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J. Clarke[_2_]

In article ,
says...

On 12/6/2010 11:01 AM, Bill wrote:
On 12/6/2010 10:50 AM, Bill wrote:
Someone suggested (I could not locate the post) that if I wanted quiet
fluorescent lights, then I should look for those with an "A" rating.

Putting that detail aside temporarily. What would be the practical
differences between a "standard electronic ballast with 20% THD"
and an "instant on electronic ballast with 10% THD"?

Bill

Continuing to research my question, evidently low THD is better for
the components in the lighting system (capacitors, etc) and may provide
lower cost in the form of the longer life for the system.

Pitiful question: If 8 fluorescent fixtures are wired in a series, and
the ballast in one of them fails, do all of the lights go out? Assume a
"modern" fixture.

They're 110 v fixtures--that means that each one needs 110v. Wiring
them in series won't do that--they have to be in parallel.

As for a ballast failure, if the ballast failure trips the breaker they
all go out, if it doesn't then they shouldn't unless you hit some kind
of patholigical case.

If the answer to the question above is yes, this suggests "Lew's wiring
design" should be wired (using 12-3 THHN) with 2 parallel (pairs of)
circuits with 4 fixtures wired on each row. Then if one of the
fixtures went bad only two of them would go out (either the ones on L1
or the ones on L2 in a given row). Is this correct?

Alternately, I could use 8 parallel circuits, 2 for each row, then if
one of the luminaire's failed, then it, and it alone, would go out.

Taking a step back, this is equivalent to wiring 2 branch circuits each
wiring 8 fixtures in parallel. Everything basically correct?

That would be the way to do it if you want to be able to kill the power
to one set and still have the other available.