Real Power: The actual power in Watts or K-Watts in AC or DC Circuits Apparent power: The Power in Inductive or Capacitive Circuits have Phase Lag & Lead measured in Volt Amperes VA or Kilo Volt Amperes KVA
Apparent power is measured in volt amperes, true power is measured in watts. Apparent power includes both real (or true power) and reactive power. So yes, if you are looking at both the apparent power and true power for a specific load, the apparent power will be greater than or (if the load is purely resistive) equal to the true power.
AnswerApparent power will always be greater than true power except at unity power factor, when the two are equal.
True power divided by apparent power is the formula for power factor.
The ratio of apparent power to true power is called 'admittance', expressed in siemens. Admittance is the inverse of impedance.
The vector-relationship between apparent power, true power, and reactive power is represented by a right-angled triangle, whose hypotenuse represents apparent power and whose adjacent represents true power. Since power factor is defined as 'the ratio of true power to apparent power', you will find that this ratio corresponds to the cosine of the angle between them.
Apparent power is the vectorial sum of the true power and reactive power. In this case, the total reactive power is the difference between 7200 var and 3600 var -i.e. 3600 var.So you can now use the equation,(apparent power)2 = (true power)2 + (total reactive power)2,to determine your answer.
There are various ways in which you can determine the reactive power (in reactive volt amperes) of a load. From the practical point of view, you can use a voltmeter and an ammeter and use the product of their readings to determine the apparent power (in volt amperes) of the load, and a wattmeter to determine the true power (in watts) of the load, then find the vectorial difference: (reactive power)2=(apparent power)2-(true power)2
Power factor is:the ratio of true power to apparent powerthe ratio of resistance to impedancethe ratio of the voltage across a circuit's resistive component to the supply voltagethe cosine of the phase angleetc.
No. The volt ampere (V.A) is the unit of measurement of apparent power. Power factor is true power (expressed in watts) divided by apparent power (expressed in volt amperes).
to put out the power fector you have to divided apparent power with true power.AnswerYou can determine the true power of any load using a wattmeter. To find the apparent power, you use a voltmeter to measure the supply voltage and an ammeter to measure the load current, and multiply the two readings together.If you then want to go on to find the power factor, then you divide the true power by the apparent power. If you want to find the reactive power you use the following equation:(reactive power)2 = (true power)2 x (apparent power)2
to put out the power fector you have to divided apparent power with true power.AnswerYou can determine the true power of any load using a wattmeter. To find the apparent power, you use a voltmeter to measure the supply voltage and an ammeter to measure the load current, and multiply the two readings together.If you then want to go on to find the power factor, then you divide the true power by the apparent power. If you want to find the reactive power you use the following equation:(reactive power)2 = (true power)2 x (apparent power)2
The ratio of apparent power to true power is called 'admittance', expressed in siemens. Admittance is the inverse of impedance.
Apparent power is the vector sum of a load's true power and its reactive power. If you draw a 'power diagram', the phase angle will be the angle between the true power and the apparent power. If true power is fixed, then increasing the phase angle will result in a greater value of apparent power.
The current's power factor is the true power divided by the apparent power. The Apparent Power is the volts multiplied by the amps. In this example, the ratio would be 200/253, or approximately .79.
The vector-relationship between apparent power, true power, and reactive power is represented by a right-angled triangle, whose hypotenuse represents apparent power and whose adjacent represents true power. Since power factor is defined as 'the ratio of true power to apparent power', you will find that this ratio corresponds to the cosine of the angle between them.
Power factor is truepower divide by apparent power.
It's actually cos phi, where the Greek letter, 'phi', is the symbol for phase angle -the angle by which a load current lags or leads the supply current in an a.c. system (the Greek letter, 'theta', is used for the displacement of instantaneous values of current or voltage from the origin of a sine wave).The reason why power factor is a cosine requires you to understand the relationship between apparent power, true power, and reactive power. Apparent power is the vector sum of true power and reactive power, and can be represented, graphically, by the so-called 'power triangle'. In the power triangle, true power lies along the horizontal axis, reactive power lies along the perpendicular axis, and the apparent power forms the hypotenuse, and the angle between true power and apparent power represents the phase angle. By definition, power factor is the ratio between true power and apparent power, and this ratio corresponds to the cosine of the phase angle.From this, we can conclude that true power = apparent power x cos phi, where 'cos phi' is the 'factor' by which we must multiply apparent power to determine true power -i.e. the 'power factor'.
The equation for power factor is PF = True power in watts/Apparent power in Volt Amps.
Use a wattmeter, as it only reads 'real power' of your load. Use an ammeter and a voltmeter, and the product of the two readings will give you 'apparent power' of your load. Since apparent power is the vector sum of real power and reactive power, use the following equation to find the reactive power of your load: (reactive power)2 = (apparent power)2 - (real power)2
Apparent power is the vectorial sum of the true power and reactive power. In this case, the total reactive power is the difference between 7200 var and 3600 var -i.e. 3600 var.So you can now use the equation,(apparent power)2 = (true power)2 + (total reactive power)2,to determine your answer.