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Hypothesis

If we increase the voltage delivered to an accessory in a DC circuit we will also increase the current delivered to that accessory.

History

Ohm's Law implies this by the formula I = E / R, where:

I = Current in Amperes

E = Voltage in Volts

R = Resistance in Ohms

If you think about what Ohm's Law really tells you, it's pretty simple - Voltage giveth and resistance taketh away. So, really what this means is that for an automotive electrical system with a fixed voltage level (for the sake of this discussion, let's say that's 14 Volts DC), accessories being run from it either run optimally or less than optimally. Now, if we consider our hypothesis, we could argue that if we decrease voltage to a given accessory by increasing the resistance in the circuit we will also decrease the current delivered to it.

Who can prove or disprove this?

And, before you ask me to provide more information, you have all that you need to solve the problem. Spell it out in a way that the laymen can understand. Or, better yet, do a video. Show me. If nobody can do this, I'll do a video myself. So, either way, you get the profit of learning. You'll learn far more if you solve this one yourself though.

Let's all learn a little about science here and how to apply it to our favorite hobby.

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I don't know about proving or disproving, but I can give my answer and explain why.

If you decrease voltage going to an accessory, you are not going to decrease the current available to it, but you will decrease the amount of current the flows through it. Current "available" is not determined by voltage, but current the passes through is.

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For me, rewriting the formula helps me understand this. You said I = E/R but I understand this better by looking at all versions of the equation. So I = E/R or E = I*R or R = E/I. Voltage is either the product OR the numerator in all equations. This tells me at least, just using math, that you can't simply say as voltage goes up so does current. That assumes resistance stays the same. If voltage goes up, either current increases, resistance decreases, or some combination of both.

 

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I'll say current goes down as voltage goes up, or that current goes up as voltage goes down. If an accessory needs 100w to operate we can look at P=VA since this is DC power. Lets use 100w for instance, 100w/14v=7.1A. If we decrease that voltage to 12, 100w/12v=8.3A. I do not believe that an accessory has a fixed resistance, but it will vary as needed to accommodate any fluctuation in voltage.

 

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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.

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It's yes and no. In the case of simple items such as bulbs where the resistance is more or less fixed (not going to get into temperature-resistance changes) - a reduction in voltage will reduce the current to the device. It's in the equation.

Motorized items like starters, power windows, HVAC are a bit more complicated since you have inductance to worry about, it's not necessarily 1:1, but they still follow the same trend. Reduce voltage, the current will drop, and so will the performance of your chosen accessory.

It's also worth noting that the power applied to simple accessories and motorized items increases exponentially with voltage, P=v2/R. That's why you need to be careful when running any sort of higher voltage system. It's easy to push components way beyond their intended operational capability. Let's take 18v vs. 14.4v. It's only 3.6v more, but in following the equation listed, will provide a 65% increase in power being supplied to a given component.

Now when we start talking about electronics with regulated power supplies, it gets a bit different. LED/LCD displays, TCM/BCM/ECM, etc have a bit of an inverse relationship. They're designed to provide constant performance across varying voltages. In that case, as voltage goes down, current draw goes up to maintain the same power envelope (wattage).

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I have a variable voltage 50A power supply sitting on my desk. I should just rip out a window motor just to test this lol

 

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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.

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I have a variable voltage 50A power supply sitting on my desk. I should just rip out a window motor just to test this lol

Keep me posted if you do I'm curious of the results. Does it maintain a fixed current-voltage ratio as does a resistor? Logarithmic? Exponential?

Edited by SnowDrifter
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Hypothesis

If we increase the voltage delivered to an accessory in a DC circuit we will also increase the current delivered to that accessory.

History

Ohm's Law implies this by the formula I = E / R, where:

I = Current in Amperes

E = Voltage in Volts

R = Resistance in Ohms

If you think about what Ohm's Law really tells you, it's pretty simple - Voltage giveth and resistance taketh away. So, really what this means is that for an automotive electrical system with a fixed voltage level (for the sake of this discussion, let's say that's 14 Volts DC), accessories being run from it either run optimally or less than optimally. Now, if we consider our hypothesis, we could argue that if we decrease voltage to a given accessory by increasing the resistance in the circuit we will also decrease the current delivered to it.

Who can prove or disprove this?

And, before you ask me to provide more information, you have all that you need to solve the problem. Spell it out in a way that the laymen can understand. Or, better yet, do a video. Show me. If nobody can do this, I'll do a video myself. So, either way, you get the profit of learning. You'll learn far more if you solve this one yourself though.

Let's all learn a little about science here and how to apply it to our favorite hobby.

I agree with DubNDodge. The higher your voltage is, the less current is needed to power a circuit, and vice versa.

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How about an industrial 3 phase motor that can be run at either 240vac or 480vac. The data plate shows less current at the higher voltage. This can be verified with a clamp meter. I have also seen power connections burn up on equipment designed for 230vac that was run on 208vac because that's all we had.

Edit: this is @ 60hz

Edited by ShadeTreeMechanic

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One time my dad bought a craftsman air compressor. Where we wanted to put it was just out of reach of the 120v outlet. All we had was a 100ft extension cord. We just wanted to get it going so that's what we used. It would kick on for 10 seconds and then trip the 20A main breaker in the breaker panel. One would think that the long cord would add resistance and lower the current. We moved it closer and plugged it in direct. No problem. The 100ft cord is still good and still in use for other things.

I think it is a natural tendency to try to apply dc theory to ac. With ac, the phase relationship of the voltage vs current needs to be taken into account.

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)

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