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'!To convert high voltage (HV) amps to low voltage (LV) amps, you can use the formula: HV amps = LV amps x (LV voltage / HV voltage). By rearranging the formula, you can calculate LV amps by dividing HV amps by the ratio of HV voltage to LV 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.
It depends on the context. Power is calculated as the product of voltage and current (P = V x I). In this case, 7.2 volts at 2 amps would result in 14.4 watts of power, while 2 amps alone does not directly indicate power without knowing the voltage.
There are zero kW in 32 amps. Watts are the product of amps times volts. Once you find the voltage of the system multiply it times 32 amps and then divide that answer by 1000. This will give you the answer in kW.
watts = amps (times) voltage watts (divided by) voltage = amps 140 (div by) 120 = 1.66 140 (div by) 125 = 1.12
To convert high voltage (HV) amps to low voltage (LV) amps, you can use the formula: HV amps = LV amps x (LV voltage / HV voltage). By rearranging the formula, you can calculate LV amps by dividing HV amps by the ratio of HV voltage to LV voltage.
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.
The number of watts for 12.5 amps is 12.5 times 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.
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.
Can not do it without knowing the voltage I = E/R. Amps = Voltage/Ohms.
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.
It depends on the context. Power is calculated as the product of voltage and current (P = V x I). In this case, 7.2 volts at 2 amps would result in 14.4 watts of power, while 2 amps alone does not directly indicate power without knowing the voltage.
The amount of current (amps) that pylons, or transmission towers, carry can vary significantly depending on the voltage of the transmission lines and the specific design of the system. High-voltage transmission lines can carry anywhere from hundreds to thousands of amps. For example, a typical high-voltage line operating at 500 kV might carry around 1,000 to 2,000 amps. Overall, the exact current is determined by the needs of the electrical grid and the load being supplied.
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.