Watts of Power are not 'drawn', it just exists, or is generated by electrical interaction in the circuit.
Amperage is 'drawn' by a load.
The equation you're looking for is: P = VI, so P = (10)(120) = 1200 Watts, or 1.2 kW.
If you have a Three-Phase circuit, the eq changes to: P = VI(square root of 3).
1.73 is usually good enough to use as a value for the sqr 3.
Divide the voltage into the wattage and your answer will be the amperage
Wattage/divided by voltage= Amperage
It is calculated a 10 amps times 120 volts, that is 1200 watts.
volts times amps equals watts so it would be 1200 watts
The formula you are looking for is W = A x V.
real P= V * I *cos(phase angle between V and I)for purely resistive loads or DC voltages this equals real power P=V*I = 120*5= 600Wattsfor not pure resistive loads you'd have to measure the phase angle between Voltage and Current to get real power.However, at home, the utility company charges for Complex power = V*I.So you'd still pay for V*I.
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.
Energy meters monitor the supply voltage and the in-phase component of the load current -in other words, they read true (real) power not apparent power which, when time is taken into account (i.e. by the rotation of the meter's disc), results in a measure of energy consumed.
A 120 v circuit would supply 120 v to both resistors if they are in parallel, which is 120/100 amps into a 100 ohm load, and 120/80 amps into am 80 ohms load, which totals up to 2.7 amps, so the total power is 120x2.7 watts or 324 watts.
nominal 600 watts 610 Watts
Power = Voltage x Current. So it will be 60W of power consumption, in your case.
The formula you are looking for is W = A x V.
real P= V * I *cos(phase angle between V and I)for purely resistive loads or DC voltages this equals real power P=V*I = 120*5= 600Wattsfor not pure resistive loads you'd have to measure the phase angle between Voltage and Current to get real power.However, at home, the utility company charges for Complex power = V*I.So you'd still pay for V*I.
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
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.
The power used up by any electrical load is . . .(voltage across the load) x (current through the load) or (voltage across the load)2/(resistance of the load) or (current through the load)2 x (resistance of the load). These are all completely equivalent, and you have your choice of which oneto use, depending on which numbers you know or can measure.
Power (W) = Current (I) X Voltage (V)Therefore a system drawing 150 Amps at 10 VoltsP=150X10P=1500 wattsor 1.5 kWAnswerWithout wishing to be pedantic, power is not 'consumed' by a load such as a starter motor. Power is simply a 'rate', the rate at which the load is consuming energy. You cannot 'consume' a rate, therefore, you cannot 'consume' watts! So your question should be rephrased to ask 'What is the power of a starter motor?', or words to that effect.
The power used up by any electrical load is . . .(voltage across the load) x (current through the load) or (voltage across the load)2/(resistance of the load) or (current through the load)2 x (resistance of the load). These are all completely equivalent, and you have your choice of which oneto use, depending on which numbers you know or can measure.
The active power of an inductor is zero. As we know, the active power is the result of product of supply voltage and in-phase component of load current. But the load current in pure inductive load lags supply voltage by 90 degrees. So there is no component of load current that is in-phase with the supply voltage. Therefore, the active power in inductive reactance is zero.
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.