I have  a 36v DC system and 24volt light, what resistor would I use to
cut the 36v down to 24v. and how do you figure this out?

Hi gezwez!!

First of all you must remember some basic formulas:

Ohm's Law:
E = I.R ; I = E/R ; R = E/I 

Joule's Law:
P = I^2.R = I.E = E^2/R 


Now we can start to work:

Call Rl the resistance of your lamp and R? the unknown resistance of
the desired resistor.
In order to cut the voltage you must connect the resistor and the lamp
(which is a resistor too) in series, so the current through each of
the resistors is the same:
I = Il = I?

The voltage drops across the resistors must add up to the total voltage supplied:

36V = Vl + V? = 24V + 12V (this is what you want).

Since V = I.R, then:

24V = I.Rl then I = 24V/Rl
12V = I.R? then I = 12V/R?

Now we have:
24V/Rl = 12V/R? which results in R? = Rl/2.

This means that whatever be the resistance of the lamp, if you connect
in series to the lamp a resistor with one half of that resistance, the
voltage dropped in the resistor will be 12V, this will result in 24V
for the lamp.

And this is the answer...but I can hear you telling me that you donīt
know the resistance of the 24V lamp. Here you have two solutions:
The first one is to measure the resistance of the lamp with a
multimeter or an ohmmmeter and then connect in series a resistor which
have one half of the resistance measured. See "Using a multimeter":
http://www.doctronics.co.uk/meter.htm

The other way is using the Joule's Law:
P = I.E [watts]

Since I is the same for both resistors and the voltage of each one is known then:
Pl = I.24V then I = Pl/24V
P? = I.12V then I = P?/12V

Now we have:
Pl/24V = P?/12V which results in P? = Pl/2 .

How this last result can help you?
If you have a 24V lamp with a power of 10 watts for example, you must
connect in series another lamp with the following caracteristics:
12V lamp with a power of 5W (note that this new lamp must be a 12V
one) this ensure you that the resistance of this new lamp is a half of
the resistance of the original one that you have.

For references visit the following pages at Department of Physics,
University of Guelph:
"OHM'S LAW":
http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.html

"RESISTORS IN SERIES":
http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.series.html

"DC CIRCUITS":
http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.html 


I hope this helps you. Please, if you feel that the answer is unclear,
incomplete or you need further assistance on this topic use the
request for answer clarification. I will gladly respond your requests.

Best regards.
livioflores-ga

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Request for Answer Clarification by gezwez-ga on 14 Jan 2004 16:23 PST

I didnt think you could measure a light bulb resistance with a ohm
meter because the bulb wasnt ohmmic? Also if I mearsure volts acros
the bulb what is the volt or the resistance? lets say we start with
12v and acrose the bulb is 10v what is my resistance? how do you
figure that out?

---

Clarification of Answer by livioflores-ga on 14 Jan 2004 21:03 PST

Hi gezwez!!

Regarding to your comment about if it is possible to measure the lamp
resistance I must tell you that the resistance of a resistor is a
physical property that is independent of the voltage with it is
measured, but, as our friend notsofast-ga said, it varies with
temperature.
When you measure the resistance of a lamp with a multimeter you use a
low voltage, that means that you obtain the resistance of the "cold"
lamp, so this will lead you to some error in the final result. I tried
to answer the question as a homework exercise, that is giving you an
aproximated but simple answer.
At this point, if you want an exact answer I think that you can leave
the "measure the resistance" solution.
But the other solution gives you a proper answer, I think, because we
are using in the calculations the power of the lamp, and this value is
a running value, that is this is the power in watts generated by the
lamp when it is working. But to use this solution you must know the
power of the lamp, normaly indicated in some part of the bulb. If you
connect a 12V lamp which its power is one half of the 24V lamp in
series, it will drop 12V if the circuit is connected to a 36V supply.
If you still want to know the resistance of the lamp when it is
working and donīt know its power, you can try, connecting batteries in
series until reach the 24V (for example 16 1.5V AA batteries) and
conect to this power source the lamp, turn it on and measure the
current, with this you will have a way to calculate the power
generated by the lamp when it is working and the resistance of the
lamp when it is working ("hot" resistance") by using the following
formulas:
R = E/I = 24V / I 
P = E.I = 24V . I 

Please refer to the following pages
"Using a multimeter":
http://www.doctronics.co.uk/meter.htm

"Basic circuits: what is current?":
http://www.schoolscience.co.uk/content/3/physics/circuits/circh1pg3.html


I hope this clarify the point, please continue using this feature if
you feel that it is necesary.

