Determination of Generator Output

We spun the generator with an electric motor at a constant 1,750 RPM. With no load we obtained 120 Volts DC on a single coil, using number 14 wire which is typically rated at 14 Amps. (There are a total of 10 coils in the generator.) We then loaded the generator coil until our output voltage had dropped by half to 60 Volts. (The output voltage dropping to half of the 'no load' voltage when a load is applied, indicates that the external load resistance and the internal supply (generator) resistance are equal.) At this point the generator was producing 20 Amps at 60 Volts on a single coil. This indicated that the generator was loaded to its maximum, useful load - fifty percent of the energy dissipated internally (96% Mechanical, 4% heat) and 50% dissipated externally (100% heat in our test). Decreasing the resistance of the load resistor below this value would only cause more energy to be dissipated in the generator rather than external to it. External load resistance should never be less than the internal resistance of your 'power supply'. It can be equal to or greater than but never less than internal resistance. Using Ohms Law, we determined our coil resistance was {R=V/I} 60/20 = 3 Ohms. Since we intend to operate the generator at 120 Volts at all loads, not just 60 Volts, this would increase our available current (I) as well. As the resistance of the coils remains essentially constant, we have 40 Amps available to use under full load when maintaining 120 Volts! Ohms Law verifies this: {I = V/R} 120V/3Ω = 40 Amps. This is twice as much as we can make use of under full load while using a number 14 wire. This gives us a very respectable headroom of 100%!

Since we will be putting two 120 Volt AC inverters* on the generator, 180 degrees out of phase with each other, each utilizing 5 coils, our internal resistance at 120 Volts will be .6 Ohms or at 240 Volts, 1.2 Ohms. Because of this, we will get a maximum, short-term output of 200 Amps x 2 at 120 Volts (rated at 100 Amps x 2 continuous) or at 240 Volts, a maximum short-term output of 200 Amps at 240 Volts (rated at 100 Amps continuous), or any combination of the above.

In other words, what we have is a 24kW 120/240V generator able to produce 48kW for short periods of time. Using Ohms Law to verify this: {W = VxI} 120V x 200A = 24,000 Watts or 240V x 100A = 24,000 Watts, under normal full load, or for short time periods, 120V x 400A = 48,000 Watts or 240V x 200A = 48,000 Watts. This time limit has not yet been determined, but with good cooling it would be minutes and not seconds as with most generators. The heat from the generator coils is only 4% of total output Watts. That would be about 1,920 Watts at 48kW, displaced over an area of 537.5 square inches or 3.5 Watts per square inch at maximum load (48kW). Under normal full load (24kW) heat output would only be 1.75 Watts per square inch! Typically, normal use is around 1.2kW, making the heat output only .09 Watts per square inch, leaving the generator cold to the touch.

*Using two inverters at 180 degrees out of phase with each other allows us to have 120/240 Volts and we can obtain 50 or 60 Hz simply by the use of a switch. Also, it will be very simple to connect multiple generators together in series to increase total output, since the timing signal can be derived from one unit, thus driving all the inverters in perfect sinc with each other.