Let me know if my explanation isn't clear.
Oh, it's clear allright..
The correct answer was:
""John, you have accidentally applied a dc analysis to an ac circuit..let me explain...
Using DC circuit theory, the two loads are in parallel, equalling 5 ohms, and that is in series with the wire which is one ohm, for a total of 6 ohms. When the amplifier puts out 6 volts, the current splits evenly between the two equal loads, and the amplifier sees a load of 6 ohms at all times.
For the circuit we are talking about, there are two distinct paths that two frequencies will take...so the analysis requires examination of each independently.
First, let's apply 11 volts AC low frequency (peak, average, rms...it makes NO difference, so let's talk peak..)..to the circuit. The tweeter has no involvement (as the capacitor isolates it), so the amplifier is putting 11 volts into a circuit which has a 10 ohm load and a 1 ohm wire, so the woofer circuit sees one ampere..
Next, repeat this calcuation for the tweeter circuit..the woofer doesn't see anything, 11 volts hf produces 1 ampere in the tweeter circuit.
Now, if we add both signals, the amplifier is putting out 22 volts, each load sees one ampere, and the wire sees 2 amperes (summation).
So calculation of the powers results in:
1.. The amplifier is delivering 22 volts peak.
2. The amplifier is delivering 2 amperes peak.
3. The wire is seeing 2 amperes peak.
4. Each load is seeing 10 volts peak.
5. Each load is seeing 1 ampere peak.
Let's calculate power.
Load total power is 10 + 10, or 20 watts peak.
Wire total power is 4 watts peak. (2 amperes and 1 ohm)
Total power dissipated between the loads and wires is...24 watts peak.
The amplifier however, it has 2 amperes and 22 volts. For a total of 44 watts peak.
If we say lossless wires, it becomes 20 watts load, and 40 watts amp..
(This disparity is why biamping is so good, btw..)
That was the correct explanation...
Let me know if my explanation isn't clear.
Hmmm. you missed the correct explanation, given the numbers and explanation I used..
(sorry... it was a test..) The answer was quite simple, I set you up to correct me, and I was more than happy to take the "mistake hit". Your response means:
1. Either you do not understand the circuit very well, so provided floobydust.
2. You are posting the work of others and do not understand it.
3. You are posting in haste, and you missed the obvious.
Personally, I'd prefer to think that it was #3. Unfortunately, this statement:
If you want to calculate the average power and assume RMS values for the currents, then that will work out as you say. But my equation does not deal with RMS currents or voltages, but with instantaneous values.
Is in conflict with my assumption that you erred as a result of haste. As, using a DC analysis cannot result in the correct numbers, regardless of rms, peak, average, phase, whatever (correct me if I'm wrong here)...so saying that "it will work out as I say..well.."hmmmm."
Can you put some numbers into that explanation for us to verify it does indeed work? I'll post it again..
Yes, you are missing the fact that the filters are there. All the voltages and currents are instantaneous values. So there IS a component in the filters that you need to calculate as well. This will be the 20 watts you are missing.
Remember, only the average power in the filters is 0, not the instantaneous power!
If you want to calculate the average power and assume RMS values for the currents, then that will work out as you say. But my equation does not deal with RMS currents or voltages, but with instantaneous values.
To find the average power you need to plug in a concrete function for Ia and Ib (say a sine) and then integrate the resulting expression to obtain the average over time.
Cheers, John
ps...The best part about missing the obvious is the crow ya gotta eat..Believe me, I am no stranger to that..you should meet my co-workers..
