From what I understand resisters are measured in ohms.

What im after is a guide that tells me how much a resister will imact a circuit (ie. cut the voltage down by).

For example, I have a light thats running from a 9V battery, but the light only needs 6V -- therefore what resister/s should I put in the circuit to downgrade it to 6V.

Resistors don't lower voltage, they lower current. A good guide is www.the12volt.com

they can lower either voltage or current...depending on if the resistance is in parallel or series in the circuit. didnt you guys do year 10 science? :P

work out the resistance total circuit using an ohm-meter.

Whatever this value is you need to slap a resister of 1/3 total resistance of the circuit. (connected in a straight line). Theoretically this should load 1/3 of the circuits max-providable voltage into that resister(3V), leaving 2/3s (6 volts) free for the globe and remaining wiring, which will loose neglibable voltage in pissy circuits.

http://www.howstuffworks.com/

Here's my shoddy ascii art of a circuit diagram of whats called a "potential divider"

 --------------------------
|                   |           |
|                   |           |
|                  R1         Vo1 (out) = 6V
|                  |            |
Vi (9V)          |            |
|                  |-----------
|                  |            |
|                  R2          |
|                  |            Vo2 = 3V
|                  |            |
|                  |            |
--------------------------
Vi = Source of 9V

As trancer was saying:
R1 will have a resistor value of 2/3
R2 will have a resistor value of 1/3

Valid resistors you can use:
R1 : 620ohms (approx)
R2: 310ohms (approx)

You will need to choose the correct resistance values for your intended circuit to be able to supply enough current.

So now go back to V = IR and see if you're resistance values are sufficient.

EDIT: woweee, lotsa &nbsp's

OK guys, from an electrical engineer (albeit that I work with High Voltage electricity transmission)

Volts(V) = amps(A) x resistance(ohms)

Power(W)=Volts(V) x amps(A)

For your particular question MAGNA,

6V lamp, say 6W (should be stamped on the side, I use 6W to make the calculations easier), will draw 1 amp (6W/6V=1A).

To drop your battery voltage by 3V, with 1A flowing in the circuit, you would need a 3ohm resistor, which by the way is a fairly small resistor with a large amp draw. That resistor will also be required to be rated at 3W (3Vx1A=3W), now there would probably be one that is a 5W rating, so that would be OK.

So, what you need to do is work out how much current the lamp will draw from its wattage(W), and match your resistor to use 3V at that current.

Hope this helps. What is the application?

hey I just started work as an Electrical Engineer :D

:D we could start our own little Frat~~ AM-elec eng Frat!

oh and MAGNA, the rest of the guys have explained this pretty well!

Yeah - I saw some stuff about VI = R blah on the internet but it wasn't explained that well. The above stuff is brilliant!

..and to the guy that asked if I did Year 10 science, I was in the "Advanced Science" class (don't ask me how) but we didn't do anything like this! Damn public schools! :cry:

hey I just started work as an Electrical Engineer :D
I just passed my 15 year anniversary at TransGrid - where the electricity is at! Gotta love getting more long service leave!

OK, I was a cadet engineer, they put me through Uni, so I am not that old :roll: , they include the time I spent at uni as service for the company.

I went to a public school, but they taught us basics of resistors, caps, transistors in Yr10 science... Didn't get more in depth until Y11 Physics though. I actually think it has nothing to do with being a public school. In actualty I think kids in private school do worse because they can afford to live on a diet of coke whilst the lowly pub-eds smoke cones all day lol...

[quote:016ce8ba59="Phonic"]hey I just started work as an Electrical Engineer :D
I just passed my 15 year anniversary at TransGrid - where the electricity is at! Gotta love getting more long service leave!

OK, I was a cadet engineer, they put me through Uni, so I am not that old :roll: , they include the time I spent at uni as service for the company.[/quote:016ce8ba59]

15 years wow :shock: , I'v only been working for 2 months :) , I don't really work with high voltage transsmition lines, The place I work at mainlly works with industrial automation, so I'm currenttly training in PLC programming and control circuit design, not bad considering I only finnished an Adv. Dip in Engineering(electronics/computer sys) hehehe ahhhh connections can't live without them :D

There is a another way you can go and that's by using a 3 pin regulator chip very stable output voltage, available from dick smith or jaycar. Depending on your current draw (divide the wattage by voltage that'll give you current eg 12W bulb divided by 6V = 2 amps) provided it isn't above 1amp this chip will work nicely.

The part number of the chip will be 7806 where 78 is the chip series and 06 is 6V, Dick smith may only have the 7805 5V 1A.
There are other types with higher current limits but it would be probably be easier to go with a resistor

If you want to know more let me know.

If the equipment you are trying to power is something more complicated than just a light bulb, then I always recommend getting a voltage regulator instead because advantages are: they can handle higher loads, the output does no fluctuate when the input isn't giving a nice DC current.

That way you don't blow things up if for some reason you start getting voltage spikes.

one of the great things about this site is thst we have so many different people on this site from all different walks of life and know so much different stuff, its great!!

Just thought I'd chuck this in for the hell of it.

Adding Resistors in Series and Parallel

Series Resistors
RT = R1 + R2 + R3 etc
Where RT is the resulting resistance.

Parallel Restistors
RX = 1/R1 + 1/R2 + 1/R3 etc
Where 1/RX is the resulting resistance.