It's easier to answer you question the other way around, that is "Why does the load current fall with an increase in power factor?"Before power-factor improvement, the load current is the phasor (or vector) sum of the load's resistive (IR) and inductive currents (IL).Power-factor improvement is achieved by adding a capacitor in parallel to the load so, after power-factor improvement, the load current becomes the phasor sum of the resistive current (IR), inductive current (IL), and the capacitive current (IC).Since the inductive current and capacitive current are displaced from each other by 180 degrees (i.e. are in antiphase), the the phasor sum of IR + IL +IC will be smaller than the phasor sum of just IR and IL.Hence, the supply current reduces as the power factor improves.
The general equation for power in a BALANCED three-phase load is as follows:P = EL IL x power factor...where EL and IL are the line voltage and line current, respectively.So all you have to do is to manipulate the equation to make the line current the subject, and insert your figures.
When the frequency of Parallel RL Circuit Increases,XL increases which causes IL (current through inductor) decreases. Decrease in IL causes It (It=Il+Ir) to decrease,which means by relation IT=Vs/Zt ,the Zt (Total Impedance) Increases.
voltage=400 volts current=17.1 ampere assumed total load since IL=KVA X 1000x0.8/400 therefore kva=17.1x400/1000x08 =8.55kva Gen set
47.18 Amps.Another AnswerThe equation for the total power of a balanced three-phase load is:P = 1.732 x VL IL x power factor, where the subscript, L, represents line values.So if we change the subject of the equation, for the line current, we have:IL = P / (1.732 x VL x power factor)IL = 32 000 / (1.732 x 440 x 0.89)IL = 47.18 A
Depends on the three phase voltage. 54 Kilowatts = 54,000 Watts. 54,000 Watts / square root of 3 / voltage. So a three phase, 54KW, 480 Volt load equals 64.95 Amps (54,000/1.732/480 = 64.95)AnswerThe equation for power of a balanced three-phase load is as follows: P = 1.732 EL IL x power factor...where EL and IL are line values.So, to find the current drawn by the (balanced) load: IL = P / (1.732 EL x power factor)Unfortunately, as you have not supplied a value for power factor, your question cannot be answered.
A phase current is the current passing through a phase, whereas a line current is the current flowing through a line.In the case of a balanced delta-connected load, IL = 1.732 IP. In the case of a balanced star-connected load, IL = IP.For unbalanced loads, these relationships don't hold true, and must be individually calculated.
If I have 1 KW In 3 Phase it will give 1.54 A and In single phase it will give 4.6 AFor cosF 0.9V 415 3 phV 240 1 phIt seems the reason is because the current is carried on more wires. Also, remember that if wattage stays constant, then as voltage increases, current decreases.AnswerIt really depends on the load. Are you assumining the three-phase load to be the same as the single-phase load or, as it is likely to be in practice, three times the value of the single-phase load?But, in either case, the single-phase current will not be double the the three-phase (line) current!The equation for the load current supplying a single-phase is: I = P / (E x power factor)The equation for the line current supplying a balanced three-phase system is: IL = P / (1.732 x E x power factor)If you insert real figures into these equations, (240 V for the single-phase voltage and 415 V for the three-phase line voltage) then you will find that, when the three-phase load is threetimes that of the single-phase load, the supply currents will be exactly the same. On the other hand, if you assume that the three-phase load is exactly the same as the single-phase load, then you will find that the three-phase line current will be one-third that of the single-phase current.
Jesse White
To reduce the chiller load which is going exceed
Not if the IL suspension is still in effect.
A. George Pradel is the current mayor of Naperville, IL.