nominal 600 watts 610 Watts
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.
heavy load
The relationship between kVA and W varies according to the nature of the load.The volt ampere is the unit of measurement for the apparent power of a load, obtained by multiplying the voltage across that load by the current through it.The watt is the unit of measurement for the true power of a load. This is a measure of the rate at which energy is dissipated by the resistive component of the load, and is obtained by multiplying the voltage across that load, by the current through it, and by its power factor. Power factor is the cosine of the phase angle between the voltage across the load and the current through it.So, for a purely resistive load, where the voltage and current are in phase, the apparent power and the true power are the same, so 1 VA = 1 W (or, in terms of the actual question, 1 kVA = 1 kW). For a purely reactive circuit, where the current leads or lags the voltage by 90o, 1 VA = 0 W.Actually, there is no such thing as a purely reactive circuit, so a situation where there is 1 VA but 0 W is only theoretical.
hot load occurs when the power is high and the cold load occurs when yhe power is low
A kilowatt is an unit of true power in an AC circuit -as measured by a wattmeter. A kilovolt ampere is an unit of apparent power in an AC circuit, which is the product of the voltage across a load by the current through it. The relationship between the two is: kilowatt = (kilovolt ampere) x (power factor of load)
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.
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.
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.
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.
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 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.
Non-reactive power (watts) is current times voltage, so P = IV. Dimensionally, current is coulombs per second, voltage is joules per coulomb, so current times voltage is joules per second, which is the same as power in watts.