I have a 3-phase, 30 kW, 480 V AC heating element.

I’m trying to utilize resistance as an aid for predictive failure.

Using Ohm’s law calculation 480 times 480 equals 230400. 230400 divided by 30000 W equals 7.68 Ω.

But when I measure the element it reads double that: 17.2 Ω.

Where am I making an error?

  • \$\begingroup\$ Welcome! Does it have three or four terminals? Which did you measure across? \$\endgroup\$
    – winny
    Jan 24 at 18:51
  • \$\begingroup\$ It has 3 terminals. L1,l2,l3 \$\endgroup\$
    – Sans307
    Jan 25 at 15:36
  • \$\begingroup\$ I measured on the element between l1-l2,l2-l3,l1,l3. The resistance im getting on multimeter is 15.3. But calculation on the ohms law states i should be seeing 7.68 \$\endgroup\$
    – Sans307
    Jan 25 at 16:06

1 Answer 1


If this really is a three-phase heating element, there are two likely ways it could be connected (as well as some less-likely ones that I won't discuss). Image from the Wikipedia article Y-Δ transform, which you should also read:

delta and wye connected resistor networks


There is a 10 kW, 480 V coil connected between each pair of phases. Each coil's resistance is:

(480 V)²/10 kW = 23.04 Ω.

Measuring across any two terminals gives the parallel-series combination of (23.04 Ω+23.04 Ω)‖23.04 Ω = 15.36 Ω, exactly twice what you calculated.

Note: The symbol "‖" means "in parallel with". Please read one of the many online articles on calculating parallel resistances (example).

Wye (star):

There is a 10 kW heater from each phase to a common (neutral) connection. This neutral is not brought out of the heater because no current needs to flow through it in a balanced system. the voltage between any phase and the neutral is 480 V/√3 ≃ 277 V. Each heater's resistance is:

(277 V)²/10 kW = 7.68 Ω

Measuring across any two terminals gives the series combination of 7.68 Ω+7.68 Ω = 15.36 Ω, the same as above. (When only connecting two terminals, no current flows through the third coil so it has no impact on the measured resistance.)

Now, why did you measure 17.2 Ω? My guess is that it is due to manufacturing tolerances. Did you measure between all three possible combinations of terminals? Was there some variation?

The resistance of heating coils does tend to increase with temperature, but unless you measured this device while hot, this effect would not explain the discrepancy in resistance.

  • \$\begingroup\$ Hi, the delta calculation you have above i just have a question. 23.4 plus 23.4 equals 46.8. 46.8 divided by 23.4 is two? When i measure l1 to l2 on element disconnected i get 15.4. Same reading between l2-l3 \$\endgroup\$
    – Sans307
    Jan 25 at 16:30
  • \$\begingroup\$ @Sans307 I've updated the answer to explain the "‖" symbol. (It's not division.) \$\endgroup\$
    – Theodore
    Jan 25 at 17:43
  • \$\begingroup\$ Thank you, one more question on same matter but different voltage. I calculated utilizing the formula inserted above. Can you see where im making an error or explain please. Heating element 3phase 208 volts. 208 devided by 1.73 equals 120v. 120v times 120v devided by 10000watt equals 1.44. 1.44 plus 1.44 equals 2.88. when i measure L1 to L2 i get 9.1 resistance and same for all three legs am i off on the calculation? \$\endgroup\$
    – Sans307
    Jan 25 at 19:28
  • \$\begingroup\$ @Sans307 Is this 208V heater also rated 30 kW total? \$\endgroup\$
    – Theodore
    Jan 25 at 20:30
  • \$\begingroup\$ @Sans307 I would expect a 208V heater that measures 9.1Ω between legs would be about 10 kW total. \$\endgroup\$
    – Theodore
    Jan 25 at 20:40

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