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Author Topic: 12 VOLT BAYONET BASE LAMP QUESTIONS  (Read 5374 times)
DominicMazoch

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« Reply #15 on: May 12, 2010, 07:11:50 PM »

The 24 V might have more resistance, but the difference would be so small I don't think it is going to a big difference.

I used a 28 V bulb on an O22 switch, and it seemed to work.  A 14 V bulb runs so hot I was worried it would melt the lantern.
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Joe Satnik


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« Reply #16 on: May 13, 2010, 10:33:49 AM »

Dom,

How did you power your O22 switches?  If constant power, not track power, what was the voltage that you used?

Lee,

Perhaps I should rephrase my previous question on lamp resistance. 

Start with an original lamp. 

If you wanted to design a new lamp that would use the same amount of electrical power, yet be supplied with twice the voltage, how would the resistances of the original ( = Ro ) and new ( = Rn ) lamp filaments compare?

Hint:  Use equation P = V x V/R, 

Hope this helps.

Sincerely,

Joe Satnik
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DominicMazoch

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« Reply #17 on: May 13, 2010, 10:10:57 PM »

I use about 14 V on the side plug on the switch.
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Joe Satnik


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« Reply #18 on: May 14, 2010, 12:00:39 PM »

Using algebra,

Po = Pn   Original power same as new power.

Po = Vo x Vo/Ro   power is V squared over R

Vn = 2 x Vo  New voltage is twice original

Pn = 2Vo x 2Vo/Rn = Po = Vo x Vo/Ro  substitute

4 x Vo x Vo/Rn = Vo x Vo/Ro  combine twos

4/Rn = 1/Ro   divide out (cancel) Vo squared from both sides

1 x Rn = 4 x Ro  Cross multiply

Rn = 4 x Ro  Drop 1 x (redundant)

So, same electrical power, @ twice the voltage and half the current, the resistance is quadrupled.  (X 4)

Hope this helps. 

Sincerely,

Joe Satnik
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Joe Satnik


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« Reply #19 on: May 14, 2010, 12:12:29 PM »

Dom,

What are the O22 lamp part numbers, 28 and 14 volt?

Joe Satnik
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DominicMazoch

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« Reply #20 on: May 16, 2010, 11:37:52 PM »

Right now i am using a Shack bulb.  Only problem is the bulb is taller than the usual O22 bulb.

But Lionel did sell a line of replacement bulbs for 18 volts because of TMCC/L.  That might be a source.
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phillyreading

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« Reply #21 on: May 17, 2010, 12:13:12 PM »

Using algebra,

Po = Pn   Original power same as new power.

Po = Vo x Vo/Ro   power is V squared over R

Vn = 2 x Vo  New voltage is twice original

Pn = 2Vo x 2Vo/Rn = Po = Vo x Vo/Ro  substitute

4 x Vo x Vo/Rn = Vo x Vo/Ro  combine twos

4/Rn = 1/Ro   divide out (cancel) Vo squared from both sides

1 x Rn = 4 x Ro  Cross multiply

Rn = 4 x Ro  Drop 1 x (redundant)

So, same electrical power, @ twice the voltage and half the current, the resistance is quadrupled.  (X 4)

Hope this helps. 

Sincerely,

Joe Satnik

Joe,

I am totally lost when it comes to algebra! I may know some geometry but that is about it, mainly basic math. Also I think that you are trying to describe an electronic circuit and how it works rather than a basic electric circuit.

Basically when you increase resistance in a circuit you increase the power consumed. So the point that I was trying to get across is that a lower volt light bulb should let you have more power to the motor, this will be noticed with a lower watt transformer.

Lee F.
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Joe Satnik


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« Reply #22 on: May 17, 2010, 12:33:20 PM »

Lee,

You said:

"Basically when you increase resistance in a circuit you increase the power consumed."

No, just the opposite. 

Power = V x V / R

If the Voltage on a load stays the same, and its resistance increases, the power decreases.

/R means "divide by R".

Hope this helps.

Joe Satnik

 
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