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The thing you're asking for is 4 quadrant control. What that means is, the motor and controller can implement 4 different modes: Motor going clockwise, controller applying clockwise drive (forward) Motor going anti-clockwise, controller applying anti-clockwise (reverse) Motor going clockwise, controller applying anti-clockwise drive (braking forward) ...


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The images you gave are IMO a kind of roundabout way of the following: simulate this circuit – Schematic created using CircuitLab This circuit is equivalent to the one shown in your figures. And calculating the Thevenin-equivalent of the boxed part will get you your answers: $$e_T = \frac{R_{B1}}{R_{B1}+R_{B2}}\cdot V_{CC}$$ $$R_T = R_{B1}//R_{B2}$$ \$...


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You really can't at your level, not without precise machining and alignments. Your best bet is to rectify your voltage and boost it with a converter if you're looking for more voltage, or parallel them AFTER the diodes for more current. This way the diodes on the outputs will prevent backfeeding the other alternator.


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If the alternator shafts are coupled together, the rotor windings could be connected in series as shown below. Obviously the three terminals of the neutral connection on one alternator would need to be accessible. This is not something that can be done with the alternators driving loads. They would need to start together and have their fields adjusted if ...


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For a servo position control loop then using a tacho-generator can reduce the speed at which the motor approaches the optimum demanded position. This means that there is less overshoot because the mass of the "thing" being moved is less likely to significantly overshoot the demand point because it doesn't have as much momentum/energy. Because a ...


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A motor is a backwards generator. So flip all the specifications around. That motor has a no-load RPM of 13500RPM when 6V is applied. So for you to get it to produce 6V, no-load, you must spin it at 13500RPM. Spin it at 1/3 the speed and you will produce 1/3 the voltage. Can you pull that off manually? Probably not. Pick a motor that spins the lowest ...


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The speed (RPM) at any instant will determine the voltage and for AC also the frequency. The current drawn by the load will determine the torque required. You can design a voltage converter that will convert whatever voltage it receives to the voltage that you want. The available output current will be governed by conservation of energy. Mechanical power in (...


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As usual, you need to understand your system. What generator do you use? Do you have any gear? What load is on your generator? Basically, you will get all the power you have on input, minus the losses. The losses include friction, heat on the generator, and other parts. Probably for each point on the speed chart the losses will differ, which means you may be ...


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I would say, 'Go for the 3 phase generator' for the following reasons: Absolutely no changes would be required to be made in the existing wiring. Back-up power would also be ensured for 3 phase loads which may be installed in the future.


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If you have a 3-phase generator, you need a 4-pole changeover switch to switch between the grid and the generator, switching all 3 phases and neutral. Never allow the generator and grid to be connected at the same time. If you have a single phase generator that is large enough to drive all the loads, then set up a 4-pole changeover switch to change between ...


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How would the single phase generator be connected in this scenario? If you have a few critical loads that you want to power with a generator during an outage, a single-phase generator would be ok if those loads are all on one phase or if they can conveniently re-connected to one phase. However you connect the generator, you must have either contactors or a ...


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It sounds like the voltage and frequency of the temporary generator power were both overly high. Grid-tied solar power systems measure the frequency and voltage of incoming power and do not supply their own if it is not within expected ranges. This is for a number of reasons that can be generally described as "if there is a problem, don't make it worse&...


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The standby mains voltage and frequency would not be as stable as the grid.Where would exported solar energy go if the neighbours were not drawing much power ?.The voltage waveform would probably have more harmonics present .This is why the power is not as good .Also the power is expensive to generate .


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Given an AC generator with 2 ohms internal resistance, the answer would be to use a step-down transformer, say, 4.5:1 in voltage, or 20:1 in impedance, to step that 2 ohm impedance down to 0.1 ohm at the secondary. To generate 190 amperes through 0.3+0.1 = 0.4 ohms, you need 76V at the secondary, or 342V on the primary, and only 42A from the generator. This ...


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437V is accounting for the voltage drop [across the generator resistance]. 437V is what is required to push 190A through 2.3 Ohms, some of which is the load and some of which is the generator's resistance. 57V is the voltage you would need not accounting for the voltage drop since If the generator had no resistance you would only need the generator to ...


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