Actually, if you connect the above described LED to 220 V, it will immediately burn out then the circuit will be "open" and the resistance will be "infinite".
220 volts alternating current
Yes, 220 volts is in the same voltage classification as 230 volts.
The load is a resistive load and as such it is governed by Ohm's law. Current is directly proportional to the voltage and inversely proportional to the resistance. As the voltage goes down so does the current. 2200 watt heater at 220 volts = 10 amps. R = E/I = 220/10 = 22 ohms of resistance in the heater. Now take the 2200 watt heater and using the same formula and at 110 volts. I = E/R, Amps = Volts / Resistance. 110/22 = 5 amps. As you can see ohm's law holds true, the current is inversely proportional to the resistance and as the voltage goes down so does the current. To answer the question, yes a 220 volt heater will run on a 110 volt circuit but at a reduced wattage. W = A x V = 5 x 110 = 550 watts.A 2200 watt heater at 220 volts would draw 1/2 the current of a 2200 watt heater at 110 volts.A 2200 watt heater at 220 volts is 22 ohms of resistance. Resistance would not change with the voltage, current would only be 5 amps (a resistor is a current limiter, it will only let more current through if you apply more voltage not less); but the wattage would only be 550 watts. This would only give you a fourth of the power this heater was designed for! The heater and the wires would have less heat.
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.
No. It would have to be wired for 220 volts and would have to be rated for 220 volts and would have a different configuration so that a typical 120 v plug wouldn't fit the outlet.
U = RxI so 11x20=220 volts
If the Peak to neutral voltage is 220 volts, the root mean square voltage is 155.6 volts (sqrt(220)).
1.25 A
No.
Ohm's Law: Resistance is voltage divided by current Power Law: Power is current times voltage Combining them gives: Resistance is voltage squared divided by power 220 volts squared divided by 100 watts = 484 ohms. Note that this is hot resistance. If you measure the bulb in the cold state, you will get an entirely different, smaller, value, due to the extreme temperature coefficient of the filament. Independently of that, since you ask for peak voltage, that means you are talking about an AC voltage source. We have to assume a sinusoidal waveform, and that the 220 volts was the RMS value. In this case, the peak value is simply the RMS value multiplied by the square root of 2, i.e. 0.707..., making the peak value 311 volts.
I(current) = V(voltage)/R(resistence) Example : 220 V / 5000 Ohm = 0.044 A (Ampère) = 44mA
The formula you are looking for is R = E/I
The formula you are looking for is R = E/I.
220 ohm's
any one of the three line to neutral is 220 volts
At 110 volts it is 0.8 amps. At 220 it is 0.4 amps. I=E/R. I=amps.E=volts R=resistance.
A 100 watt 220 volt light bulb (or anything consuming 100 watts on 220 volts) draws 100/220, or .45 Amps. It will also have about 220²/100, or 484 ohms resistance. A 60 watt 220 volt light bulb (or anything consuming 60 watts on 220 volts) draws 60/220, or .27 Amps. It will also have about 220²/60, or 807 ohms resistance.