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To answer this question a voltage needs to be stated. Wire is sized by the amount of amperage the load takes. W = Amps x Volts. Amps = 650/ volts.
It depends on the voltage of the compressor. Two horse power (electric) is 1492 watts, but watts are volts times amps, so you need to know the voltage. Since the motor is an inductive load, you will also need to know the power factor, so as to compensate for true vs apparent power.
You need to rewire to a lower voltage at panel or get a transformer that steps down 277 to 110 volts. Make sure wire is sized for new load as well as breaker.
To reduce 12VDC to 7VDC you need a voltage divider in the ratio of 7/12, such as two series resistors of values 5KOhms and 7KOhms. Of course, this "trite" answer (the resistor divider) depends on how much power you need to supply, and how much source to load current regulation ratio you need. You might need to follow this up with an class C amplifier, compensating for the 0.6 volt drop across the base-emitter junction. You might need an OP-Amp controlled emitter follower, with a zener reference. For high current applications, you might even need a switching regulator.
Yes. You can use a voltage divider. Say, for instance, one 1KOhm resistor in series with a 3KOhm resistor. Connect the 3k resistor to the 48 volts and connect the 1k resistor to ground. The 1k resistor will have 12 volts acress it. These resistors need to be at least 1 watt each as they are going to dissipate 0.576 watts and get warm. Now, if you attempt to pull power from the 1k resistor, note that regulation will be poor because the impedance of the load will go in parallel with the 1k resistor and change its value.
Watts = Volts x Amps So you need to know the current. This is only for resistive load.
No, you will need to obtain an adapter to change the 120 volts to 230 volts. The adapter must be sized to the load wattage of the 230 volt appliance. To find the load wattage multiply the amperage times the voltage of the appliance. Once this is assessed, the size of the adapter or transformer will be of equal or greater value than that of the connected appliance. The adapter will be rated in VA or KVA depending on the needed load amperage.
You need to have the amperage to determine how many volts you get out of 20 watts.
To answer this question a voltage needs to be stated. Wire is sized by the amount of amperage the load takes. W = Amps x Volts. Amps = 650/ volts.
You will need to take the resistance of the load into account if you are going to design a voltage divider. The resistance of the load can completely change the voltage ratio of a voltage divider if not factored into the calculation. you can measure or read R(load), then R(needed) = 0.8 R(load)
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Assuming a resistive load, the continuous current flowing would be 600/220 = 1.36 amps. The resistance of the load is 220/1.36 = 162 ohms. If you have a 200 ampere hour battery that only supplies 24 volts you can't run your 600 watt device that is designed to run at 220 volts. For sake of argument, say your load is an incandescent light bulb designed to work at 24 volts. If you attached the battery it would try and draw 600/24 = 25 amps and the resistance of the load would be about 1 ohm. You need to match the voltage source to the load requirements. CAVEAT - This example assumes that if a 24 volt battery was used that the 600 watt device was made to work for 24 volts. It is not the same load that would be for a 600 watt device at 220 volts. The problem is that the hypothetical question asked does not match reality.
To answer this question a current value needs to be given. 125000 is the product of amps times volts. I = W/E = 521 amps. So to bring 240 volts up to 125 KVA you need the circuit load to draw 521 amps.
Wire sizes are governed by the amperage the wire is to carry. To answer this question the load current is required.
Ohm's Law states Volts = Amps x Resistance. You would need to apply 600 volts across 3 ohm load to have 200 Amps flow in circuit. Not sure what you are really asking and why you mentioned 2 gauge.
You need to use this formula I = kW/1.73 x Volts x % Eff x pf.
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