My Ohm question...

Fret Wilkes · November 29th, 2006, 12:38 PM

I beleive that 2, 8 ohm cabinets in parellel yield 4 ohms.

What happens when you run a 4 ohm and and 8 ohm cabinet in parellel?

Thanks,
Fret

Jim.Collins · November 29th, 2006, 12:43 PM

If you have one load, r1, and another load, r2, and wish to find the total load, R, when the two loads are in parallel, the formula is:

1/R = 1/r1 + 1/r2

1/R = 1/4 + 1/8

1/R = 3/8

R = 8/3

R = 2.67

In general, the formula is:

1/R = 1/r1 + 1/r2 + ... + 1/rn

Fret Wilkes · November 29th, 2006, 12:59 PM

Quote:

Originally Posted by Jim.Collins

If you have one load, r1, and another load, r2, and wish to find the total load, R, when the two loads are in parallel, the formula is...

Thanks Jim MOST helpful!

Fret

BrassRat · November 29th, 2006, 04:47 PM

Fret,

I've noticed a couple of things playing with this formula that I don't necessarily think are intuitive:

1) any two resistances in parallel always aggregate to less than the smaller of the two (since the current has alternate paths - at least that is the way I conceptualize it).

2) if the values of the individual resistances are identical (like in your 2x8 ohm example), then R = r/n. For example, 4 x 8-ohm speakers in parallel will give you a 2 ohm load.

BTW, inductance (for inductors in parallel) and capacitance (for capacitors in series) behave according to the same math.

Fret Wilkes · November 29th, 2006, 04:49 PM

Quote:

Originally Posted by BrassRat

Fret,

I've noticed a couple of things playing with this formula that I don't necessarily think are intuitive:

1) any two resistances in parallel always aggregate to less than the smaller of the two (since the current has alternate paths - at least that is the way I conceptualize it).

2) if the values of the individual resistances are identical (like in your 2x8 ohm example), then R = r/n. For example, 4 x 8-ohm speakers in parallel will give you a 2 ohm load.

BTW, inductance (for inductors in parallel) and capacitance (for capacitors in series) behave according to the same math.

Thank you sir!

November 29th, 2006, 12:38 PM	#1 (permalink)
Fret Wilkes Tele-Holic Join Date: Mar 2003 Location: Harmony, R.I. Age: 51 Posts: 955 28 Photos	My Ohm question... I beleive that 2, 8 ohm cabinets in parellel yield 4 ohms. What happens when you run a 4 ohm and and 8 ohm cabinet in parellel? Thanks, Fret __________________ See, I didn't reconize ya when ya first drove up till ya stuck yer hand out and wove then I seen right off who ya was and knowd ya. Epiphany: The Ignore List can make the TDPRI and even BETTER place to be!

November 29th, 2006, 04:47 PM	#4 (permalink)
BrassRat Tele-Meister Join Date: Apr 2006 Location: Arlington, VA Age: 43 Posts: 269 6 Photos	other useful R tips... Fret, I've noticed a couple of things playing with this formula that I don't necessarily think are intuitive: 1) any two resistances in parallel always aggregate to less than the smaller of the two (since the current has alternate paths - at least that is the way I conceptualize it). 2) if the values of the individual resistances are identical (like in your 2x8 ohm example), then R = r/n. For example, 4 x 8-ohm speakers in parallel will give you a 2 ohm load. BTW, inductance (for inductors in parallel) and capacitance (for capacitors in series) behave according to the same math. __________________ "It's not your job to be as confused as Nigel." Last edited by BrassRat; November 29th, 2006 at 04:48 PM. Reason: spell check

November 29th, 2006, 12:43 PM	#2 (permalink)
Jim.Collins Tele-Meister Join Date: Feb 2005 Location: Pleasanton, CA Age: 54 Posts: 468	If you have one load, r1, and another load, r2, and wish to find the total load, R, when the two loads are in parallel, the formula is: 1/R = 1/r1 + 1/r2 1/R = 1/4 + 1/8 1/R = 3/8 R = 8/3 R = 2.67 In general, the formula is: 1/R = 1/r1 + 1/r2 + ... + 1/rn