calculation for resistance
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Forum Name: General Discussion
Forum Discription: General Mobile Electronics Questions and Answers
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=39975
Printed Date: May 31, 2024 at 9:22 PM
Topic: calculation for resistance
Posted By: joeyng
Subject: calculation for resistance
Date Posted: September 29, 2004 at 1:25 AM
Hi, I have no electronic background. trying to do a project. anyone know the calculation how to find out: 1) what resistance is required to turn a 3.6v 900mAh battery to supply 3.6v 300mAh? 2) what resistance is required to turn a 4.8v 2000mAh battery to supply 3.7v 800mAh? just resistance will do the job or do it require anything else? any help will be appreciated.
Replies:
Posted By: dxav
Date Posted: September 29, 2004 at 10:27 AM
Ohm's Law (on this forum, look at the menu on the left) Under basics, click : https://www.the12volt.com/ohm/ohmslaw.asp
Basically, you can work V = IR to any situation. V (or E) is Voltage, I = Current, and R = Resistance.
For power, you can also substitute elements of Ohm's Law
P = I^2R (Power = Current [square] x Resistance)
Good Luck,
DXAV
Posted By: joeyng
Date Posted: September 29, 2004 at 6:06 PM
Thanks for reply. I read about the equation, but I'm too new to this stuff, really dunno how to subsitute the formula. I need to change both Voltage and Amp. Can you please show me the correct calculation to convert (4.8v 2000mAh) to (3.7v 800mAh)? Thanks in advance
Posted By: dxav
Date Posted: September 29, 2004 at 8:32 PM
Well, I just noticed you are not reall asking for a simple resistor value. The label of your 'current' is actually a current rating per hour. This looks like a charging circuit requirement. Where did you get these values, and what are you trying to do?
If there were no hours on that label, it would be as follows:
1) what resistance is required to turn a 3.6v 900mAh battery to supply 3.6v 300mAh?
R = V/I
R = 3.6V/.3A (you would need to convert the amperage to the same accuracy as the voltage)
R = 12 Ohms
But with your situation, you are working with rates, which is different.
For your other example, the easiest would be to use a voltage divider (2 series resistors) to step down the voltage to 3.6V, using the final value of the current to decide the R2, and make R1 dependant on your R2 value.
Probably doesn't make too much sense, huh? You may want to do a little electrical theory research. I can help, but it's not as easy to try to explain it to someone.
DXAV
Posted By: joeyng
Date Posted: September 29, 2004 at 9:04 PM
Thanks again, I think I misused mAh in my previous post. I am planning to use 3 x 1.2 volt 2000mA rechargeable batteries to drive a 1 watt Luxeon LED rated at 3.42 volt 350mA. (It have to keep the Amp at or below 350mAh) I think I did something wrong with the battery calculation.  I am also planning to build another one with 3 watt Luxeon LED later. I think the specification is 3.7 volt 700mA
Posted By: joeyng
Date Posted: September 29, 2004 at 9:09 PM
Also, I have problem doing the very basic What's the actual value for a 1R5 resistance? is it 1.5ohm or 15ohm?
Posted By: dxav
Date Posted: September 30, 2004 at 7:11 AM
You may have misused the value, which makes more sense now. You have the known values, now it should be a simple divide like I showed you above.
1R5 as in a part number value? Vendors list caps, res, and inductors differently. I usually see an R in a part number to mean a decimal place.
I will elaborate more later if I get the chance.
DXAV
Posted By: hurtado_roberto
Date Posted: October 03, 2004 at 3:17 AM
I forgot my basic circut calculations but isn't current dependent on whatever is down the line and not the source? A resistor only drops the voltage but maintains the same current in a series configuration right. In a parallel configuration more current will flow trough the resistor with less risistance. Correnct me in the parts I'm wrong. I conducted an experiment in my electronics class were we connected a light bulb in series with a power source. We maintained the same current and manipulated the voltage. The more voltage the brighter it got. I think the 2000mA is the rating of the actual mAh since that's what some of the batteries we drained in another lab turned out being. I think I confused myself even more now. Can someone help clear this up. I guess I'm sleepy ------------- Poly Dollies
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