If the resistance of the circuit remains the same, yes. E = I x R Formulas for you to use are E/I = R, E being voltage and I being current or amps.
AnswerNot necessarily. For example, electricity transmission is only possible because, for a given load, the higher the voltage, the lower the resulting load current. So the answer is that it depends on the load'!At what voltage? When you know the voltage then, to get the amps those kilovolt-amps contain, you simply divide the kilovolt-amps by the voltage.
It would depend on the voltage, because at a higher voltage a larger voltage drop can be tolerated for the same percentage voltage drop. That means a thinner wire could then be used. And 339 amps is a VERY high current for 120/240 volt system- more than double the peak demand for most houses. Please check your question, and give us the voltage used.
Watts = Amps x Volts. Amps = Watts/Voltage. Amps = 2500/apply voltage here.
Impossible to answer 2 amps could carry a lot of power if the voltage is high, and 7.2 volts could carry a lot of power if the current is high.
If the voltage in your vacuum is 120 then you divide the Wattage by the Voltage to get 3 amps.
Voltage.
Amps can not give you a kilowatt with out a voltage being applied to the question. Watts = Amps x Volts. Amps = 1000/ Volts.
At what voltage? If you know the voltage then, to get the amps those kilovolt-amps contain, you simply divide the kilovolt-amps by the voltage.
At what voltage? When you know the voltage then, to get the amps those kilovolt-amps contain, you simply divide the kilovolt-amps by the voltage.
The number of watts for 12.5 amps is 12.5 times the voltage.
High voltage loads is a high electrical discharge that can result to electrical breakdown. Example: High power amplifier vacuum tubes or particle beams. High current loads are the loads that can have peak current greater than 10 amps. Example: Motors, solenoids or nitinol wire.
It would depend on the voltage, because at a higher voltage a larger voltage drop can be tolerated for the same percentage voltage drop. That means a thinner wire could then be used. And 339 amps is a VERY high current for 120/240 volt system- more than double the peak demand for most houses. Please check your question, and give us the voltage used.
Watts = Amps x Volts. Amps = Watts/Voltage. Amps = 2500/apply voltage here.
Can not do it without knowing the voltage I = E/R. Amps = Voltage/Ohms.
Impossible to answer 2 amps could carry a lot of power if the voltage is high, and 7.2 volts could carry a lot of power if the current is high.
High voltage and low resistance would cause high amps. E = I R or I = E/R. PS E would be electromotive force measured in volts. R would be resistance measured in ohms. I would be current flow measured in amps.
Look on the light bulb for the voltage and the power in watts. Then divide the watts by the voltage and that gives the amps. Some CFL bulbs also state the current as well as the voltage and power, which is because they can have a poor power factor.