, 2023 at 5:15:54 AM UTC-4, Phil Allison wrote:> Mad Ricky wrote: > --------------------------- > > > Phil Allison wrote: > > > > > > > > > > > Clive Arthur wrote: > > > > > > > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > > > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > > > > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > > > > > > > > > No heat losses and clean current waveform. > > > > > > > > > > Whacked out power factor. > > > > > > > > > > > ** LOL - what complete nonsense. > > > > > > > > So, another topic we've found that Phil A. knows nothing about. > > > ** FFS you do LOVE making idiotic assumptions. > > > > > > Using a cap in series to get 1/2 power produces a PF of 0.7, which is not "wacked out" - whatever that means. > > > The 3 posted ways to half the power (using a diode, triac or cap) have the same PF, making nonsense of your claim. > > > > > > PF = true power / VA where V and A are rms values. > > > > > > The true power, is 1500W in each case. > > > The rms current value is also the same since the load is a fixed resistance. > > > The current wave is the same in the load and the supply each time. > > I should also point out that your analysis is faulty, > ** No it fucking is not. > > > > The current is in phase with the voltage on the resistor, and so, out of phase with the line voltage. > > This means the product of the RMS of the current and the RMS of the line voltage are higher than the power in the resistor. > > The excess power is flowing in and out of the capacitor, out of phase with the real power. > > This extra power causes line loses significantly larger than if a real load were being powered. > > > ** FFS you just over explained how the VA becomes greater than the true watts. > > Take you long to Google that ?As usual. You can't understand what is being said, so you use ridicule, rather than trying to discuss it. We all know you are just a profane loser. You can't even use Google to find some web pages that will explain it to you. Hell, I gave you a link to one page that explains it pretty well. I guess you are just too lazy to even try to understand. Whatever. It's not like you will ever do any work where you actually need to know electronics. -- Rick C. +-- Get 1,000 miles of free Supercharging +-- Tesla referral code - https://ts.la/richard11209
half wave rectifier
Started by ●June 29, 2023
Reply by ●July 1, 20232023-07-01
Reply by ●July 1, 20232023-07-01
On 30/06/2023 23:08, Ricky wrote:> On Friday, June 30, 2023 at 5:31:50 PM UTC-4, TTman wrote: >> On 30/06/2023 16:39, Ricky wrote: >>> On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote: >>>>>>> P = E^2 / R >>>>>>> >>>>>>> If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power. >>>>>>> >>>>>> No, you don't understand this english stuff. Twin heaters are rare in the UK >>>>> >>>>> Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That's not exactly "standard" practice anywhere. >>>> Correct. Not standard. Read my original post that explains why. >>>>> >>>>> If you have mechanical limitations, that's what you have. Electrically, this makes sense. >>>> No mechanical limitations. >>>>> >>>>> You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level. >>>>> >>>> No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating? >>> >>> You don't design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job. >> I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 = >> 20 Ohms ish, give or take a bit. So, back in the day, to halve the >> power, one had to double the resistance. Still with me ? So double 20 = >> 40. Correct? So what I need is a 20 ohm resistor capable of dissipating >> half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts >> will go in heat into the resistor, the other 1500 watts in the 3kW >> heater element will go to heat the water. >> Have things changed so much in modern technology that the basic >> equations have been repaced with garbage ? P=ZxI and V=I/R back in the day. >> So what are these new equations that tell me I need 600 Ohms at 700 watts ? >> >>> >>> https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA > > > The equations are all the same. But you have to apply them correctly. Maybe I don't understand your setup, but my mental image is a resistance heater built into a pot of some sort. You want to heat the pot at half power. Your calculations are figuring the power in the entire load. Only the power in the pot heater is useful. The power in the added resistors is waste heat, no? > > My calculations give you half power in the pot heater. Please remember that the added resistor will cut both the voltage and the current to the pot heater. If you are trying to do something else, I have no understanding of it at this point. >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... -- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●July 1, 20232023-07-01
On Saturday, July 1, 2023 at 3:38:59 AM UTC-4, Phil Allison wrote:> Mad Ricky wrote: > --------------------------- > Phil Allison wrote: > > > > > > > Clive Arthur wrote: > > > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > > > > > No heat losses and clean current waveform. > > > > > > Whacked out power factor. > > > > > > > ** LOL - what complete nonsense. > > > > So, another topic we've found that Phil A. knows nothing about. > ** FFS you do LOVE making idiotic assumptions. > > Using a cap in series to get 1/2 power produces a PF of 0.7, which is not "wacked out" - whatever that means. > The 3 posted ways to half the power (using a diode, triac or cap) have the same PF, making nonsense of your claim. > > PF = true power / VA where V and A are rms values. > > The true power, is 1500W in each case. > The rms current value is also the same since the load is a fixed resistance. > The current wave is the same in the load and the supply each time.Capacitor and inductor series reactive elements are out of the question because the 1500 W heater draws about 10 A- making those components very pricey. A second-hand variac should be doable, but in this VA range it will be kind of big. If all he wants is a 1500 W immersion heater, Walmart carries one for USD$9.99. For some reason aliexpress brings up ridiculous search results.> > > > ....... Phil
Reply by ●July 1, 20232023-07-01
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.
