Current flow through voice coils

[GreenMachine70 10+ year member]

Ok guys i've been pondering this thought for like two hours now, and i'm quite sure i've had a complete and total brain fart. The question is:

Q 1. Will the voice coil see less current flow (amps) at the same power level (watts) with i higher impedence (ohm's)?

For reference

V=volts (electromotive force)

P=power(watts)

R=resistance(ohm's)

I=current(amps)

Lets use a dual 2ohm 15" driver for and example and it will be given 3000 watts rms. Thus the amp will see a 1ohm load

I=Sq root (P/R)

I=sq root (3000/1)

I=54.77amps

now this is for the driver wired at 1ohm it would theoredicly see 54.77 amps @ 1ohm

V=P/I

V=3000/54.77

V=54.77volts

So the driver would see 54.77 amps @ 54.77volts wired at 1ohm

For example #2 lets use a 15" dual 1ohm driver at the same power level(3000w rms) This time the amp will see a 2ohm load

same formulas

I=sq root(P/R)

I=sq root(3000/2)

I=38.72amps

So @ 2 ohm with the same power level its seeing only 38.72amps. thats 16.05amps less than @ 1ohm

V=P/I

V=3000/38.72

V=77.47volts

So now with thw increase in resistance we've seen a decrease in current and an increase in voltage. With higher voltage level it requires less current to see the same output... theoredicly

So in theory I driver wired at a higher ohm load will see less current @ the same power levels. And a decrease in current will cause less of a thermal strain on your coil, making it harder to "smoke" the voice coil? I guess what i'm trying to say is if your sending less curent through your woofer wont it make it less likely to blow? given that everything else is ideal like charging/electrical system, clean non distorted signal etc...

Will someone tell me i'm right? i'm sick of thinking

Thanks

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[dragnix 5,000+ posts]

i'm sick of people typing too much //content.invisioncic.com/y282845/emoticons/crap.gif.7f4dd41e3e9b23fbd170a1ee6f65cecc.gif

 

[GreenMachine70 10+ year member]

i'm sick of people typing too much //content.invisioncic.com/y282845/emoticons/crap.gif.7f4dd41e3e9b23fbd170a1ee6f65cecc.gif

Its really not much just spaced out...

 

[Haunz 10+ year member]

So in theory I driver wired at a higher ohm load will see less current @ the same power levels. And a decrease in current will cause less of a thermal strain on your coil

NO......

 

[ultimate157 5,000+ posts]

NO......

Correct.

Even though you are lowering the total amperage in the circuit by going from parallel to series wiring, each coil will still be seeing the same amount.

Say you have a DVC 4 ohm driver and want to run it on 400 watts.

If you wire it to 8 ohms, series wiring, then total current running through the circuit is 7.07 amps. Since the wiring is in series, then the current through both components in the circuit is equal to the total current.

If you wire it to 4 ohms, parallel wiring, then total current in the circuit is upped to 14.14 amps. However, since they are wired in parallel, then each coil still has 7.07 amps running through it.

Edit: Using your example you are actually increasing the current through the coils by running the sub in series. Each coil carries the total amperage of the circuit.

In the parallel example, the coils are seeing equal amounts of current (since they are the same resistance) so 54.77/2 = 27 amps.

 

[vehementSPL 10+ year member]

Correct.
Even though you are lowering the total amperage in the circuit by going from parallel to series wiring, each coil will still be seeing the same amount.

Say you have a DVC 4 ohm driver and want to run it on 400 watts.

If you wire it to 8 ohms, series wiring, then total current running through the circuit is 7.07 amps. Since the wiring is in series, then the current through both components in the circuit is equal to the total current.

If you wire it to 4 ohms, parallel wiring, then total current in the circuit is upped to 14.14 amps. However, since they are wired in parallel, then each coil still has 7.07 amps running through it.

Edit: Using your example you are actually increasing the current through the coils by running the sub in series. Each coil carries the total amperage of the circuit.

In the parallel example, the coils are seeing equal amounts of current (since they are the same resistance) so 54.77/2 = 27 amps.

Agreed. I asked this ? about a month ago and honestly JUST NOW I realized I felt stupid .. the reason being. A 100 watt light bulb can get HOTT. At 120v only draws .83 amps of current which you would think would be NO HEAT> But can generate a lot of heat. wattage equals heat. Even if they made a 100 watt bulb that operated on 1000volts. Still would get just as hott. But would only draw .08 amps lol. again I just NOW thought of this analogy ... oh well I am an idiot at times

 

[GreenMachine70 10+ year member]

Agreed. I asked this ? about a month ago and honestly JUST NOW I realized I felt stupid .. the reason being. A 100 watt light bulb can get HOTT. At 120v only draws .83 amps of current which you would think would be NO HEAT> But can generate a lot of heat. wattage equals heat. Even if they made a 100 watt bulb that operated on 1000volts. Still would get just as hott. But would only draw .08 amps lol. again I just NOW thought of this analogy ... oh well I am an idiot at times

Thats the same reason why i kept doubting myself, but i guess we all have our "tard" moments. Thanks for clearing this up guys.

 

[Immacomputer 10+ year member]

Actually, you're correct that decreasing current lowers heat but in the case of a sub, wiring the coils in series or parallel won't make a difference as ultimate157 said.

But that's exactly why power plants step up the voltage like crazy when leaving the plant. But, in that case, the resistance is set by the transmission line and the real difference is how much power is lost in the load.

Power is power though. If you want your sub to perform like it is getting 100watts, you will need 100watts. The current and voltage is pretty much irrelevant as long as you're getting the correct power.

 

[MyFartsStink 10+ year member]

voltage is E not V //content.invisioncic.com/y282845/emoticons/fyi.gif.9f1f679348da7204ce960cfc74bca8e0.gif

 

[ultimate157 5,000+ posts]

But that's exactly why power plants step up the voltage like crazy when leaving the plant. But, in that case, the resistance is set by the transmission line and the real difference is how much power is lost in the load.

Power plants step up the voltage for 2 reasons. If they were to run low voltage but high current, then they would need massive cables to carry the current.

The second reason is to overcome the high resistance of the long distances that the power lines span in between power stations.

 

[GreenMachine70 10+ year member]

voltage is E not V //content.invisioncic.com/y282845/emoticons/fyi.gif.9f1f679348da7204ce960cfc74bca8e0.gif

E or V, both are used for representing voltage, when i was in school we used V.

 

[Kyle_Keating 10+ year member]

amps don't matter, its power or better yet, energy transfer.

P=IV

relativity speaking, you need both high volts (potential energy) and high amps (amount of charge(electrons) moving per unit of time) to have high power. Think of the electrons as marbles, then roll them down a street that has a slight tilt and let them hit your foot after 100 yards, no big deal. Now drop them 100 yards in the air (high voltage, same current) and let them hit your foot. Big deal!

Also, the current, resistance (impedance) and power for that matter are non-linear with respect to frequency so they are change all the time, every split second, the only thing that remains constant is the voltage unless you're playing only 1 frequency... a la burp test? or what-have-you. What really matter is how much voltage you need for your specific subwoofer(s) and how they are wired. Most amps will produce different voltage maximums for different resistive loads. The voltage is called your "gain" which depends on on that setting and your input voltage from your headunit too.