The Shop > Electronics & IC Programing
Water Heater Monitoring
vtsteam:
--- Quote from: awemawson on January 05, 2015, 02:56:57 AM ---Double the voltage and quadruple the power as wattage is 'I squared R' R stays the same and I is proportional to V so the wattage on 240 is four time that on 120
--- End quote ---
:scratch: Andrew, isn't Sparky already measuring (not calculating) the actual increased current through the 240V circuit? So power equals measured current times known voltage....yes?
Not sure I'm explaining this well :(......Sparky's not trying to estimate what would happen if he doubled voltage through a circuit. He's trying to correct a mistaken assumption about what the voltage was in a circuit where the current is known.
(I hope I've got that right........tricky, this! :scratch:)
DavidA:
Sparky,
Perhaps you can provide a sketch of this set up.
At first I thought that you were using the coil as an inductive pick up with the the heater supply lead (one side only) running through it.
But now I'm not sure.
Dave.
vtsteam:
Yes, David, and Sparky are you measuring short circuit current for your pickup loop without any resistor in series?
awemawson:
:scratch: Andrew, isn't Sparky already measuring (not calculating) the actual increased current through the 240V circuit? So power equals measured current times known voltage....yes?
Steve, from my re-reading of his post no, he seems to be pro-rata'ing his 12.2 W/mA figure despite these being mA at twice the voltage. But we need him to clarify
sparky961:
Sorry for my lack of clarity, guys. I was trying to give only facts to avoid skewing the discussion with any of my own mistaken assumptions. Seems I was a bit lean on the details.
I'm using an inductive pickup, 100 turns of wire on a toroid to 1 turn carrying the current to the water heater. The voltage is 220 or close thereto (I'll verify the exact number to fine tune my calculations later.) I missed stating that in the original post, but I was thinking it would make sense to double the displayed value (problem #1?). I figured if this got me close to the nameplate value it would validate my assumption. It didn't.
My setup is the pickup loop with no resistor (problem #2?), connected to my DVOM on 200mA current setting.
I'm almost embarrassed to have to even ask this because I should be able to figure it out. But AC and power has always been a tough area for me.
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