Sylvania Silverstar headlights

I just bought the 9006XS 's (low beam) for my PT Cruiser. They look a
little whiter and brighter than the 2.5 year old originals. They are 55
watts and the originals were 51 watts, I think. Has anyone else here got
them? What do you think? DS could you explain what you think about them?

Thanks in advance;
SRG

 

He probably won't answer, as I (and probably a few others) have written to
him regarding this subject. He's dead-set against them according to his
website.

I like them though. The light is whiter, but that only refers to color temp
and not brightness. They only last a year according to Sylvania, maybe
longer depending on usage.

 

A google search on this newsgroup will reveal DS's opinions of the
SilverStar (generally negative). I've also seen a lot of complaints on
various forums on their short life.

Bill Putney
(To reply by e-mail, replace the last letter of the alphabet in my
adddress with the letter 'x')

 

I already *did* answer. Extensively.

I answer all my e-mail.

DS

 

In addition to my earlier, long response about why Silverstars aren't
better (let me know if it doesn't show up for you -- I might be having
trouble with my news host), note that there is no wattage difference
between your originals and the Silverstars. 51w is the European rating
(at 12v), 55w is the US rating (at 12.8v).

DS

 

The light is *bluer* (not "whiter") and dimmer.

 

I may be off on my Ohm's Law, but does not a load draw more current at a
lower voltage? Give me a reality check on this Dan.

Richard.

 

Think about it - if that were true, then at 0 volts, the device would
pull more current than at 13.8 volts, and at 100 volts, it wouldn't pull
much current at all. Doesn't make sense does it.

For ohmic devices (passive devices like resistors), ohms law is ohms =
volts/amps, so amps = volts/ohms. For a given resistance, amps go up as
volts go up.

Light bulbs are not linear because their resistance changes with
filament temperature. The hotter the filament, the higher its
resistance, which is why you get inrush current at initial turn on. The
result is that you won't get as drastic a change in current with a given
change in voltage, but still, you do get increase in current with
increase in voltage (i.e., it's less than a factor of 1).

Bill Putney
(To reply by e-mail, replace the last letter of the alphabet in my
adddress with the letter 'x')

 

No. I=V/R. Assuming that the load resistance is constant, lowering V
will lower the I (current). I suspect you are thinking about keeping
the power constant. It is true that to maintain the same power, more
current is required as the voltage is reduced since power is the product
of current times voltage. However, maintaining constant power would
require an "intelligent" load that could vary its resistance in response
to changes in available voltage.


Matt

 

Yep; thanks for getting my tired old brain back into focus on this point.

Richard.