Thanks, I could easily add a resistor across the CR3110 leads. Based on the chart it looks like a 2K resistor should bring the voltage down to about 8vac. Does that sound right?
Here's how I would approach it. Your water heater is 3380w at 240v. That's 14amps. Plugging that into the formula V = 14*2000/3100 = 9vac output. That should be OK since its less than spec-ed 10vac max. I would use a 1K resistor V = 14*1000/3100 = 4.5vac.
Going back a couple of weeks to your original post, the circuit has the resistor on the DC side of the bridge. I don't know if or how that's going to affect the formula and resulting voltage. I'd put the resistor on the AC side so you know exactly what's going into the bridge and therefore out of it.
The CR3110 on my baseboard heater circuit outputs about 9vac. Do you think I should still add a burden resistor?
"What happens if the burden resistor is left off or opens during operation? The output voltage will rise trying to develop current until it reaches the saturation voltage of the coil at that frequency. At that point, the voltage will cease to rise and the transformer will add no additional impedance to the driving current. Therefore, without a burden resistor, the output voltage of a current transformer will be its saturation voltage at the operating frequency."
Well, there's that.
Yes, size the burden resistor based on the amps being drawn and the required output voltage. This is a good article about CTs and burden resistors. Quote above is from that article.
http://powerelectronics.com/mag/power_expl...nt_transformer/