Matt Steiner

[Matt Steiner]

I had thought that with my follow-up post that I understood it. Here’s my logic:

Using the “zen trick” we can determine that load 1 (forgive me, I forgot the load labels) which is connected to the detector, when we “become the load” we realize that we are connected on the left and right ultimately to the same branch that loads 2 and 4 are connected to. Therefore, with no resistance indicated, we would divide the 120v among loads 1, 2, and 4. That would make all three loads equivalent, correct? Also if the detector switch was open, that would mean that no power at all is going to that branch.

Please let me know if I missed anything.

[Matt Steiner]

Or perhaps a lesson that covers the circuit configuration that happens in this question in particular? Either would be helpful, because I am not sure what I am missing. I have an inkling of it, in that the parallel circuit with the detector switch, if redrawn in series (hopefully I’m on the right track here) would mean that the 1, 2, and 4 loads are all part of the same circuit. With that being said we would split the voltage drop among those three, as they are part of the same branch, and the “parallel” we’re seeing between those three loads is not an independent parallel. However, we also have no relative terms for each load, so they would be divided equally, correct??

[Matt Steiner]

Susan,

Do you happen to have the title of the webinar that relates to this question? I am also struggling with it at this time, and would love to see a better explanation before I give the Midterm my second try. However, the Office Hour recordings do not specify by date, so I am unable to locate the correct video.

Thank you,

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