Reply by ●July 1, 20232023-07-01
On Saturday, July 1, 2023 at 8:33:42 AM UTC-4, TTman wrote:> On 30/06/2023 23:08, Ricky wrote: > > On Friday, June 30, 2023 at 5:31:50 PM UTC-4, TTman wrote: > >> On 30/06/2023 16:39, Ricky wrote: > >>> On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote: > >>>>>>> P = E^2 / R > >>>>>>> > >>>>>>> If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power. > >>>>>>> > >>>>>> No, you don't understand this english stuff. Twin heaters are rare in the UK > >>>>> > >>>>> Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That's not exactly "standard" practice anywhere. > >>>> Correct. Not standard. Read my original post that explains why. > >>>>> > >>>>> If you have mechanical limitations, that's what you have. Electrically, this makes sense. > >>>> No mechanical limitations. > >>>>> > >>>>> You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level. > >>>>> > >>>> No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating? > >>> > >>> You don't design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job. > >> I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 = > >> 20 Ohms ish, give or take a bit. So, back in the day, to halve the > >> power, one had to double the resistance. Still with me ? So double 20 = > >> 40. Correct? So what I need is a 20 ohm resistor capable of dissipating > >> half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts > >> will go in heat into the resistor, the other 1500 watts in the 3kW > >> heater element will go to heat the water. > >> Have things changed so much in modern technology that the basic > >> equations have been repaced with garbage ? P=ZxI and V=I/R back in the day. > >> So what are these new equations that tell me I need 600 Ohms at 700 watts ? > >> > >>> > >>> https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA > > > > > > The equations are all the same. But you have to apply them correctly. Maybe I don't understand your setup, but my mental image is a resistance heater built into a pot of some sort. You want to heat the pot at half power. Your calculations are figuring the power in the entire load. Only the power in the pot heater is useful. The power in the added resistors is waste heat, no? > > > > My calculations give you half power in the pot heater. Please remember that the added resistor will cut both the voltage and the current to the pot heater. If you are trying to do something else, I have no understanding of it at this point. > > > It's absolutely clear to me, and no doubt to many others on here that > you have no idea.I don't know why you need to be rude, just because you don't understand what I'm saying. You keep saying you want to cut the power in the 3 kW pot heater, and I'm showing you how to do that. Instead of saying you don't under stand how adding a smaller resistor does that, you try to insult me. How does that make sense?> 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...I already did, but here it is again. P = E^2 / Rh, you want to reduce the voltage on the heater by adding a resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the added resistor. So, how large does Rs need to be for the power in Rh to be half the original value (1,500 W vs. 3,000 W)? So, now we have Ph = Eh^2 / Rh. Since Ph is 1/2 * P from above, 2 * Eh^2 / Rh = E^2 / Rh. Multiply by Rh / Eh^2 to get, √ 2 = E / Eh = 1.414, Eh = E * 0.707. So now, we know the voltage needed on the heater for it to dissipate half the original power. Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh). 0.293 Rs + 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414. Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 6.6861 ohms. Was there any part you didn't follow? I would point out that anyone familiar with using dB to compare power levels has likely discovered that -3 dB is half power, but 0.707 of the voltage and current. Most designers simply know this off the top of their heads. I take it you don't use such calculations very often. -- Rick C. +-+ Get 1,000 miles of free Supercharging +-+ Tesla referral code - https://ts.la/richard11209
Reply by ●July 1, 20232023-07-01
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? -- Rick C. ++- Get 1,000 miles of free Supercharging ++- Tesla referral code - https://ts.la/richard11209
Reply by ●July 1, 20232023-07-01
On 7/1/2023 8:33 AM, TTman wrote:> On 30/06/2023 23:08, Ricky wrote: >> On Friday, June 30, 2023 at 5:31:50 PM UTC-4, TTman wrote: >>> On 30/06/2023 16:39, Ricky wrote: >>>> On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote: >>>>>>>> P = E^2 / R >>>>>>>> >>>>>>>> If you string two heaters in series, you have the same voltage, >>>>>>>> but twice the resistance, so half the power. >>>>>>>> >>>>>>> No, you don't understand this english stuff. Twin heaters are >>>>>>> rare in the UK >>>>>> >>>>>> Not sure what you are talking about. You want to rig a diode to >>>>>> create half wave AC to drive a heater. That's not exactly >>>>>> "standard" practice anywhere. >>>>> Correct. Not standard. Read my original post that explains why. >>>>>> >>>>>> If you have mechanical limitations, that's what you have. >>>>>> Electrically, this makes sense. >>>>> No mechanical limitations. >>>>>> >>>>>> You could add a resistance in series. It would need to be about >>>>>> 30% of the resistance of the heater. The