Volts = Amps x Resistance
Therefore
Amps = Volts / Resistance
Wattage is volts times amps. To find amps when you know volts and watts, divide wattage by voltage. In this case, it would be 0.5 amps.
P=VI=V^2/R P=12x12/6=24W
60
nominal 600 watts 610 Watts
defination of suddenly applied load
NO LOAD:--- as it is very clear from the word itself that it is no load that means no load has been applied to the machine..i.e.the output terminal of machine is open circuited FULL LOAD:-- the maximum load value that can be applied to the machine at which it can work without damaging any part of the machine is called machine working at full load.
what is the full load amps for 2.4 hp motor at 460 volts ?
You cannot use a capacitor as a 'power saver' or, more accurately, 'energy saver'! A capacitor may improve the power-factor of a load, and this may reduce the value of its load current, but this does not reduce the energy consumed by the load. For a residence, a so-called 'power save' capacitor is nothing more than a rip-off.
P=VI=V^2/R P=12x12/6=24W
The formula for this question is I=P/E Where I = Current P = Watts E = Volts Therefore applying this formula: I = 1500 / 120 = 12.5
Power is consumed whenever a load is connected to the distribution supply panel.The load is usually controlled by a switch, contactors for motors or breakers located in the distribution panel. Load on line power is consumed, load off line no power is consumed.
The actual energy consumed in load is inductive load
Volts cause current to flow through the load. The current is measured in amps, and the volts multiplied by the amps gives the power in watts.
Assuming a pure resistive load, the current will be 240/8 = 30A. Watts = Volts x Amps so power is 30 x 240 = 7200 watts or 7.2 KW.
It is a voltage (potential) applied to a load that causes a current to flow through the load. Ohm's Law encapsulates this principal and states Volts = Current x Resistance. In your example, the applied voltage would be 200 volts.
Watts is the product of amps x volts. To find the wattage a voltage needs to be stated so that it can be multiplied with the amperage value.
Assuming a pure resistive load Ohm's Law tells us 240 / 8 = Current = 30 Amps. Watts = Volts x Amps so 240 x 30 = 7,200 Watts = 7.2 KW.
A: the rms value will be169 volts add a capacitor and no load 240 volts and the average will be 153 volts
Watts = Volts x Amps So you need to know the current. This is only for resistive load.
First of all the power consumed is only dependent on the load (eg. any appliance) connected to the source. A load will always draw its rated power. If you have increased your voltage to twice then the current drawn by the device will become half but the power consumed will remain same.the power consumed is given by:P= V*I* cos(fi)here for a given load P(power), cos(fi) are constants.Then if V becomes 2V then current will be I/2.