I have a 2000W, 240V electrical heating element that I need to move. Connecting it to where it needs to be makes it draw, together with 2 other elements, more current than the fuse will allow (16A). Is there a way for me to add a resistor somewhere such that the heating element will draw less current?

Edit: The two other elements have a wattage of 1200W and 800W. The three elements are used to heat a large room. I don't mean to put two of these elements in series, just to reduce the heat produced by the 2000W element by making it consume less, if possible.

Ideally, I'd like to cut the power of the 2000W element about by half. Yes, I could disconnect the small 800W element, but I like it, it's the nicer one and it's close to the couch :)

  • 3
    @JPhi1618 you misunderstand basic electricity. Power = V*I = I^2/R . For constant source voltage, I = V/R , so increasing R decreases I and thus reduces power. – Carl Witthoft Jan 7 at 20:18
  • 2
    If you put this element in series with another of the elements the current, as well as the heating power delivered, will decrease significantly. If you tell us more about the situation (the wattage of the other two elements and what the heat is used for) we might be able to help analyze whether that would make sense to do. – Greg Hill Jan 7 at 20:26
  • 1
    @CarlWitthoft, what I'm getting at is that if you have a resistor that is capable of the power levels required by even a 5A heating element, it's going to need to dissipate a lot of heat. – JPhi1618 Jan 7 at 20:30
  • 2
    Just so you understand, (and others), while a heater "is essentially a resistor" its a pretty small value. The "short" it creates (by design), allows the heat to be spread over the element. If you intend to pull 1000 watts through the element your added resistor is going to be hugh or it'll burn out. You basically need a 1200 watt resistor. Just buy a smaller heater. – noybman Jan 8 at 4:26
  • 2
    a diode instead of a resistor will block half the power while only dropping a few watts (~0.7*I) on the component itself; much better (safer, cheaper, easier) than a resistor to half the power. – dandavis Jan 8 at 17:13

Let's explore what can be done with series combinations. We'll begin by working out the resistance of each of the elements. V=IR and P=VI, so P=V^2/R and thus R=V^2/P. The 800W element is 72 ohms, the 1200W is 48 ohms, and the 2000W is 28.8 ohms.

First, to directly answer your question, could a resistor be added in series with that 2000W element to make it heat less? Suppose we wanted to get 1500W from the combination of the element and the resistor. We'd need the combined resistance to be 240^2/1500=38.4 ohms, and the 2000W element contributes 28.8 of this, so another 9.6 ohms are needed. How much power would that new resistor need to safely dissipate? Well, the 1500W power at 240V means a current of 1500/240=6.25 amps. Then the power in that new 9.6 ohm resistor is I^2*R or 375W. It's not impossible to find a resistor capable of handling that power, but it's not exactly easy either.

OK, what would happen if two of your heaters were wired in series? Suppose we pick the 2000W and the 800W. The total resistance is 100.8 ohms and the power would be 571W. Added to the full-power 1200W unit, you'll have 1771W of heating. If the 2000W and the 1200W are in series then it's 76.8 ohms and 750W, for a room total of 1550W. Finally, if the 2000W is allowed to run full power while the 800W and 1200W units are in series, they'll have a resistance of 120 ohms and power 480W. The room total would be 2480W.

One other option.. you could consider powering any of the heaters from 120V instead of 240V. It'll deliver 1/4 of the nameplate power if you do that.

  • Right on! Another option is a variac, but these aren't necessarily cheap either, and depending on the heater, it may not work. A simple resistive element type would work, but a new (smaller) heater is still smart and safe. – noybman Jan 8 at 4:30
  • It looks like the needed resistor would be more expensive of a full replacement heating element with a lower power. – Jean-Philippe Pellet Jan 9 at 15:10

Get another 2000W unit. Seriously.

Place it where practicable and useful, and place it in series with the existing 2000W unit.

Together they are now individually 500W units, and 1000W together. I believe that's the number you wanted to hit.

I mention this because resistive heaters are naturally perfectly inexpensive. Talking about for-permanent-installation baseboard heaters, here in the US a 2000W unit is a mere $50+tax. So this is a very economical way to do what you want, assuming there is not some mad discrepancy between units.

The gory details

When doing series connections with commercial products, I prefer to match up same-size.

Now, if you want all the numbers, here they are.

 Voltage  Power  Current Resistance Conductance
 E or V     P       I        R         G
 240 V   2000 W   8.33 A   28.8 Ω   .0347 Siemens
 240 V   1200 W   5.00 A   48.0 Ω   .0208 S
 240 V    800 W   3.33 A   72.0 Ω   .0139 S

 240 V   1000 W   4.17 A   57.6 Ω   .0694 S
 120 V    500 W   4.17 A   28.8 Ω   .0347 S
 240 V    750 W   3.13 A   76.8 Ω   .0130 S

 240 V   3200 W  13.33 A   18.0 Ω   .0556 S practical circuit limit
 240 V   3840 W  16.00 A   15.0 Ω   .0667 S absolute circuit limit

When placing units in series, add resistances.

Your calculation for the 2000+1200W heater in series is correct: 28.8 + 48 ohms = 76.8 ohms giving 750W.

All three heaters together are 16.6 amps. The change I suggest will drop this to 12.5 amps, giving 3000W of heat in the room. This is within the 13.33A circuit limit after the 125% derate for heaters is being applied. Your country may not require this derate, but apparently, your fuse does.

  • 2
    The wiring used for these units if in series has to handle the full 16.7 Amps, all the way to/through the units – noybman Jan 8 at 4:37

You can add a resistor, but please don't . In the first place it wouldl have to be capable of handling a lot of voltage (without breakdown) and a lot of current (without melting). If you put it in the wrong place in the circuit you'll create "floating" voltages: the heater is designed so one end of the element is at ground, as is the chassis and so on. Further, if the heater has any controls on it, they will not "understand" that there's a resistor in series (whatever you do, don't go parallel!) and trouble may ensue.

If you need all 3 heaters running simultaneously then you must run a new line, on a separate fuse / breaker to support the necessary peak current.

  • Thanks for your answer. So basically you're saying that if I have a 2000W heating element, I'm stuck with it and there's no easy way to lower its consumption and the heat it produces? – Jean-Philippe Pellet Jan 7 at 20:32

It depends on the circuit of the heater. If it is without any fan, i.e. a silent one, a simple 10A-Diode (on a heatsink) in series with the heater will - as already mentioned in a comment - reduce the power by factor 2.

If there is a fan in the heater, this simple method, f.i. to insert a diode in the plug of the heater, must not be realised. Instead, the diode must only be in the parallel current path of the heating element - again in series with this heating element. It could be complicated if a part of the heating wire is used as resistor for the motor.

In any case the diode must not change the current through the motor. In that way the fan's current is neither reduced nor rectified - both could be harmful to the motor and/or unit (not running at all, sometimes stopping, running rough etc., dependent on the type of motor). As a plus, the heat sink for the diode can be omitted if the diode is cooled by the fan.

A parallel bridging 10A-switch (short cutting the diode) may set the unit to the normal status as before. That way, it can be switched to 2000W or 1000W just like most hairdryers reduce the power output by factor 2.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.