so after first start-up : no smoke

6,04 V on the heating circuit

I have 245 V AC on the output of my transformer but 273 V DC at the valve with a 10K resistor (4w)

I'm a noob with the ohm law…

I've a 15k and a 22k resistors (both 4w), which is the better replacement one in order to get a proper voltage at the valve?

Thanks for your help

If you measure VDC before the drop resistor, you can check exactly how much VDC you are dropping with your installed 10K resistor.

I will assume its 340VDC before the drop, and you measured 273VCD after, so the drop = 67V.

**Ohms law: Voltage(V) = current(I) x resistance(R).**

Calculating I

We know the voltage, V (67) and the resistance, R (10K), so we can fill in ohms law to find the unknown I.

67(V) = I x 10000(R) or 67(V)/10000(R) = I

I = 0,0067 ampere (6,7ma)

Calculating VFor a target value of 250V, you will want a voltage drop from 340 to 250 = 90V.

Calculating R

Now that we know I and V, we can find R

90(V) = 0,0067(I) x R or 0,0067(I)/90(V) = R

R = 13433.

I have tried and had no luck getting close to the expected value calculating it like this, but I am not sure if its because of current draw changes, bad math skills (I hope someone will correct me if thats the case) or something third. I would just install the 15K resistor and see what values I get.

If you try to calculate it, remember to measure VDC before the drop resistor to insert the correct value instead of the 340V

Gustav