half wave rectifier

Sorry, my big mistake :)

A big f*ck-off Variac is what you need, if you can find/borrow one.

What nonsense. If you time slice the power using a diode, you don't need to use any equation other than 1/2 P. Do you have some other image of what the OP wants to do?

OK. 3 kWh half the time is 1.5 kWh. Does the -3 dB half power point play a role with kW (eg not kWh)?

Danke,

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 don't think your inverter is going to like facing a resistive 3kW load every alternate half cycle. Things may break and magic smoke comes out. It is also probably against UK electrical code to do this with a high power load.

It is one thing to run an electric blanket at tens of watts in configurations R+R+diode for the lowest heat, R+R, R, R||R

But is quite another to ask it to supply a load that effectively unbalances the local mains to the extent of 3kW. There are commercial smart switches that divert solar panel electricity to the immersion heater whenever the sun shines and the water is below temperature.

With UK feed in tariffs today this is generally worthwhile unless you are an early adopter on a very favourable ancient tariff.

When you put the second 18R in series the current halves as does the voltage drop across each one so that the power in each of the two resistive loads is then 1/4 of the original 3kW. Total 1.5kW.

He will be wasting 50% of the heat though unless the second load resistor is also an immersion heater inside the hot water tank.

- I presume it was a typo for 6x 1R. BTW I get it as 7.5R 650W

I took his intention being to reduce the load presented to the mains by a factor of two not the power dissipated in the water tank by half.

Only the OP knows what he actually wants to do but he could end up with a very expensive mess by using a big power diode if the grid tie inverter isn't very robustly built! He would be much better off with a properly designed diverter given the lack of knowledge shown already.

I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 =

1. 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 ?

OdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA

Correct. So that the power taken from the grid, or solar or battery will end up at around 1500 Watts.

I know a bit about electronics, in fact a lot in one area but not mains power stuff. My ignorance wants to know why adding an appropriate series diode would likely damage the inverter.? Is it because the load effectively becomes discontinuous?

I have one of those- I-Boost. My mad professor brain was thinking of a cheap alternative. By all accounts, ill founded.Thx.

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.

C = 1/ ( 2.pi.50.18) = 175uF

No heat losses and clean current waveform. .... Phil

Whacked out power factor.

--------------------

...... Phil

So, another topic we've found that Phil A. knows nothing about.

It is not uncommon to add capacitance to mitigate inductive loads, restoring the power factor to close to 1.0. I've never heard of adding inductors to mitigate capacitive power factors. I guess that's because there are so few large capacitive loads, and they would likely be countered by other inductive loads.

The reason a power factor not close to 1.0 is undesired, is that it means the current is flowing out of phase with the voltage, resulting in larger current flow than is needed to transmit the "real" power, with excess power losses in the wiring. The power company doesn't like losing power in their wires. They only bill for the power delivered to your meter, and only the real power at that.

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.

....... Phil

You ignored using a series resistor, which has zero impact on the power factor.

The utility does not care about YOUR real power. They care about THEIR imaginary power which wastes power in the power line. With a power factor of 0.7, the power company would see significantly higher line losses.

"For example, if the load power factor were as low as 0.7, the apparent power would be 1.4 times the real power used by the load." Current would be 1.4 times higher, with the line losses doubled. No, the power company does not like that!

"Utilities typically charge additional costs to commercial customers who have a power factor below some limit, which is typically 0.9 to 0.95."

Both quotes from Wikipedia.

I should also point out that your analysis is faulty, but it will be harder to explain it to you, since it involves complex math, which I'm guessing you don't understand.

While the current in the cap and resistor are the same, also the same as the current in the line, the voltage from the line is not on the resistor. The resistor voltage is E = I * R. However, the voltage on the cap is 90° out of phase with the current. The complex sum of the capacitor voltage and resistor voltage are equal to the line voltage. This is equivalent to the three sides of a right triangle, where the hypotenuse is the line voltage, the cap and resistor voltages are the other two sides with the right angle between them.

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.

If this doesn't make sense to you, try reading some source on power factor, or maybe you need to learn more about complex math?

Good luck,