Hi,
I have a Peavey Duel 212 with, well, two 12" drivers, funnily enough.
The drivers are 8 ohms connected in series which makes a 16 ohm load.
So, to increase the amps flexibility a bit I tried wiring each speaker individually and inputting them into the speaker outs in parallel, two 8 ohm loads would then make a 4 ohm load when using both speaker outputs on the amp.
Odd thing though, when I measure the resistance of the speakers wired in series it measures 16 ohms, however, each speaker on its own only measures 0.8 ohms.
I figured I was just reading things wrong and wired one speaker up and gave it a go with the amp switched for an 8 ohm load. It was unusually quiet, so I figured probably best not to continue, something was clearly not right.
I've rewired speakers cabs before and have never come across this. How can the series resistance of both drivers together measure 16 ohms, but the drivers individually only measure 0.8 ohms?
I look forward to being educated.
I have a Peavey Duel 212 with, well, two 12" drivers, funnily enough.
The drivers are 8 ohms connected in series which makes a 16 ohm load.
So, to increase the amps flexibility a bit I tried wiring each speaker individually and inputting them into the speaker outs in parallel, two 8 ohm loads would then make a 4 ohm load when using both speaker outputs on the amp.
Odd thing though, when I measure the resistance of the speakers wired in series it measures 16 ohms, however, each speaker on its own only measures 0.8 ohms.
I figured I was just reading things wrong and wired one speaker up and gave it a go with the amp switched for an 8 ohm load. It was unusually quiet, so I figured probably best not to continue, something was clearly not right.
I've rewired speakers cabs before and have never come across this. How can the series resistance of both drivers together measure 16 ohms, but the drivers individually only measure 0.8 ohms?
I look forward to being educated.