It's been a while since I studied ohms law and voltage drops/power.
I have a question as to the formula for calculating how much power each driver gets in a series circuit if the impedance is different in each driver.
Say: 1000W into a series circuit> first driver is 8 ohm> second driver is 4 ohms...
Total Impedance is naturally 12 ohms but how is the power calculated? Would it go 1000 / 3 = 333W... 333W going to the 8 ohm driver and 666W going to the 4 ohm driver?
What about 1000W series circuit> first driver 8 ohm> sedond driver has DVC of 1 ohm each (each in series makes it 2 ohm)...
1000W/10 ohms=100W... 200W going to the 8 ohm driver and 800W going to the DVC driver?
(my logic = 4x the impedance... 4 times less power???)
What's the formula to calculate how many watts each driver gets?
I have a question as to the formula for calculating how much power each driver gets in a series circuit if the impedance is different in each driver.
Say: 1000W into a series circuit> first driver is 8 ohm> second driver is 4 ohms...
Total Impedance is naturally 12 ohms but how is the power calculated? Would it go 1000 / 3 = 333W... 333W going to the 8 ohm driver and 666W going to the 4 ohm driver?
What about 1000W series circuit> first driver 8 ohm> sedond driver has DVC of 1 ohm each (each in series makes it 2 ohm)...
1000W/10 ohms=100W... 200W going to the 8 ohm driver and 800W going to the DVC driver?
(my logic = 4x the impedance... 4 times less power???)
What's the formula to calculate how many watts each driver gets?












If a driver had a ZERO impeadance (i.e. a shorted coil), then the voltage at it's termals would be the the same on the + & - leads. Essentially, no voltage drop. As you increase impedance, voltage differential would increase. Since the power consumed by the driver is volts * amps, and since the amps are the the same if the drivers are in series, then the driver with the highest voltage drop is going to be consuming the highest power.

