Applying a negative or positive sign to power factor is an obsolete method of describing whether it is 'leading' or 'lagging'. We no longer do this. These days, a 'leading power factor' indicates that the load is capacitive and the load current is leading the supply voltage, and a 'lagging power factor' indicates that the load is inductive, and the load current is lagging the supply voltage. Having said that, a 'negative' power factor is also the mathematical consequence of 'negative power' -i.e. the direction of power when it is fed from the load back to the supply (e.g. when the grid feeds energy into a generator, causing it to 'motor'). In this case, the power factor isn't really negative, but simply appears to be so.
Yes of cours1 kw =1.25 kva wich mean6.5 kw =8.12 kvaif you have generator 8.12 kva it give you 6.5 kw=============================Answer #2:1 kw does not necessarily = 1.25 kvaThe relationship between KW and KVA depends on the nature of the load you'repowering, and is called the "power factor". It describes how closely the voltageand current peaks coincide in time. If the load has any inductive or capacitxivereactance, then the voltage and current waveforms become separated in time.The "KVA" is the product of the full voltage and full current without regard fortheir "phase difference", but the "real" power in KW is the KVA multiplied by thecosine of the phase angle. So if there's any inductance or capacitance present,then the KW is less than the KVA. But if the load is pure resistive, then thevoltage and current on the line are in phase, the angle between them is zero,and the KW and KVA are equal.When everything is just exactly perfect, and there is no reactance on the load orthe line, then your 6.5 KVA generator can just exactly supply 6.5 KW of load, withnothing to spare. More commonly, of course, a generator with somewhat morethan 6.5 KVA capacity is required in order to supply 6.5 KW of 'real' power.
Normally refers to a poly phase motor starter that monitors current draw in all three legs and interrupts all legs. the most popular use "heater-melting alloy" technique, magnetic current monitoring is also popular but IMHO thermal alloy offers superior protection in high starting load applications.
Yes, if the power factor (cosine of the phase-angle between voltage and current)is ' 1 '. In order for that to be true, the total load impedances on the line have tobe pure resistive, with zero reactance.All of this stuff applies only on an AC line. On a DC line, voltage and current arealways in phase, the power factor is 1, and KVA = KW .
because when we r applying a certain load the deflection also increasing..))
load
It is a transformer with No load attach to it.
2 to 5% of full load current
No load current is energizing current. This is effectively "lost" power, power used in the transformer to energize the core. It, therefore, should be small!
is it primary current ?
To calculate the no load current from transformer & core loss is also calculated.
No load current, in a transformer for example, is the current necessary for exciting the transformer. If you wish to keep it energized, and you need to keep it energized at full voltage, there is nothing you can do to reduce this other than replace the transformer with one that has lower no load current. If you are referring to a different piece of equipment, you may need to specify what you are meaning by "no load current".
When the secondary of a transformer is opened, there is no longer any load on the transformer. There will be some current flowing in the primary winding, which is needed to induce the voltage in the secondary. This primary current is referred to as the "no load" current, and is indicative of the core losses in the transformer.
The secondary load current will change. This, in turn, will cause the primary current to change (the primary current being the phasor sum of the [IS (Np/Ns)] and the primary current (Io).
No load current depends on the design of the transformer, and what voltage it is energized at. It will typically be below 1% of full load, and can be significantly below 1% for utility sized transformers.
The secondary current is determined by the load, not by the transformer. For example, if the secondary voltage is 50 V and the load is 100 ohms, then the secondary current will be 0.5 A. If the load is 25 ohms, then the secondary current will be 2 A. It is important that a continuous secondary current doesn't exceed the rated secondary current of the transformer.
The current flowing through a transformer's secondary is the current drawn by the load, so it will be exactly the same as the current flowing through your induction motor -assuming that is the load. Don't really understand the point of your question!