For a three-phase, three-wire, system there are three conductors called 'line conductors', and there is a voltage between any pair of line conductors, so there are three voltages.
For a three-phase, four-wire, system there are four conductors: three 'line conductors' and a 'neutral' conductor. So there are three line voltages (voltages between lines) and three phase voltages (voltages between any line conductor and a neutral conductor).
There is no 'total voltage' in a three-phase system. There are three line voltages and three phase voltages.
Generator coils generate the voltage, motor coils use the generated voltage.
Line-to-line voltage in a three-phase system is calculated using the formula ( V_{LL} = \sqrt{3} \times V_{LN} ), where ( V_{LN} ) is the line-to-neutral voltage. This relationship arises because the line-to-line voltage represents the voltage difference between two phases, while the line-to-neutral voltage is the voltage from a phase to the neutral point. If you have the line-to-line voltages directly measured, you can also use the voltage differences between any two phases to determine the line-to-line voltage.
depending on 1phase or 3phase voltage motor , the amphere requirement is found outfor 3phase AC motor use this formula: (HP x 746)/(1.73 x E x Eff x PF)for 1phase AC motor : (HP x 746) / (E x Eff x PF)
The term, 'unbalanced system' refers to an unbalanced load. Under normal circumstances, an unbalanced load leads to unbalanced line currents. The line voltages are determined by the supply and remain symmetrical, even when the load is unbalanced. As your question refers to a 'line to neutral' voltage (i.e. a phase voltage), you must be referring to a star (wye) connected load, in which case the phase voltage (line to neutral voltage) is 0.577 (the reciprocal of the square-root of 3) times the line voltage (line to line voltage).
There is no 'total voltage' in a three-phase system. There are three line voltages and three phase voltages.
Ohm's Law - V = IR.
A-for star connectionE(line)=1.73E(Phase)I(line)=I(Phase)&B-for delta connectionE(line)=E(Phase)I(line)=1.73I(Phase)
It is not advisable to run lower voltage motor on a higher voltage.
This is a voltage drop question. A voltage must be stated to answer this question.
Generator coils generate the voltage, motor coils use the generated voltage.
Your question is not clear. Are you asking how you can obtain a single-phase supply from a three-phase system? If so, then it depends on the type of three-phase supply and the voltage you require. For example, a three-phase, three-wire, system, single phase is obtained between any two line conductors. For a three-phase, four-wire, system, single phase is obtained either between any two line conductors, or between any line conductor and the neutral conductor. In the case of the latter, two voltages are available, a line voltage and a phase voltage, with the line voltage being 1.732 x larger than the phase voltage.
50 kVA is 16.667 kVA per phase and you divide that by the phase voltage. Current = 16667 / 277 = 60 Amps
Line-to-line voltage in a three-phase system is calculated using the formula ( V_{LL} = \sqrt{3} \times V_{LN} ), where ( V_{LN} ) is the line-to-neutral voltage. This relationship arises because the line-to-line voltage represents the voltage difference between two phases, while the line-to-neutral voltage is the voltage from a phase to the neutral point. If you have the line-to-line voltages directly measured, you can also use the voltage differences between any two phases to determine the line-to-line voltage.
You can use either. I assume the 240 volts is the line to neutral voltage, and the 415 is the line to line voltages. They will both give you the correct answer. It is important to note these values are RMS, not 0 - peak voltages, but this may be beyond your question. The equations below are for calculated from RMS values (both voltage and current).If you are using a L-L voltage, P = I*V*sqrt(3)If you are using the L-N voltage, 1-phase power P = I*V (for the power in a single phase, for all three, multiply by 3), or 3phase power P = 3*I*VYou will get the same answer, since the L-N voltage is (1/sqr(3)) times the L-L voltage.
A voltage is applied to a signal line. The voltage of the line changes gradually from 0 to +V. The "edge speed" is the rate of change of voltage of the line. A voltage is applied to a signal line. The voltage of the line changes gradually from 0 to +V. The "edge speed" is the rate of change of voltage of the line.
depending on 1phase or 3phase voltage motor , the amphere requirement is found outfor 3phase AC motor use this formula: (HP x 746)/(1.73 x E x Eff x PF)for 1phase AC motor : (HP x 746) / (E x Eff x PF)