main element would >>>>>> dissipate half of the original heat. The heat in this added >>>>>> resistance would be about 22% of the heat dissipated in the >>>>>> original heating element. One nice thing about this is that you >>>>>> can tailor the heat in the main element to other than half the >>>>>> original heat level. >>>>>> >>>>> No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating? >>>> >>>> You don't design much electronics, do you? You need about 600 ohms >>>> at 700 watts. Six 1 ohm, 150 watt resistors will do the job. >>> I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 = >>> 20 Ohms ish, give or take a bit. So, back in the day, to halve the >>> power, one had to double the resistance. Still with me ? So double 20 = >>> 40. Correct? So what I need is a 20 ohm resistor capable of dissipating >>> half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts >>> will go in heat into the resistor, the other 1500 watts in the 3kW >>> heater element will go to heat the water. >>> Have things changed so much in modern technology that the basic >>> equations have been repaced with garbage ? P=ZxI and V=I/R back in >>> the day. >>> So what are these new equations that tell me I need 600 Ohms at 700 >>> watts ? >>> >>>> >>>> https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA >> >> >> The equations are all the same. But you have to apply them >> correctly. Maybe I don't understand your setup, but my mental image >> is a resistance heater built into a pot of some sort. You want to >> heat the pot at half power. Your calculations are figuring the power >> in the entire load. Only the power in the pot heater is useful. The >> power in the added resistors is waste heat, no? >> >> My calculations give you half power in the pot heater. Please >> remember that the added resistor will cut both the voltage and the >> current to the pot heater. If you are trying to do something else, I >> have no understanding of it at this point. >> > 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... >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. Ed
Reply by ●July 1, 20232023-07-01
> > >> 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... > > I already did, but here it is again. > > P = E^2 / Rh, you want to reduce the voltage on the heater by adding a resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the added resistor. So, how large does Rs need to be for the power in Rh to be half the original value (1,500 W vs. 3,000 W)? > > So, now we have Ph = Eh^2 / Rh. Since Ph is 1/2 * P from above, 2 * Eh^2 / Rh = E^2 / Rh. > > Multiply by Rh / Eh^2 to get, √ 2 = E / Eh = 1.414, Eh = E * 0.707. > > So now, we know the voltage needed on the heater for it to dissipate half the original power. > > Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh). 0.293 Rs + 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414. > > Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 6.6861 ohms. > > Was there any part you didn't follow?None of it. You're over thinking it. Rh = Rs. Rh = ~18 Ohms. therefore Rs = 18 ohms.Half the power = twice the resistance.Simple. 1500W into Rh and 1500W into Rs.> > I would point out that anyone familiar with using dB to compare power levels has likely discovered that -3 dB is half power, but 0.707 of the voltage and current. Most designers simply know this off the top of their heads. I take it you don't use such calculations very often. >-- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●July 1, 20232023-07-01
> > 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. > EdToo 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. -- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●July 1, 20232023-07-01
On Saturday, July 1, 2023 at 6:09:30 PM UTC-4, TTman wrote:> > > > > >> 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... > > > > I already did, but here it is again. > > > > P = E^2 / Rh, you want to reduce the voltage on the heater by adding a resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the added resistor. So, how large does Rs need to be for the power in Rh to be half the original value (1,500 W vs. 3,000 W)? > > > > So, now we have Ph = Eh^2 / Rh. Since Ph is 1/2 * P from above, 2 * Eh^2 / Rh = E^2 / Rh. > > > > Multiply by Rh / Eh^2 to get, √ 2 = E / Eh = 1.414, Eh = E * 0.707. > > > > So now, we know the voltage needed on the heater for it to dissipate half the original power. > > > > Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh). 0.293 Rs + 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414. > > > > Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 6.6861 ohms. > > > > Was there any part you didn't follow? > None of it.Yes, that's what I figured.> You're over thinking it. Rh = Rs. Rh = ~18 Ohms. therefore > Rs = 18 ohms.Half the power = twice the resistance.Simple. 1500W into Rh > and 1500W into Rs.I'm really sorry you were not willing to try to understand the math. Unfortunately, such simplistic talking about the problem, is not remotely the same as math. Once you learn how to do the math, come back and I'll be happy to discuss this with you. Meanwhile, you will find others agree with me, and not you. Try to recognize when there is something you need to learn. -- Rick C. +++ Get 1,000 miles of free Supercharging +++ Tesla referral code - https://ts.la/richard11209
