snafu Posted October 13, 2016 Author Report Share Posted October 13, 2016 OK. Let's go back to post 1 and look at Ohm's Law. If we're going to prove / disprove the hypothesis I've laid out, this is an excellent place to start. I = E / R (see first post for what each stands for if you don't know already) For the sake of simplicity, let's assume that we'd like to apply voltage to a 10,000 Ohm (10k) resistor and do the math on that. Let's consider the following voltages: 12.6V (nominal voltage of a fully charged battery) I = 12.6V / 10,000 Ohms I = .00126A 13.8V (healthy charging system) I = 13.8V / 10,000 Ohms I = .00138A 14.4V (optimistic) I = 14.4V / 10,000 Ohms I = .00144A So, we can see that the formula does indeed support our hypothesis. So, now it's time to prove it - and that's the fun part. 1 Quote Tony Candela - SMD Sales & Marketing Email me at [email protected] to learn about becoming an SMD Partner! Link to comment Share on other sites More sharing options...
snafu Posted October 17, 2016 Author Report Share Posted October 17, 2016 Long ago, I found this image in a textbook and used it in my presentations. Now, you can buy a t-shirt with it. Love it, and it's absolutely relevant here. 1 Quote Tony Candela - SMD Sales & Marketing Email me at [email protected] to learn about becoming an SMD Partner! Link to comment Share on other sites More sharing options...
deathcards Posted October 17, 2016 Report Share Posted October 17, 2016 OK. Let's go back to post 1 and look at Ohm's Law. If we're going to prove / disprove the hypothesis I've laid out, this is an excellent place to start. I = E / R (see first post for what each stands for if you don't know already) For the sake of simplicity, let's assume that we'd like to apply voltage to a 10,000 Ohm (10k) resistor and do the math on that. Let's consider the following voltages: 12.6V (nominal voltage of a fully charged battery) I = 12.6V / 10,000 Ohms I = .00126A 13.8V (healthy charging system) I = 13.8V / 10,000 Ohms I = .00138A 14.4V (optimistic) I = 14.4V / 10,000 Ohms I = .00144A So, we can see that the formula does indeed support our hypothesis. So, now it's time to prove it - and that's the fun part. doesn't the resistor have to change to stay at 10,000 ohms at 14.4v versus at 12.6v? Quote skar sk2500.1 0 gauge power and ground kunukonceptz alpine HUvxi65 components on BA gt-275new build log -> http://www.stevemeadedesigns.com/board/topic/150642-project-d-kon-deathcards-build-log/#entry2148821 2 x-15 sundowns singer alt, odyssey bat, and maxwell ultra caps Link to comment Share on other sites More sharing options...
_paralyzed_ Posted October 17, 2016 Report Share Posted October 17, 2016 fricken just tell us already Quote Link to comment Share on other sites More sharing options...
ShadeTreeMechanic Posted October 17, 2016 Report Share Posted October 17, 2016 It's been answered about 10 times already. Lol Quote 91 C350 Centurion conversion ( Four Door One Ton Bronco) 250A Alternator (Second Alternator Coming Soon) G65 AGM Up Front / Two G31 AGM in Back Pioneer 80PRS CT Sounds AT125.2 / CT Sounds 6.5 Strato Pro component Front Stage CT Sounds AT125.2 / Lanzar Pro 8" coax w/compression horn tweeter Rear Fill FSD 5000D 1/2 ohm (SoundQubed 7k Coming Soon) Two HDS315 Four Qubes Each 34hz (Two HDC3.118 and New Box Coming Soon) Link to comment Share on other sites More sharing options...
snafu Posted October 17, 2016 Author Report Share Posted October 17, 2016 fricken just tell us already Obviously, you've missed the entire point of this thread. Quote Tony Candela - SMD Sales & Marketing Email me at [email protected] to learn about becoming an SMD Partner! Link to comment Share on other sites More sharing options...
snafu Posted October 17, 2016 Author Report Share Posted October 17, 2016 OK. Let's go back to post 1 and look at Ohm's Law. If we're going to prove / disprove the hypothesis I've laid out, this is an excellent place to start. I = E / R (see first post for what each stands for if you don't know already) For the sake of simplicity, let's assume that we'd like to apply voltage to a 10,000 Ohm (10k) resistor and do the math on that. Let's consider the following voltages: 12.6V (nominal voltage of a fully charged battery) I = 12.6V / 10,000 Ohms I = .00126A 13.8V (healthy charging system) I = 13.8V / 10,000 Ohms I = .00138A 14.4V (optimistic) I = 14.4V / 10,000 Ohms I = .00144A So, we can see that the formula does indeed support our hypothesis. So, now it's time to prove it - and that's the fun part. doesn't the resistor have to change to stay at 10,000 ohms at 14.4v versus at 12.6v? A resistor is a fixed variable. So, a 10,000 Ohm resistor is a 10,000 Ohm resistor as long as it's used within the design specifications. For example, you can't use a 1/2 watt resistor where you may need to dissipate 5 watts across it. Quote Tony Candela - SMD Sales & Marketing Email me at [email protected] to learn about becoming an SMD Partner! Link to comment Share on other sites More sharing options...
mothra Posted October 18, 2016 Report Share Posted October 18, 2016 (edited) resistors oppose current flow not voltage. you'd have to use a substantial value of resistance to lower voltage. example would be using a resistor to get rid of amp turn on pop in Ford vehicles. to alter voltage with resistors you'd have to use it in a voltage dividing network type of scenario. Edited October 18, 2016 by fasfocus00 Quote if nothing changes, nothing changes You don't know what you don't know, till you don't know Link to comment Share on other sites More sharing options...
DubNDodge Posted October 18, 2016 Report Share Posted October 18, 2016 Long ago, I found this image in a textbook and used it in my presentations. Now, you can buy a t-shirt with it. Love it, and it's absolutely relevant here. You will appreciate this then too 3 Quote '01 Dodge Stratass Sealed Trunk Build Log 2008 Honda Fit Sport Build Log On 10/3/2013 at 10:00 AM, ROLEXrifleman said: Anyone who says they knew everything they wanted out of life at 19 can go suck a bag of dicks cause they are lying to themselves or brought up in a cult. Link to comment Share on other sites More sharing options...
snafu Posted November 1, 2016 Author Report Share Posted November 1, 2016 The momentum here was lost. I'm trying something new here in an effort that those that follow this thread may learn something that otherwise would have not been learned. It's not about me asking a question and then giving you the answer. It's not about looking the answer up in the back of the book before you do the problem. It's about solving the problem to the best of your ability and whether you get it right or wrong you learn something in the process. If I give you the answer, nobody benefits. Still want the answer? Quote Tony Candela - SMD Sales & Marketing Email me at [email protected] to learn about becoming an SMD Partner! Link to comment Share on other sites More sharing options...
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