111 | | My resistor bed consist of of 4 x 2R2 (WH10) in series resulting in 9 Ohm, PSU does not like this low load conditions or is simply broken which would make my 2nd dead power supply. The new PSU is only 12V which is not heating up properly. According to [https://en.wikipedia.org/wiki/Ohm%27s_law Ohm's law] (I = V/R or R = V/I or V = IR) and my resistor rating i get 1.3A, which in result would yield only (W = VI) 15.6W heating power, not sufficient since my total resistor capacity is 40W. Which would give me W = VI combined with V = IR becomes W = IRI. W=40, R=9, makes I = 2A. Which in result would require V to be 40W / 2A = 20V. |

| 111 | My resistor bed consist of of 4 x 2R2 (WH10) in series resulting in 9 Ohm, PSU does not like this low load conditions or is simply broken which would make my 2nd dead power supply. The new PSU is only 12V which is not heating up properly. According to [https://en.wikipedia.org/wiki/Ohm%27s_law Ohm's law] (I = V/R or R = V/I or V = IR) and my resistor rating i get 1.3A, which in result would yield only (W = VI) 15.6W heating power, not sufficient since my total resistor capacity is 40W. Which would give me W = VI combined with V = IR becomes W = IRI. W=40, R=9, makes I = 2A. Which in result would require V to be 40W / 2A = 20V. To fix this I will need to change the resistor bed layout. Instead of 4 in series. Put two in parralel and next two pairs in serie. |