E=I/R (Voltage=Current/Resistance). Lets take a 2-channel (50W per channel@4ohm) amplifier. When bridged, the wattage is increased to 100. Is the resistance cut in half or the current doubled? In other words, do amplifier companies claim that wattage based on speaker resistance changing as a result of bridging, or does current actually double?


Modified by nolimits at 8:04 PM 10/21/2003

 

that's a confusing *** question, but i think the company rates their amps at how many watts both channels put out at 2 ohms, i'm not 100 percent sure but thats what i think, if i read your question right?

 

Sorry, I know it's confusing...proves I'm confused! I know factory rating on amps can be deceiving anyway, but if you have an amp that puts out an actual 50W rms per channel (2 channel) @4 ohm...then bridged would it put out 100W @ 2ohm? Or 200W @ 2ohm?

 

Think about it. The speaker (load) doesn't change, so why would the resistance change?

 

The load actually does change but not because of the bridging the load changes with frequency. That is why they say nominal impedence. Any time you hear impedence you should think time dependence. Frequency relates to time. A resistor is not time dependent and does not change its resistance with frequency.

The power increases because the voltage doubles when you bridge an amp you use both voltage rails.

for non technical (and even some technical) people I just tell them the impedence is cut in half. It is easier for them to understand and makes it easy for them to figure out power output.

 

I stand corrected. I forgot about that characteristic of a speaker. I was just thinking back to my days in digital electronics class.

Thanks for the refresher.