Ohms Law

[bdawson72 0]

I'm slightly unsure as to what section to put this topic in so mods please move this if you see fit.

I'm a first semester electrical student and I'm pretty familiar with the ohms law wheel and its formulas. But, for some reason I fall short when trying to calculate the total impedances of a certain car audio circuits.

My question is what formulas do you guys use to figure out all of the different ohm loads that can be used w/ corresponding VC configurations.

Ex. 2 Dual 2 Subs

What are all of the ohm loads that can be reached.

I understand that anyone can go 2 the rockford fosgate wiring wizard and find a solution. But I'm more interested in how these values are derived.

[Acidburn 10]

Series: R1 + R2 + R3 + ... + RN = Rtotal

Parallel: 1/R1 + 1/R2 + 1/R3 + ... + 1/RN = 1/Rtotal

[bdawson72 0]

Series: R1 + R2 + R3 + ... + RN = Rtotal

Parallel: 1/R1 + 1/R2 + 1/R3 + ... + 1/RN = 1/Rtotal

Yea, i knew those 2.....But how is it that whenever i see people listing impedances that they have figured they have upwards of 4 different ones?

[Acidburn 10]

those are all the combinations that one could get with a certain set of coils

2 SVC or 1 DVC driver will have 2 options (series or parallel)

2 DVC drivers will have 3 options (series, series or series, parallel or parallel, parallel)

1 QVC driver will have 3 options (series, series or series, parallel or parallel, parallel)

[bdawson72 0]

I guess i will toy around with the formulas later

[DevilDriver 28]

Series: R1 + R2 + R3 + ... + RN = Rtotal

Parallel: 1/R1 + 1/R2 + 1/R3 + ... + 1/RN = 1/Rtotal

These two formulas are the only ones that can be used.

So, the question is: how can you possibly have 4 or 5 different possibilities when wiring multiple subs?

Well, quite simply: break it down.

Let's say there are 4 dvc subs. 2 of them, we will wire in series. The other 2, we will wire in parallel. Then we will wire each sub in the group to the other the same way we wired their voice coils, ie. the first group will be wired to each other in series, the second group will be wired to each other in parallel. Lastly, we will wire the two groups together.

You need to calculate each part of the circuit on it's own. To calculate the total resistance, we must know the resistance of each group first. To know the resistance of each group, we must know the resistance of each sub. To know the resistance of each sub, we must know the resistance of voice coil. If you change the way a particular step is wired, then you change the final resistance.

edit: That made sense when I first typed it out, though I'm not sure it still does. Please post if you need further clarification.

[///M5 2,833]

These two formulas are the only one that can be used.

wasn't sure whether to put this in my sig or just clarify that you meant

[juddspaintballs 0]

i did fairly well on this portion of my first engineering circuits test 3 years ago. then i fell asleep in class after that section and my circuits grade went to crap. BUT, at least i remembered how to do all this stuff and it's pretty darn simple with subs

[DevilDriver 28]

These two formulas are the only one that can be used.

wasn't sure whether to put this in my sig or just clarify that you meant

Haha.

Touche, sir.

[X-OvrDistortion 0]

Remember, when wiring in parallel, you final impedance will always be lower then you lowest resistance in the circuit. That is an easy way to make sure if you did your calculations right.

[Adrian_D 520]

Remember, when wiring in parallel, you final impedance will always be lower then you lowest resistance in the circuit. That is an easy way to make sure if you did your calculations right.

? detail please.

as far as i know, dcr is lower than the actual impedance.

too bad the drawings i made for VJ dissapeared during the data loss...

makes sense to me Neil. i always have hard times explaining these things

[X-OvrDistortion 0]

OK. Say you have a 5ohm load and a 10ohm load in parallel, you final ohm load will be less then 5 ohms.

1/5 + 1/10 = answer (reciprocal equation) 1/answer = impedance

.2 + .1 = .3

1/.3 = 3.333

It is just a fast way to make sure that you did your match correct.

Edited April 15, 2007 by X-OvrDistortion

[bdawson72 0]

I'm glad to see that this has turned into a technical discussion

[///M5 2,833]

Remember, when wiring in parallel, you final impedance will always be lower then you lowest resistance in the circuit. That is an easy way to make sure if you did your calculations right.

In parallel only and you needed to say lowest impedance in the circuit not lowest resistance.

[X-OvrDistortion 0]

Remember, when wiring in parallel, you final impedance will always be lower then you lowest resistance in the circuit. That is an easy way to make sure if you did your calculations right.

In parallel only and you needed to say lowest impedance in the circuit not lowest resistance.

That is why I said, and I quote "when wiring in parallel".

[///M5 2,833]

Remember, when wiring in parallel, you final impedance will always be lower then you lowest resistance in the circuit. That is an easy way to make sure if you did your calculations right.

In parallel only and you needed to say lowest impedance in the circuit not lowest resistance.

That is why I said, and I quote "when wiring in parallel".

Since impedance is not resistance I thought I should clarify.

[X-OvrDistortion 0]

Since impedance is not resistance I thought I should clarify.

Don't make me smack you boy!!

[ANeonRider 3]

How did Ohms Law make it to 2 pages?

[bdawson72 0]

no clue