ClintOHenry
New Member
I have some old 1.5 volt, 30 ma incandescent lights I want to put on my 12 volt line. How do I calculate the resistor needed?
dropping voltage
- Thread starter ClintOHenry
- Start date
Ohms Law is what you need to use.
https://en.wikipedia.org/wiki/Ohm's_law
Here is a site that does the math for you...
http://www.ohmslawcalculator.com/ohms-law-calculator
But to answer your question you need a resistor that will drop 10.5 volts (12 volt supply minus the 1.5 volt bulb) at 30 mA.
Starting with E = I x R where E is the voltage, I is the current in amps and R is the resistance in ohms. (If I is current in milli-amps then R is resistance in kilo-ohms)
Rearrange to R = E / I and solve.
10.5 / 30 = 0.35 kilo-ohms or 350 ohms.
https://en.wikipedia.org/wiki/Ohm's_law
Here is a site that does the math for you...
http://www.ohmslawcalculator.com/ohms-law-calculator
But to answer your question you need a resistor that will drop 10.5 volts (12 volt supply minus the 1.5 volt bulb) at 30 mA.
Starting with E = I x R where E is the voltage, I is the current in amps and R is the resistance in ohms. (If I is current in milli-amps then R is resistance in kilo-ohms)
Rearrange to R = E / I and solve.
10.5 / 30 = 0.35 kilo-ohms or 350 ohms.
RCJunction
New Member
[Edit: ninja'ed by fcwilt - he's a machine, I tell ya...]
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"We are the Borg. Your biological and technological distinctiveness will be added to our own. Resistance is futile."
ClintOHenry
New Member
Thanks, if I put 10 of these in do I add their ma together and get 300ma?
Yes you can do that assuming you put the bulbs in parallel.
Doing it that way would require a 35 ohm resistor rated at 5 watts (3.15 actually but 5 is probably the closest you will easily find).
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Iron Horseman
Well-Known Member
You could do that but I would not recommend it. Besides getting the necessary high wattage of the resistor, if one of the bulbs would go out, then the amps would drop and the resistance would be too low. If I've done my math correctly the remaining bulbs would get 2.55v which could burn them out quickly. I would recommend putting one resistor with each bulb. In general resistors are cheaper than bulbs.
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gregc
Apprentice Modeler
wiring each 1.5V bulb with a resistor using a 12V supplies drops most of the power across the resistor. If each circuit is drawing 30 ma from a 12V supply, each circuit is using a 0.36 (12 * .03) W. The bulb itself is only using 45 mW (1.5 * 0.03). 8 bulbs would require 2.88 W (8 *.36).
another approach is to wire 8 bulbs in series and have a single circuit. Now 8 bulbs uses only .36W.
Of course if one bulb goes out, that all do. But a volt meter can be used to quickly isolate the problem bulb.
