Resistance device uses 1.0 kW on standalone usage load measurement. I tried using diodes in series to half the power rate to 500 W, Wondering whether to use a 220V to 110 V step down or reduced the resistance coils and then find a way to set the optimum power saving current for it, or better get the power saving induction cooker. (Safety measures was taken, Variable resistor and reduced P=IV
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4\$\begingroup\$ You are aware that if you reduce the power usage of your cooker, you won't be able to get it as hot and it will take longer to cook things? If you're looking at reducing the power usage of a resistive element, just put another resistor in series, that will increase total resistance, which will reduce total current. This can be shown by rearanging V=IR and P=IV formular. \$\endgroup\$– PuffafishCommented May 23, 2023 at 15:09
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1\$\begingroup\$ No -- that would also force power dissipation in the additional resistor - while the total power would be lower, likely you would need a resistor capable of dissipating 375 W. Not only is that not practical, it would make your cooker LESS efficient. \$\endgroup\$– jp314Commented May 23, 2023 at 16:35
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\$\begingroup\$ You might need a slow-blow fuse. \$\endgroup\$– Criticizing Israel not allowedCommented May 23, 2023 at 17:40
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\$\begingroup\$ I suppose you are also like that, and you are willing to mess around with stuff to make it work. If your cooker has two heating elements in parallel, if you re-wire them to series then the total power will be 1/4 \$\endgroup\$– Criticizing Israel not allowedCommented May 23, 2023 at 17:45
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\$\begingroup\$ If you want to save electrical energy, a steamer pot is also an option. They have a rubber seal so they keep the steam and thus, also the heat inside the pot. You heat up the pot for five minutes, then turn off the stove and it continues cooking by itself. Usually you have to let off the steam when the pot is still hot to avoid the vegetables becoming squishy. It's also great for cooking meat until it is tender. \$\endgroup\$– JankaCommented Jul 9, 2023 at 13:13
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2 Answers
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Diodes with the appropriate (current surge) rating should work with a resistive heating element.
Note the diodes will dissipate ~ 10 W, and will need good heat sinking.
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Assuming you're on 220 V, 1.5 kW would be around 6.8 A. the reason it blows a 13 A fuse is that resistive heating elements have less resistance when they're cold so on startup they will draw considerably more current than when they've reached their operating temperature.
With a Simmerstat you're basically turning it on an off repeatedly so you're getting a series of current surges, this stresses the fuse which eventually blows.
Diodes will rectify the AC voltage and give you a lower average current, but they need to be able to withstand the current surges. Adding resistance will work(maybe incandescent light bulbs?), but it would need to handle a lot of power, all of which will be wasted as heat. Something like a TRIAC light dimmer might be an option, if it was designed to handle the 1500 W. The most efficient way of lowering the power would probably be an appropriately rated transformer, but at that point it may be more cost effective to just get a more efficient appliance, whilst adhering to safety measures.
