You can't convert kV (kilovolts) to current (amps) unless you know the resistance (ohms) of the load which is taking current from that voltage source.
Once you know the resistance then you can use Ohm's Law to get the amperage:
I = V / R
In words, Ohm's law is:
Current (amps) equals voltage divided by resistance (ohms)
Yes. The IEC 60071.1 Std states a diminish withstand of aprox of 1% every 100mts above 1000mts. For example: for a 13.8kV system the insulation level should be of 95kV (interior). If the installation is going to be at 1,500mts then de minimum insulation level of your equipment must be: 95kv/(1-0.01*(500/100)) = 100kV BIL. The next std value for BIL is 125kV.
The prefix 'micro' means, one millionth. So there are one million micro amps in one amp
How many Amps is the fridge pulling? Multiply the Amps by the 120V circuit you're plugging into and you'll get your Watts.
Lightning Impulse withstand test is a destructive test carried out on high voltage switchgear. It is classed as a destructive test as each shot damages the insulation and lays carbon tracks. The Energy Networks Association Technical Specification (ENA-TS) 41-36 states the BIL (basic Impulse level) Required for the Voltage rating of the piece of switchgear rated up to 36kV. For example an 11kV switchgear will have a BIL of 95kV. Switchgear which include a Vacuum circuit breaker must have a second means of isolation, ie a disconnector. This disconnector can be a two position (mains on, off) or three position (mains on, off, or earth on) disconnector operated off load which is in series with the Circuit breaker. To pass the test 15 shots at +95kV and 15 shots at -95kV are fired at all positions of the disconnector and only 2 out of the 15 (and not the last 5) shots can flash over to earth. For example if the disconnector was in the off position one side will be connected to the impulse generator and the other side will be earthed. Positive and Negative shots will be fired and the impulse can not flash through the air gap to the earthed side of the disconnector. This proves the isolation gap inside the disconnector has a BIL rating of 95kV. Therefore if the incoming side of a switchboard is live and the outgoing side is isolated for maintenance. There can be a transient impulse on the live side of the circuit and it will not flash across the isolating gap and endanger anyone working on the dead side of the circuit
Current (amps)=Watts/Volts =2000/120 =16.75 =16.75 amps
Yes. The IEC 60071.1 Std states a diminish withstand of aprox of 1% every 100mts above 1000mts. For example: for a 13.8kV system the insulation level should be of 95kV (interior). If the installation is going to be at 1,500mts then de minimum insulation level of your equipment must be: 95kv/(1-0.01*(500/100)) = 100kV BIL. The next std value for BIL is 125kV.
Multiply the vots by the amps to find the volt-amps. Or divide the volt-amps by the voltage to find the amps.
0.35897 amps = 0.35897 amps.
200ma is .200 amps or .2 amps
Amps = Watts / Volts Amps = 130000 / 480 Amps = 270.83
7 amps
.1 amps will give you .1 amps.
Amps is amps be it DC or AC.
Amps is short for ampere, a unit for current.
There are 20 million amps or 20,000,000 amps.
In 50 VA the V stands for volts and the A is for amps. Hence the formula you are looking for is 50/240 = Amps.
What is the generator amps