At 50Hz it's 50x60x60 = 180000
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.
As watts are made of volts times amps, the answer would be 0.660kw.
Single-phase ac has current that changes direction 100 times a second (if it's a 50 Hz supply). It only needs two wires to distribute a single-phase supply. Imagine it as a battery that is continually being reversed.
To calculate the per phase current when wattage is given, you first need to know the system's voltage and the power factor (if applicable). For a three-phase system, use the formula: ( I = \frac{P}{\sqrt{3} \times V \times PF} ), where ( I ) is the current per phase, ( P ) is the total power in watts, ( V ) is the line-to-line voltage, and ( PF ) is the power factor. If it's a single-phase system, the formula simplifies to ( I = \frac{P}{V \times PF} ). Ensure you convert units as necessary to maintain consistency.
A load current is a current drawn by an electrical load. In other words, it is the current flowing from the source to the load.For a single-phase system, a line current is a current flowing through the line, or 'hot', conductor, while the current through the neutral conductor is called the neutral current.For a three-phase system, the three 'hot' conductors between the load and the source are called 'lines' and, so, the currents passing through them are called 'line currents'. For a three-phase system, loads are either connected between line conductors (delta-connected system) or between each line and the neutral (star- or wye-connected system), and represent the phases -so the currents passing through the loads are called 'phase currents'.For a balanced three-phase system, the line current is 1.732 times the value of a phase current, where the phases (loads) are connected in delta. For phases (loads) connected in star (or 'wye') the line current is numerically-equal to the phase currents.
A connection can be taken between phase lines, or between one phase and neutral. Both methods give a single-phase supply. Between phases the voltage is sqrt(3) times more than between one phase and neutral. In each case the load gives an unbalanced current on the 3-phase system but the idea is to average out the unbalances over a group of single-phase loads.
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.
The line current would be the same if the motor were connected in delta. The current can be based on the rule of thumb which says 7 amps must be allowed for a 1-HP single-phase motor on 240 v. A 2.2 kW motor is three times as powerful, and on a three-phase supply of the same voltage (240/415) it would draw 7 amps.
the three phase induction motor will start run in single phase . through the remaining healthy phase, around 4.5 times normal current will flow through.Also rotor sbeed also decreases.
Power, in 'watts'.
given a balance three phase, three wires system with star-connected load for which lime voltage is 230v and the impedance of each phase is (6+j8)ohm. find the line current and power absorbed by each phase.
You can't have a three phase earth fault, you can have a phase to phase or a phase to earth fault. If you want the potential phase to earth fault current it will be your voltage times your impedance. If you want the phase to phase potential fault current then you should just double the above result.