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Re: 3-15 B Cabinet (3x15') 4ohm or 8 ohm? [message #7587 is a reply to message #7585] |
Sat, 23 September 2006 12:57 |
Optyk
Messages: 125 Registered: August 2006 Location: Texas
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If I remember my math correctly, the formula is
1/ohmage + 1/ohmage + 1/ohmage = 1/x (where x = the actual load in ohms)
Taking that further, with 16's, you'd get
1/16 + 1/16 + 1/16 = 3/16 = 1/5.33333 so x = 5.33333 ohms total load
With 8's, you'd have:
1/8 + 1/8 + 1/8 = 3/8 = 1/2.66667 so x = 2.66667 ohms total load. This would be too low of a load for your amp.
I just built my "Shure Thang" kustom/peavey/shure contraption. the speakers I put in it were two 10's at 16 ohms each and one 8 at 8 ohms. If I calculated correctly, then I'd get
1/16 + 1/16 + 1/8 = 1/16 + 1/16 + 2/16 = 4/16 = 1/4, so x = 4 ohms total load. If I'm out to lunch on all this, someone straighten me out quickly before I have a problem. So far, it's all worked perfectly. I haven't been running those internal speakers actually. I have a pair of Crate small PA speakers rated at 8 ohms each and that's what I'm actually running instead of the internals.
I guess I should also mention that all of the above pertains to speakers wired in parallel, not series. In series, I believe it's very simple, you simply total up the ohmage of the three speakers and that's your load. So, in my case, 16 + 16 + 8 = 40 ohms wired in series. Not a workable situation.
Rod
There's only two kinds of music. . . . blues and zippity doo dah.
[Updated on: Sun, 24 September 2006 15:06] Report message to a moderator
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Re: 3-15 B Cabinet (3x15') 4ohm or 8 ohm? [message #7597 is a reply to message #7590] |
Mon, 25 September 2006 19:53 |
Optyk
Messages: 125 Registered: August 2006 Location: Texas
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I had forgotten about that method, steve. Like I said, it's been a long time since I took some electronics and used those formulae on a day to day basis. Leave it to me to use the hard way. LOL
Rod
There's only two kinds of music. . . . blues and zippity doo dah.
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Re: 3-15 B Cabinet (3x15') 4ohm or 8 ohm? [message #7601 is a reply to message #7600] |
Tue, 26 September 2006 00:59 |
C4ster
Messages: 686 Registered: June 2001 Location: Mukwonago, WI (Milwaukee...
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No, product over sum only works for 2 resistors. The impedance is 2.6666 ohms. The reciprocal method is used for more than 2. However you can group 2 and then group another 2 so 8X8/8+8=64/16=4 then 4x8/4+8=32/12=2.6666, That could work for any number.
Conrad
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