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matt piper

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  • April 29, 2020 at 9:43 am in reply to: Half-Splitting #19030
    matt piper

      This has me a bit confused in reading it, Your neutral could be totally disconnected from the circuit and would still measure as a potential difference between it and L1. How are you confirming neutral is good, exactly?

      April 29, 2020 at 9:35 am in reply to: Bake element/loose connection problem #19029
      matt piper

        Actually, I think I understand now. The extra resistance is drawing less amps, so less power overall.

        April 10, 2020 at 12:38 am in reply to: Mod. 03 Unit 03: Voltage, Current, Resistance, And Power – Quiz Question #11 #18869
        matt piper

          Ok, on the loose connection problem detailed in the video, the loose connection is having it heat produced calculated by
          P=I squared * R
          6.5A squared is 42.25 and times 5 = 211Watts

          My question in this…why does the other equation for watts not work?

          P= I*E
          6.5*240= 1560

          What am I missing?

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