OHM loads

 

Silver Member
Username: Redbone15367

Mo U.S.A.

Post Number: 129
Registered: Sep-05
I know how to figure x2 but not diff loads like these. any tell me what these would be?
1.4 + .7 =
.7 + .7 + .7=
1.4 + 1.4 + 1.4=
Thanks
 

Silver Member
Username: Redbone15367

Mo U.S.A.

Post Number: 130
Registered: Sep-05
bump, help?
 

Silver Member
Username: Brynm

Prince Albert, SK Canada

Post Number: 142
Registered: Jul-05
for the same values you can cheat and divide by the number of values (.7+.7+.7)/3 = 2.1/3 = .7
three 1.4 ohm loads would be 1.4 a ohm load

for different values
(z1*z2)/(z1+z2) (for you it works out to .466 ohms)
you can use this for as many as you like however do not run two speakers like this! one speaker would be getting about 66% of the power and the other would get 33%, and would sound like crap
 

Silver Member
Username: B101

Queen City, NC USA

Post Number: 378
Registered: Sep-05
the question is NOT clear at all!

but yea, the Product/Sum rule will work!

if he wire them like that he is diffenatley creating a voltage divider rule. LOL
 

Silver Member
Username: Redbone15367

Mo U.S.A.

Post Number: 135
Registered: Sep-05
Thanks for the help. The reson for the ? is I have a mt quad.7 and a mt dual .7, that i plan on running just 1 2500d for now. Would like to keep it at .5 or higher.
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