You're right on about the speakers blowing, but,
I don't think the majority of the power would go to the 8 ohm load. I think it'd be divided evenly.
Unfortunately, yes, it would (not be divided equally, that is). *That's basic first year electronics or physics Ohm's Law calculations, BTW, I have a 4 yr degree in it. *The 8 ohm side would have less resistance to the current being dumped into it than the 24 ohm side. *
Damn, gotta get out my calculator again, assuming a 100W amp:
Total impedance of 8 ohm &#0124;&#0124; 24 ohm = 1/((1/
+ (1/24)) = 6 ohms.
There would be about 24.5V across each load, V is the same since they are &#0124;&#0124;, V=sqrt(100W*6ohms). * *
Now, the current through the 8 ohm leg = 24.5V/8ohms = 3.06A and the current through the 24 ohm leg is 24.5V/24ohms = 1.02A.
We can check this by finding the total current, 100W/24.5V = 4.08A which is equal to 3.06A + 1.02A
Now, the power dissipated in the 8 ohm leg = 24.5V * 3.06A = 75W (74.97 but I'm rounding).
The power through the 24 ohm leg = 24.5V * 1.02A = 25W.
Basic Ohm's law equations:
Rseries = *R1 + R2 + R3 + ... + Rn
Rparallel = 1/((1/R1) + (1/R2) + (1/R3) + ... *+ (1/Rn))
V = IR *(V = Voltage, I = current, R = Resistance)
therefore, I = V/R and R = V/I
P = IV (P = power)
therefore, P = Isqrd * R = Vsqrd/R *(sqrd means squared)
These are the tools you need to figure loads (impedance) and how power, voltage and current is divided in circuits. *For series/parallel resistance circuits figure the series Rs in each leg first then do the parallel calculation. *Capacitors and inductors add a new dimension, their impedance (called reactance) is frequency dependent but you don't have to worry about them in speaker load calculations. *PA spkrs with built in crossovers have caps in them but you can usually just figure the load based on the cab rating and ignore the fact that the caps are there.
Hope this is helpful, e-mail me if anybody has any questions,
(Edited by rgr at 1:32 pm on Jan. 25, 2001)
(Edited by rgr at 10:27 pm on Jan. 25, 2001)