Question about ohms law?

[the727kid 10+ year member]

E=volts but volts as in what, DC or AC (battery volts, or the volts you get when you meter speaker terminals). It seems to me that it would be AC volts.

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[KaeZoo 10+ year member]

Ohms Law equations are valid for DC or AC.

 

[the727kid 10+ year member]

That's confusing as hell. What if you want to find say wattage....volts x amps.....what voltage would I use?? The only time I seem to come back to a matching anwser is when using the AC volts.

 

[luvinthebass 5,000+ posts]

AC is alternating current so that's to speakers, DC is direct current so that is battery voltage, hope that helps ya out

 

[luvinthebass 5,000+ posts]

That's confusing as hell. What if you want to find say wattage....volts x amps.....what voltage would I use?? The only time I seem to come back to a matching anwser is when using the AC volts.

you would use the voltage from the battery or if ya measured the voltage usin the ground and power terminals at the amp you'd use that number then, at the battery is more accurate.

 

[the727kid 10+ year member]

So when do I know when to use DC and AC to figure out what I am looking for? Currently I am trying to figure out how many amperes one of my amps draw. I could do...I=P/E. Which would be say 1800/14.4(DC) which means 125 amps of draw (seems a bit high). Or I could do 1800/60(AC) which means 30 amps of draw (seems more realistic.

It seems to me AC is the sopposed voltage because randomly throwing in DC messes up the whole chart and every equation. Ex. if you use the equation I = Square Root of (P / R) with the same info, the square root of 1800/2 = 30 amperes. I.E. AC volts matches, DC does not.

Now, furthermore of an example, I want to find what Ohms I am running at. I will use the equation R = E 2 / P.

With AC volts, 60squared/1800 = 2ohms. With DC volts, 14.4 squared/1800= a VERY low impendence.

Also, with the equation E = Square Root of (P x R), the square root of 1800x2 = 60, not 14.4. Now with this equation, I would have to be running 103 watts of power at 2 ohms to get a voltage of 14.4.

What am I missing?

 

[knukonceptz 10+ year member]

Watts = Volts * Amps

1800 = 14.4 * X

1800/14.4 = X

X= 125A

Plug it back in, works just fine. You are missing efficiency of the amp though

 

[Thieroff 10+ year member]

http://www.bcae1.com

/thread

CT

 

[KaeZoo 10+ year member]

So when do I know when to use DC and AC to figure out what I am looking for? Currently I am trying to figure out how many amperes one of my amps draw. I could do...I=P/E. Which would be say 1800/14.4(DC) which means 125 amps of draw (seems a bit high). Or I could do 1800/60(AC) which means 30 amps of draw (seems more realistic.

Where are you getting 60 (AC)??? Your battery doesn't produce AC power. Your alternator does, but it's converted to DC before it's used anywhere in the vehicle. The unrealistic number is probably the 1800, but if that's the right wattage number then 125 amperes of current wouldn't be out of line.

It seems to me AC is the sopposed voltage because randomly throwing in DC messes up the whole chart and every equation. Ex. if you use the equation I = Square Root of (P / R) with the same info, the square root of 1800/2 = 30 amperes. I.E. AC volts matches, DC does not.

Why are you randomly throwing in DC? Your circuit is either using DC or AC. The amplifier draws DC current through the power wire, and outputs AC at the speaker terminals. Why are you mixing the two?

Now, furthermore of an example, I want to find what Ohms I am running at. I will use the equation R = E 2 / P.With AC volts, 60squared/1800 = 2ohms. With DC volts, 14.4 squared/1800= a VERY low impendence.

Also, with the equation E = Square Root of (P x R), the square root of 1800x2 = 60, not 14.4. Now with this equation, I would have to be running 103 watts of power at 2 ohms to get a voltage of 14.4.

What am I missing?

I think you're just confusing yourself. If you know two components of Ohm's/Watt's laws, you can determine the others, and the formulas are the same whether you're working with AC or DC. But you have to start with real-world measurements, and not mix up two different circuits.

First example: you want to figure out amperage through the power wire, using a wattage rating of 1800 and a voltage of 14.4VDC. If these numbers are correct, then using I=P/E, 125 amps is the correct amperage. (Leave aside that you're assuming output power is equal to input power, which isn't true because amplifiers aren't 100% efficient). Your 60VAC reading at the speaker terminals isn't relevant to the current draw at the main power input. On the other hand, let's use the same equation to find the amperage at the OUTPUT. I=1800/60 so you're running 30 amps through the speaker wires. Same equation, both answers are correct though one is for DC and one is for AC. You're just assuming that the DC answer is wrong because the 125 amps answer doesn't "feel" right to you, but why do you think amplifiers require much larger power wire than speaker wire?

Second example, calculating impedance from power and voltage. Of course the DC calculation gives you a crazy answer, because 14.4VDC isn't present at the speaker terminal. Garbage in, garbage out: if you plug in the wrong number, you'll get the wrong answer. But guess what: if your amp was producing 1800 watts, and only 14.4 volts were present at the speaker terminals, you can bet your speakers would have to be a VERY low impedance.

Bottom line, in every example you're mixing up power line voltage with speaker terminal voltage, then blaming the DC answer because it doesn't seem right. For power line calculations, use 14.4VDC. For speaker output calculations, use 60VAC. In both cases, your answers will be correct if your other numbers are accurate.

 

[the727kid 10+ year member]

"For power line calculations, use 14.4VDC. For speaker output calculations, use 60VAC."

Thanks, I get it now. That's all I needed to hear to understand it lol. I see what I was doing wrong now. Appreciate it.

 

[luvinthebass 5,000+ posts]

"For power line calculations, use 14.4VDC. For speaker output calculations, use 60VAC."
Thanks, I get it now. That's all I needed to hear to understand it lol. I see what I was doing wrong now. Appreciate it.

lol it's okay you're learning , that shit was confusing as hell to me too man when i was figuring it out.