Hey folks. I've searched PE along with the web and have not found a definite answer to my question. I know that 1w/1m and 2.83v/1m is the same at 8 ohms, but that the 2.83v/1m at 4 ohms equates to 2w/1m, effectively doubling the power and resulting in a rating 3db higher than if the same 4 ohm driver was measured at 1w/1m. So, I have 3 questions...

1. If speaker A is rated at 93db at 1w/1m(4 ohms), and speaker B is rated 96db at 2.83v/1m(4 ohms), does that mean that they are both actually 93db at 1w/1m?

2. If speaker A is rated 93db at 1w/1m(this time it's an 8 ohm) and speaker B is rated at 96db at 2.83v/1m(this one a 4 ohm) am I correct in thinking that these 2 drivers are both actually 93db at 1w/1m as well?

3. Now, assuming the amplifier powering these speakers doubles it's ouput power when increasing it's load from 8 ohms to 4 ohms, as many amps do, does that mean that speaker B from question 2 will actually play 3 db louder than speaker A in question 2?

This seems like it should be simple, and many of you probably think it is, but I've read so much about it in the last 24hrs that all the info is actually confusing my thinking as opposed to helping. I'm hoping someone here can help set this straight. Thanks.

1. If speaker A is rated at 93db at 1w/1m(4 ohms), and speaker B is rated 96db at 2.83v/1m(4 ohms), does that mean that they are both actually 93db at 1w/1m?

2. If speaker A is rated 93db at 1w/1m(this time it's an 8 ohm) and speaker B is rated at 96db at 2.83v/1m(this one a 4 ohm) am I correct in thinking that these 2 drivers are both actually 93db at 1w/1m as well?

3. Now, assuming the amplifier powering these speakers doubles it's ouput power when increasing it's load from 8 ohms to 4 ohms, as many amps do, does that mean that speaker B from question 2 will actually play 3 db louder than speaker A in question 2?

This seems like it should be simple, and many of you probably think it is, but I've read so much about it in the last 24hrs that all the info is actually confusing my thinking as opposed to helping. I'm hoping someone here can help set this straight. Thanks.

## Comment