What is the maximum power a DC dummy resistive load can take, if I connect two 50 ohm 2kW power resistors in parallel? Is it 2kW or 4kW?

  • 1
    \$\begingroup\$ What is the exact application? Do yo have a picture of the dummy loads and what you want to connect them to? \$\endgroup\$ – jippie Jan 4 '13 at 14:50
  • \$\begingroup\$ @DaveTweed kW is correct, KW is not. :-) \$\endgroup\$ – exscape Jan 4 '13 at 15:37

Two resistors in parallel will consume twice the current for a given voltage, and will collectively dissipate twice the power, 4kW.

Make sure that the two resistors are far enough apart so that they don't heat each other; otherwise, their individual ratings will have to be derated, and the total power capacity will be less.


To add to Dave's answer:

The 4kW rating is correct if you add the two 50Ω 2kW resistors in parallel, just be aware that the combined parallel resistance will now be 25Ω, not 50Ω. This means assuming your source voltage remains the same, your load will draw twice as much current (if you want a 4kW 50Ω load, you need 2 * 2kW 100Ω resistors in parallel)

  • \$\begingroup\$ Only true when the output stage of the power supply has a near 0 ohm output impedance. I realize OP mentioned DC, but 50 ohms does remind me about RF too and those output stages usually don't like badly matched loads. \$\endgroup\$ – jippie Jan 4 '13 at 17:20
  • \$\begingroup\$ @jippie - yes, this is pretty much the same as saying "assuming the source voltage stays the same" (i.e. does not sag at all due to internal impedance) \$\endgroup\$ – Oli Glaser Jan 4 '13 at 17:27
  • \$\begingroup\$ How about if I connect both the 50 ohm power resistor in series? Is the maximum power still 4KW? \$\endgroup\$ – user17565 Jan 5 '13 at 7:24
  • \$\begingroup\$ If you connect them in series, then you effectively have a 100 ohm, 4kW resistor. \$\endgroup\$ – Oli Glaser Jan 5 '13 at 8:52
  • \$\begingroup\$ Is there any other way to build a high power DC dummy load beside using power resistor as the load? \$\endgroup\$ – user17565 Feb 2 '13 at 13:14

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