




#4434  09/25/01 07:24 PM
Re: sizing resistors

Member
Registered: 09/15/01
Posts: 816
Loc: Bergen County, NJ

If each lamp is 35W @ 12V , then the lamp current is:
I=P/E
I=35/12
I=2.92 Amps
Assuming a series circuit, the current is the same through all the lamps.
If you have 5 lamps, each dropping 12 V, the total dropped across the lamps is 12V x 5, or 60V.
If you want to run this lamp bank from a 120V source, you will need a series resistor that will drop 60V at a current of 2.92A. According to Ohm's Law:
R=E/I
R=60/2.92
R=20.5 ohms
The power dissipated in this resistor is given by:
P=ExI
P=60x2.92
P=175.2W
The nearest standard resistor would be a 200W wirewound type, which is a pretty large device. Imagine a ceramic tube about a foot long and 1.5" in diameter.
You may have a problem getting the lamps to start well with a simple resistive ballast like this, because the resistance of a filamenttype lamp is much lower when cold, resulting in a much lower drop across the lamp and a consequently higher drop across the series resistor. The lamps may remain cold while the series resistor burns up from excessive dissipation. Even if the lamps eventually heat up and run properly, you need to take this startup current surge into account when sizing the series resistor.
All in all, the use of a transformer is a better way to go. You save 175W of wasted power!
[This message has been edited by NJwirenut (edited 09252001).]

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