Philip Boerstler

[Philip Boerstler]

I think I also was initially looking at the schematic symbol (Picture) and saw it as being an obvious break…

[Philip Boerstler]

In haste of what I was doing, I made the mistake of taking the number found on the bottom section of the equation and multiplying by 10/1 — or some silly such business, thinking that this should be equivalent. (Maybe I should go take a math class again!!) Anybody have a dunce cap that I might be able to borrow? (Doh!)

[Philip Boerstler]

Also — just out of curiosity, does the line for that switch being thinner suggest that it is thinner gauge wiring? Something else? Just drawn that way?

[Philip Boerstler]

So, now that I have seen the circuit in question, I think the ability to understand this question is a bit better. In the thread above, John correctly points out that there are 3 parallel circuits (assuming the zen trick of ‘becoming’ the load and saying “what’s on my left, what’s on my right.”) What throws me — and likely anyone else looking at this is the connection of the first two lines in parallel. Does this change ANYTHING from making it a simple line drawing of
L1 — Load — Switch — N
L1 — Load — Load — N
L1 ——- Load ——- N

….And I think I just answered my own question….
L1 — Load — Switch — N
L1 — Load — Switch — N
L1 ——–Load——– N

[Philip Boerstler]

I went back to the last video in Unit 3 which helped tremendously…

[Philip Boerstler]

This thread reminded me of a documentary that I had initially seen a few years ago.

http://www.dailymail.co.uk/news/article-3126209/Man-demonstrates-maintaining-high-voltage-line-half-million-volts-pass-body.html

Faraday Cage anybody?

Then there’s this one…
https://www.youtube.com/watch?v=QqEesFaboV4

[Philip Boerstler]

This was also confusing to me. I got that the total for 240²/37Ω was 1556.76W. What did not appear to be very clear to me was how to determine the Wattage for the loads offering resistance of 5Ω and 32Ω with connection of Lines 1 and 2 (totaling 240 V AC.) And When I was looking at the equation P=I²R, I still was lost. To get A (or I) I took 240/5 = 48; tried to apply A²Ω as 48²(5)=11,520W and 240/32 = 7.5; 7.5²(32)=1800W — both well beyond my total of 1556.76W— so clearly, something was incorrect. Using W=VA, 240 x 7.5 = 1800, 240 x 48 = 11,520. Am I doing something wrong to find Amperes?

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