Regards.
livioflores-ga

---

Clarification of Answer by livioflores-ga on 14 Jan 2004 21:09 PST

Note that if you know the power of the 24V lamp, using the Joule's Law
you can determine the resistance of the lamp when it is working:
P = E^2 / R  then  R = = E^2 / P

Then a resistor (that not varies it temperature significantly when it
is connected) with a resistance of (E^2 / (2.P)) (one half of the 24V
resistance)could work.

Sincerely.
livioflores-ga

thanks for your time

Hi livioflores,
I just read your response to the question about reducing the voltage
drop across the lamp to 24 volts.  I don't disagree with with your
basic formulas but I do disagree with how you applied them.  I defy
you to determine the required series resister as you outlined in your
paragraph, (And this is the answer...).  You are forgetting that a
typical lamp has tungsten filaments and the resistance of the lamp
filaments varies with its temperature.  i.e. the resistance increases
as the voltage is applied to the lamp.

Gezwez-I can tell you how to do this if you wish.

Hi Gezwez,
I just read your last comment.  First of all, what do you mean by
"ohmic?"  I believe that most everything has an ohmic value by
definition.  And your comment about, "starting out with 12 volts and
reading 10 volts across the lamp"-for this to happen you realize that
the other 2 volts have to be dropped somewhere, most likely across the
leads to the lamp.  This means that you have real long copper leads or
some type of high resistance leads or a high resistance connection of
some sort.  But, at any rate, you cannot determine the lamp resistance
with just this information.

this note is to notsofast-ga please tell me 

I didnt know how to get in teach with you.
Haroldhokcpi@aol.com

First of all,
if a series(in line) circuit isn't what you want to use for multiple
reasons...fistly, what I believe Gezwez was tring to explain that
resistance changes with temperature(both the breaking resistor and the
filiment resistance(as well as any leads or wiring).  The variance of
the breaking resistor can be mitigated using a parallel circuit(where
2 resistors would be needed), and as long as the were one resistor
(I'll call R1) would be in series with another resistor(R2) where R2
is in parallel.  Then you have what is called a Voltage Divider
circuit...Then the ratios of the Resitors(R1/R2) must equal the ratio
of the Voltages(V1/V2) where V1 is the applied Voltage and V2 is the
Required Voltage.

That should work fairly well, and is usually a first crack at shifting
DC voltages around.  One problem with this approch is the
variation(flicker) of the filiment(or any load for that matter).  But,
you have now linearized the flicker so by using another parrallel
combination of resistors(R3,R4)...Which is generally called a
"H-bridge" because of the way the circuit is layed out where the 2
parallel branches of resistors are the legs of the 'H' and the lamp is
the cross part of the 'H'...
This works extreamly well if loads are small, but if loads are large
or effency is important you just created another problem...all the
wasted power(and money)  And another problems comes in when you want
to increase the voltage (for instance you have a 12 volt car battery
and want to drive a 48 V motor) yes you could just hook 4 car batterys
in series (4*12=48) but now you have 4 times the weight and space(not
disirable)...Now what do you do?...
fortuantly about 50 years ago the transistor was invented and power
electronics soon followed...and boost, buck, boost-buck, etc. DC to DC
converters evolved...I'm getting long winded here but my point is
there are superior ways to convert power from one voltage to
another...and don't settle for the "series" configuration suggested.

kmc

Firstly resistores do not reduce Voltage, they "resist" current,
secondly to reduce voltage you need to use a transformer with the
right amount of wireing turns, example to reduce 24 volts to 12 the
primary coil needs 1/2 as many turns as the secondary coil (like 2000
wraps on the primary and 1000 in the secondary) this is a step down
transformer, if you turned it around you would end up with 48 volts,(a
step up transformer) this is all due to flux coupling which is another
story.