Question about wattage.

I'm trying to hook-up 3 tactile transducers wired in series to a single mono amp. The amp's power rating exceeds the recommended wattage of 2 of the transducers. The third transducer is capable of accomodating the amp on its own. I'm not sure how the power is distributed. Will the lesser transducers be ok online with the amp as long as they are in series with the higher rated transducer?
I'll assume your transducers are 4 ohms. Resistance in series adds up so putting the transducers in series makes a total of 12 ohms. This will decrease the total power available from the amp. The transducers will each take a portion of the power. An equal amount if they are the same impedance.

You need the following formulas:

Voltage = square root of power times impedance
Current = voltage divided by impedance
power = current squared times impedance

To figure it out get the rated power of the amp into 4 or 8 ohms. Take this power times the ohms and then the square root of that. This will give you the voltage rating of the amp. So if you have 100 watts into 8 ohms it will be the square root of 800 = 28 volts. 200 watts into 4 ohms gives the same answer of 28 volts.

Since you now have 12 ohms the maximum current will be 28V divided by 12 ohms = 2.3 amps. Current in series is the same so each will have 2.3 amps. Take the square of 2.3 times 4 ohms = (2.3 x 2.3 x 4) = about 22 watts for each transducer or a maximum from the amp of 66 watts.

If they are different impedances do the same calculations and you will see that the higher impedance gets more power. With one 8 and two 4 ohm you would have a total of 16. The same 28 divided by 16 = 1.8 amps. The 8 ohm would get (1.8 x 1.8 x 8) = 25 watts and each 4 ohm would get (1.8 x 1.8 x 4) = 12.5 watts.
Nice answer Herman.

Since I did not know what a tactile transducer was and for what use in audio such a device has...